A negative answer to the problem: are stratifiable spaces M 1 ?
aa r X i v : . [ m a t h . GN ] S e p A NEGATIVE ANSWER TO THE PROBLEM: ARESTRATIFIABLE SPACES M ? HUAIPENG CHEN ∗ AND BOSEN WANG ∗∗∗
Abstract.
In accordance with M -structures in paper [4], we construct astratifiable space which is not M -spaces. Introduction “Are stratifiable spaces M ?” is a well-known problem in set-theoretic topology.It comes from [3]. Ceder [3] defined M i -spaces ( i = 1 , ,
3) and proved M ⇒ M ⇒ M . It is an interesting problem whether these implications can be reversed. Borges[1] gave a characterization of M -spaces and renamed M -spaces as stratifiablespaces. Gruenhage [8] and Junnila [16] proved that stratifiable spaces are M -spacesindependently. Their results aroused people’s great interest to the problem “Arestratifiable spaces M ?”. It¯o and Tamano [15] using closed mappings got interestingresults. T.Mizokami got some important progresses on the problem in [19],[20] and[21]. Also there are many important results on stratifiable spaces commended bysurveys of Tamano [22], Gruenhage [9] and [10], Burke and Luter [2]. Wang madesome comment about the problem and called it “Problem on Generalized MetricSpaces” in [23]. In 1990, Rudin made some comment about the problem in herwell-known paper “Some Conjectures” in [14].In 2000, Gartside and Reznichenko [6] gave out C k ( P )-spaces which was com-mended by Gruenhage [11], and was researched in wide range. In 2008, Chen [4]got a new characterization of stratifiable spaces. In this paper, we use an idea ofChen [4] wholly to prove the following main theorem:Theorem 1. There exists a space ( X, τ ) which is a stratifiable space, and is notan M -space.Then, by section 8 in [22], Theorem 1 gives a negative answer to the followingquestions:-Is every M -space an M -space (Ceder [3] 1961)? .-Does any point in a stratifiable space have a σ -closure-preserving base (Tamano[22] 1989)?-Is every (closed) subspace of an M -space an M -space (Ceder [3] 1961)?-Is the closed image of an M -space an M -space (Ceder [3] 1961)?-Is the perfect image of an M -space an M -space (Burke and Lutzer [2] 1976)?-Is every stratifiable spaces a µ -space (Tamano [22] 1985)?-Is each zero-dimension submetric stratifiable space X with an M -structure an M -space (Chen [4] 2008)? ∗ the author is supported by The National Natural Science Foundation of China (No.11171162).keywords: stratifiable spaces, M -spaces, metric spaces, g -functions, unbounded setssubjclass[2010] Primary: 54G20. Secondary: 04A20, 54E35, 54E20, 54E99. ∗ AND BOSEN WANG ∗∗∗
Theorem 1 suggest some questions too in this paper.Recall that a space X is an M - space if X has a σ -closure preserving base B .Recall that a family B is a quasi-base for X if for each open set U of X and a point x ∈ U , there is B ∈ B such that x ∈ IntB ⊂ B ⊂ U . A space X is an M - space if X has a σ -closure preserving quasi-base and an M - space if X has a σ -cushionedpair-base.In this paper, the letter N denotes the set of positive integers and ω denotes thefirst infinite ordinal. h, i, j, k, l, ℓ, m, n, a, b, c, d and e are used to denote membersin ω and N . If there are signs and definitions which have not been defined in thispaper, we can see it in [9] or [22] in topology and in [18] in set theory.2. To construct a set X Following only the idea of Theorem 5.1 in [4], in order to construct a zero-dimension metric space (
X, ρ ), we construct the set X at firstly. To do it let Q = { , / , / , / , ... } = { p n : n ∈ N } be the set of all rational numbers in [0 , p n , p m ) = { q ∈ Q : p n ≤ q < p m } for p n , p m ∈ Q . Then Q = [0 , Q = { [ p n , p m ) : p n , p m ∈ Q and q n < q m } . Then Q is a base of some topology suchthat each [ p n , p m ) is a closed and open set. Denote the topology by ρ ′ . Then ( Q, ρ ′ )is a zero-dimensional metric space since ( Q, ρ ′ ) is regular T and Q = { B n : n ∈ N } is a countable base. Construction 1.
Let (
Q, ρ ′ ) be the zero-dimensional metric space.A). Let S = { q } = { } ⊂ Q and S ( q ) = { q i : i ≥ } ⊂ Q − { q } be aconvergence sequence which converges to q with q < ... < q i +1 < q i < ... < q < q < q , q i ) = { q ∈ Q : q ≤ q < q i } . Here q ∈ Q − { q } is the first number in Q − { q } with 1 − q = | − q | ≤ / q i by q i . Let S = { q i : i ∈ N } . Then S = S ( q ).B). Assume we have had a convergence sequences set S n = { q ni : i ∈ N } ⊂ Q .Pick a q ni from S n . And then pick a q i such that:(1). If q ni = max S n , let q i = inf { q ∈ ∪ i ≤ n S i : q > q ni } . Then q i ∈ ∪ i ≤ n S i with( q ni , q i ) ∩ ( ∪ i ≤ n S i ) = ∅ .(2). If q ni = max S n , let q i = 1. Then ( q ni , q i ) ∩ ( ∪ i ≤ n S i ) = ∅ .Denote n + 1 by k . Take a convergence sequence S ( q ni ) = S k ( q ni ) = { q ij : j ∈ N } ⊂ ( q ni , q i ) ∩ [ Q − ( ∪ i ≤ n S i )]which converges to q ni and q ni < ... < q ij < ... < q i < q i < q i . Here q i is thefirst number in ( q ni , q i ) ∩ [ Q − ( ∪ i ≤ n S i )] with q i − q i = | q i − q i | ≤ / k . Let S k = ∪{ S k ( q ni ) : q ni ∈ S n } = { q ij : i, j ∈ N } with S k ( q ni ) = { q i , q i , q i , ..., q in , ... } for i = 1 , , , ..., n, ... . Then S k is countable. We numerate points of S k in according NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 3 with the following order: q , → q , q , q , ... , q n , ... ւ ւ ւ ...q , q , q , q , ... , q n , ... ւ ւ ...q , q , q , q , ... , q n , ... ւ ...q , q , q , q , ... , q n , ...................................................................................... Figure 0Then S k = { q , q , q , q , q , q , ..., q n , q n − , q n − , ..., q n , ... } . Let q kl = q ni . Then l = 1 + 2 + ... + ( n + i −
2) + n = ( n + i − n + i − / n . Then S k = ∪{ S k ( q ni ) : q ni ∈ S n } = { q kl : l ∈ N } . We call it ∆ -order for convenience.Then, by induction, we may construct convergence sequences sets S , S , ..., S n , ... such that Q = ∪ n ≥ S n and S n ∩ S m = ∅ if n = m . Enumerate points of Q = ∪ n ≥ S n in accordance with ∆-order. Then Q = { q n : n ∈ ω } . We will always use thefollowing symbol throughout this paper. Q and S , S , ..., S n , ... means sets withthe ∆-order. [ q i , q ij ) means an interval with the end points q i and q ij such that q ij ∈ S n ( q i ) ⊂ S n and S n ( q i ) converges to q i . Here S n ( q i ) means q i = q n − i ∈ S n − throughout this paper.Let q = q ℓ ∈ Q = ∪ n ≥ S n and q = q mℓ ′ ∈ S m . Then ℓ > ℓ ′ by the ∆-order. Proposition 2.1.
Let Q = { q n : n ∈ ω } with ∆ -order. Then:1. Let q h ∈ Q with q h = q kl ∈ S k and q kl = q ni ∈ S k ( q n ) . Then l = 1 + 2 + ... + ( n + i −
2) + n = ( n + i − n + i − / n and h = 1 + 1 + 2 + ... + ( k + l −
2) + k > l.
2. If q ki , q kj in S n ( q k ) with i < j , then q ki = q nn i , q kj = q nn j in S n with ∆ -ordersatisfy n i < n j and q ki = q h i , q kj = q h j in Q with ∆ -order satisfy h i < h j .3. For each q h ∈ Q with q h = q kk i ∈ S k , there uniquely exists a q n ∈ Q with q n ∈ S k − such that q h = q ni ∈ S k ( q n ) and h > n in Q .4. Let q ki ∈ S n ( q k ) and q hj ∈ S m ( q h ) . Then [ q k , q ki ) ∩ [ q h , q hj ) = ∅ or [ q k , q ki ) ⊂ [ q h , q hj ) or [ q k , q ki ) ⊃ [ q h , q hj ) . Here q k = q n − k ∈ S n − and q h = q m − h ∈ S m − .5. Q = ∪ n ≥ S n and S n ∩ S m = ∅ if n = m .Proof. To prove 1 let S mk ( q a ) = { q ai ∈ S k ( q a ) : i ≤ m } with q a = q k − a ∈ S k − and a = 1 , , ..., n + i −
2, and let S nk = { q kj ∈ S k : j ≤ n } . Take q kl ∈ S k and assume q kl = q ni ∈ S k ( q n ). Then we have q kl = q ni ∈ S ik ( q n ), S lk = S n + i − k ( q ) ∪ ... ∪ S i +1 k ( q n − ) ∪ S ik ( q n ) ∪ S i − k ( q n +1 ) ∪ ... ∪ S k ( q n + i − )and l = 1 + 2 + ... + ( n + i −
2) + n by the definition of ∆-order in accordance withthe above Figure 0. Take q h ∈ Q = { q n : n ∈ ω } with ∆-order. Let q h = q kl ∈ S k and q kl = q ni ∈ S k ( q n ), and let A h = { q i ∈ Q : 0 ≤ i ≤ h } . Note that S = { q } isthe first line by the ∆-order of Q . Then q h = q kl ∈ S lk and A h = S ∪ S k + l − ∪ ... ∪ S l +1 k − ∪ S lk ∪ S l − k +1 ∪ ... ∪ S k + l − . HUAIPENG CHEN ∗ AND BOSEN WANG ∗∗∗
Then h = 1 + 1 + 2 + ... + ( k + l −
2) + k with l = 1 + 2 + ... + ( n + i −
2) + n . Pick q ni and q nj from S k ( q k − n ) with i < j = i + 1. Then we have q ni = q kk i ∈ S k and q ni = q kk i = q h i ∈ Q . Then, by the above 1, h i = 1 + 1 + 2 + ... + ( k + k i −
2) + k and k i = 1 + 2 + ... + ( n + i −
2) + n. Note q nj = q kk j ∈ S k and q nj = q kk j = q h j ∈ Q with j = i + 1. Then, by the above 1, h j = 1 + 1 + 2 + ... + ( k + k j −
2) + k and k j = 1 + 2 + ... + ( n + j −
2) + n. Note j = i + 1. Then k j = 1 + 2 + ... + ( n + i −
2) + ( n + i −
1) + n > k i and h j > h i . Pick a q h ∈ Q with q h = q ki ∈ S ik . Then, by ∆-order, thereuniquely exists q k − n ∈ S k − with q h ∈ S k ( q k − n ). Let q k − n = q m n ∈ Q for q k − n ∈ S nk − . Note that S = { q } is the first line by the ∆-order in Q . Then we have A m n = S ∪ S n + k − ∪ ... ∪ S nk − ∪ S n − k ∪ ... ∪ S n + k − with q k − n ∈ S nk − and m n = 1 + 1 + 2 + ... + ( n + k −
3) + ( k − . When q h ∈ Q with q h = q ki ∈ S ik , then h = 1 + 1 + 2 + ... + ( k + i −
2) + k .Note q ki = q nℓ ∈ S ℓk ( q n ). Then i = 1 + 2 + ... + ( n + ℓ −
2) + n > n . Then h = 1 + 1 + 2 + ... + ( k + i −
2) + k > ... + ( k + n −
2) + k > m n . Thisimplies 3.To check 4, take q ki ∈ S n ( q k ) and q hj ∈ S m ( q h ) with [ q k , q ki ) ∩ [ q h , q hj ) = ∅ .Case 1, q k = q h . Then i > j implies [ q h , q hj ) = [ q k , q kj ) ⊃ [ q k , q ki ), i < j implies[ q h , q hj ) = [ q k , q kj ) ⊂ [ q k , q ki ) and i = j implies [ q h , q hj ) = [ q k , q kj ) = [ q k , q ki ).Case 2, q k ∈ ( q h , q hj ). Then q k ∈ [ q hl +1 , q hl ) for some l ≥ j with q hl +1 and q hl in S m ( q h ). Then [ q k , q ki ) ⊂ [ q k , q k ) ⊂ ( q h , q hj ) by the definition of ( q h , q hj ).Case 3, q h ∈ ( q k , q ki ). Then [ q h , q hj ) ⊂ [ q h , q h ) ⊂ ( q k , q ki ) in the same way asCase 2.It is easy to see 5 from Construction 1. (cid:3) In order to construct g -functions of stratifiable spaces, we construct families ofclosed and open intervals in ( Q, ρ ′ ). Construction 2.
Call a family I ′ D in ( X, τ ) if I ′ is discrete in ( X, τ ), and
P.D in ( X, τ ) if I ′ is pairwise disjoint ( X, τ ).Call a set B c.o in ( X, τ ) if B is a closed and open set in ( X, τ ). Call a family I ′ c.o.D in ( X, τ ) if I ′ is D in ( X, τ ) and each B ∈ I ′ is c.o in ( X, τ ).Take Q with ∆-order and pick an a ∈ N with a > a = n + i ≥ H ( n, i ) after Section 3.1. Take q = 0 ∈ Q . Let Q a = { q } and I a = { [ q , q a ) } . Here q a ∈ S ( q ).Let Q a = { q i ∈ S : q ∈ Q a with q i = q i ∈ S ( q ) for 1 ≤ i ≤ a } . Then wehave Q a = { q j ∈ S : 1 ≤ j ≤ a } and I a = { [ q i , q ia ) : q i ∈ Q a } . Then I a isc.o.D in ( Q, ρ ′ ) and ( ∪ I a ) ∩ ( ∪ I a ) = ∅ .2. Assume we have had Q ak = { q kk j ∈ S k : 1 ≤ j ≤ a k } = { q i ∈ S k : i ≤ a k } and I ak = { [ q i , q ia ) : q i ∈ Q ak } for k ≤ l such that:(1). I ak is c.o.D in ( Q, ρ ′ ) for each k ≤ l .(2). ( ∪ j
1) = Q = ∪ n S n . Then q ∈ S n for some n ∈ N .Then q = q li ∈ S n ( q l ) for some q l ∈ S n − . If i > a , then q = q li ∈ [ q l , q la ) ∈ I an .If a ≥ i >
1, then q li ∈ Q an +1 ⊂ ∪ I an +1 . This implies ∪ I a = [0 , ∪ I an ) ∩ ( ∪ i Let [ q i , q ia ) ∈ I a with b > a . Then q i , q ia ∈ Q ∗ a ⊂ Q ∗ b . Pick a q j ∈ [ q i , q ia ) ⊂ [0 , q h , q hb ) ∈ I b with q j ∈ [ q h , q hb )by the above 1. Then ( q h , q hb ) ∩ Q ∗ b = ∅ since I b is pairwise disjoint. Then q i , q ia / ∈ ( q h , q hb ). Then q j ∈ [ q h , q hb ) ⊂ [ q i , q ia ). Then ∪{ [ q j , q jb ) ∈ I b : [ q j , q jb ) ⊂ [ q i , q ia ) } = [ q i , q ia ) since I b is a c.o.D family with ∪ I b = [0 , The proof of 4. Pick a q i ∈ Q . Then q i ∈ S n for some n ∈ N . Then q i = q n k ∈ S n ( q n ) for some q n ∈ S n − . And then q n = q n k ∈ S n − ( q n ) forsome q n ∈ S n − .... And then q n n = q n n k n ∈ S ( q n n ) for some q n n ∈ S . Let a = max { k n , ..., k , i } . Then i ≤ a . Then q i = q n k ∈ Q an ⊂ Q ∗ a . The proof of 5. It is a corollary of 3 in Proposition 2.2. (cid:3) Let Q n ’s be copies of Q with the topology ρ ′ for n ∈ N and X = Π n Q n withproduct topology. Let Q i ’s be copies of Q , X n = Π i>n Q i , B ′ n = { B n × ... × B n n × X n : B n i ∈ Q i for i = 1 , , ..., n } and B ′ = ∪ n B ′ n . Then B ′ is a countable base. Denote the topology with thecountable base B ′ by ρ . Then ( X, ρ ) is a zero-dimensional metric space.3. To construct H and G on set X Following only the idea of Condition A of M -structures of Proposition 4.7 in[4], we construct H = ∪ n ∈ N H n in X . Construction 3.1 HUAIPENG CHEN ∗ AND BOSEN WANG ∗∗∗ (A). Take Q = { q n : n ∈ ω } with ∆-order. Let H (1 , i ) = { q i } × X and H ′ = { H (1 , j ) : j ∈ N } with j = i + 1. Then H ′ has the same ∆-order as Q .(B). Assume we have had H ′ n = { H ( n, i ) : i ∈ N } .Let P ( n, i ) = { q i } × { q i } × ... × { q i n } for H ( n, i ) = P ( n, i ) × X n ∈ H ′ n and k = n + 1. We take P ( k, l ) in accordance with the following order: P ( n, × { q } , → P ( n, × { q } , P ( n, × { q } , P ( n, × { q } , ... ւ ւ ւ ...P ( n, × { q } , P ( n, × { q } , P ( n, × { q } , P ( n, × { q } , ... ւ ւ ...P ( n, × { q } , P ( n, × { q } , P ( n, × { q } , P ( n, × { q } , ... ւ ...P ( n, × { q } , P ( n, × { q } , P ( n, × { q } , P ( n, × { q } , ............................................... Figure I. ......................................................... Let P ( k, l ) = P ( n, i ) × { q j } for k = n + 1. Then l = 1 + 2 + ... + ( i + j − 2) + i =( i + j − i + j − / i . We call it a ∆-order still. Let H ( k, l ) = P ( k, l ) × X k . Let H ′ k = { H ( k, l ) : l ∈ N } . Then, by induction, we have H ′ k for each k ∈ N and call that H ′ k has a ∆-order. Let H ′ = ∪ n H ′ n . Proposition 3.1. H ′ n is a partition of X for each n ∈ N .2. If H ( n, i ) , H ( n, i ′ ) ∈ H ′ n , then ρ ( H ( n, i ) , H ( n, i ′ )) = r > or H ( n, i ′ ) = H ( n, i ) .3. If H ( n, i ) , H ( n ′ , i ′ ) ∈ H ′ with n ′ > n , then ρ ( H ( n, i ) , H ( n ′ , i ′ )) = r > or H ( n ′ , i ′ ) ⊂ H ( n, i ) .4. If H ( n, i ) , H ( n ′ , i ′ ) ∈ H ′ with H ( n ′ , i ′ ) ⊂ H ( n, i ) , then n ′ ≥ n and i ′ ≥ i .5. Let H ( n, i ) = { q i } × ... × { q i n } × X n . Then i ≥ i j for each j ≤ n .Proof. To check 1, let Q n = { q i : i ∈ ω } be the copy of Q with ∆-order and H ′ = {{ q i } × X : q i ∈ Q } . Then ∪ H ′ = X .Assume ∪ H ′ n = X with ∪ i P ( n, i ) = Π i ≤ n Q i . Let k = n + 1. Then ∪{ P ( n, i ) ×{ q j } : i, j ∈ N } = (Π i ≤ n Q i ) × Q k by induction of assumption. Then ∪ H ′ k = X .On the other hand, Take H ( n, i ) and H ( n, j ) from H ′ n with H ( n, i ) = H ( n, j ).Then P ( n, i ) = P ( n, j ). Then H ( n, i ) ∩ H ( n, j ) = ∅ . This implies 1. The proof of 2. Take H ( n, i ) and H ( n, j ) from H ′ n with H ( n, i ) = H ( n, j ). Then P ( n, i ) = P ( n, j ). Then ρ n ( P ( n, i ) , P ( n, j )) = r > i ≤ n Q i , ρ n ).Pike x ∈ H ( n, i ) and x ′ ∈ H ( n, j ). Then ρ ( x, x ′ ) = ρ n ( P ( n, i ) , P ( n, j )) = r > ρ ( H ( n, i ) , H ( n, j )) = min { ρ ( x, x ′ ) : x ∈ H ( n, i ) and x ′ ∈ H ( n, j ) } = r > The proof of 3. Note that for each H ( k, i ) ∈ H ′ k and k = n + 1, there uniquelyexists an H ( n, j ′ ) ∈ H ′ n with H ( k, i ) ⊂ H ( n, j ′ ). Then 3 is a corollary of 2. The proof of 4. It is easy to see that H ( n ′ , i ′ ) ⊂ H ( n, i ) and H ( n ′ , i ′ ) = H ( n, i )imply n ′ > n . Let n ′ = n + 1 = k with H ( n, i ) = P ( n, i ) × X n and H ( k, i ′ ) = P ( n, i ) × { q j } × X k ∈ H ′ k . Then i ′ = 1 + 2 + ... + ( i + j − 2) + i = ( i + j − i + j − / i ≥ i by the ∆-order of H ′ k . The proof of 5. We use induction on k to prove 5.A. k = 1. Take an H (1 , i ) = { q i } × X ∈ H ′ . Then i ≥ i for each i ∈ N . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 7 B. k = 2. Take an H (2 , i ′ ) ∈ H ′ . Then H (2 , i ′ ) = P (1 , i ) × { q j } × X . We canprove that i ′ = ( i + j − i + j − / i ≥ i and i ′ = ( i + j − i + j − / i ≥ j for i + j ≥ 2, and i ′ > i and i ′ > j for i + j ≥ i + j ≥ i ≥ j ≥ i + j = 2. Then i = j = 1. Then i ′ = ( i + j − i + j − / i = i = j = 1.Case 2, i + j = 3. Then i ′ = (1 · / i = 1 + i > i and i ′ = 1 + i ≥ j since j ≤ i + j = 4. Then i ′ = (2 · / i = 3 + i > i and i ′ = 3 + i > j since j ≤ i + j ≥ 5. Then ( i + j ) ≥ i + j ) and ( i + j ) − i + j ) ≥ i + j ).Then i ′ = ( i + j − i + j − / i > i + j . So i ′ > i and i ′ > j .C. Assume that H ( n, i ) = P ( n, i ) × X n = { q i } × ... × { q i n } × X n such that i ≥ i j for each j ≤ n when k = n . Let k = n + 1.Take an H ( k, i ′ ) = P ( n, i ) × { q i k } × X k . Then i ′ = ( i + i k − i + i k − / i .Note i ≥ i j for each j ≤ n by inductive assumption. Then i ′ ≥ i by the above 4.So i ′ ≥ i ≥ i j for each j ≤ n . We can prove i ′ = ( i + i k − i + i k − / i ≥ i k if i + i k ≥ (cid:3) Then H ′ = ∪ n H ′ n satisfies condition A of M -structure in [4]. Following onlythe idea of Condition B of M -structures of Proposition 4.7 in [4], we construct a g -function in X by H ′ = ∪ n H ′ n in the following Construction 3.2. Construction 3.2 A. Take H ′ n = { H ( n, i ) : i ∈ N } and an H ( n, i ) ∈ H ′ n . Note H ( n, i ) = { q i } × ... × { q i n } × X n . Let a = n + i . Take I a . Take [ q l j , q l j a )’s from I a for j ≤ i . Let J ( n, i, l ) = [ q l , q l a ) × ... × [ q l i , q l i a ) × X n + i . Let J ( n, i ) be the family of all J ( n, i, l )’s. Then J ( n, i ) is countable and c.o.Din ( X n , ρ ) with ∪ J ( n, i ) = X n . Let J ( n, i ) = { J ( n, i, l ) : l ∈ N } , H ( n, i, l ) = P ( n, i ) × J ( n, i, l ) and H ( n, i ) = { H ( n, i, l ) : l ∈ N } . Then we have the following definition. B. Definition. H ( n, i, l ) = { q i } × ... × { q i n } × [ q l , q l a ) × ... × [ q l i , q l i a ) × X n + i and H ( n, i ) = { H ( n, i, l ) : J ( n, i, l ) ∈ J ( n, i ) } = { H ( n, i, l ) : l ∈ N } . Then H ( n, i ) is a c.D family in ( X, ρ ) with ∪ H ( n, i ) = H ( n, i ). Let H n = ∪ i H ( n, i ) and H = ∪ n H n . HUAIPENG CHEN ∗ AND BOSEN WANG ∗∗∗ In the same way as the ∆-order of H ′ n , we give H n a ∆-order by the ∆-order of H ′ n as the following: H ( n, , , → H ( n, , , H ( n, , , H ( n, , , ..., H ( n, , l ) , ... ւ ւ ւ ...H ( n, , , H ( n, , , H ( n, , , H ( n, , , ..., H ( n, , l ) , ... ւ ւ ...H ( n, , , H ( n, , , H ( n, , , H ( n, , , ..., H ( n, , l ) , ... ւ ...H ( n, , , H ( n, , , H ( n, , , H ( n, , , ..., H ( n, , l ) , ............................................. Figure II. ................................... Then we can calculate the i ′ th member H ( n, i, l ) in H n = ∪ i H ( n, i ) by i ′ = 1 + 2 + ... + ( i + l − 2) + i = ( i + l − i + l − / i. Call that H n has a ∆ -order. For H ( n, i, l ) ∈ H ( n, i ), note a = n + i . Let I ( n, i ) = [ q i , q i a ) × ... × [ q i n , q i n a ) and g ( n, i, l ) = I ( n, i ) × J ( n, i, l ) . Then we have the following definition. C. Definition. g ( n, i, l ) = [ q i , q i a ) × ... × [ q i n , q i n a ) × [ q l , q l a ) × ... × [ q l i , q l i a ) × X n + i . Let G ( n, i ) = { g ( n, i, l ) : l ∈ N } . Then G ( n, i ) = { g ( n, i, l ) = I ( n, i ) × J ( n, i, l ) : J ( n, i, l ) ∈ J ( n, i ) } . Then G ( n, i ) is a c.o.D family in ( X, ρ ). Then ∪ G ( n, i ) = [ q i , q i a ) × ... × [ q i n , q i n a ) × X n = I ( n, i ) × X n by ∪ J ( n, i ) = X n . Let G n = ∪ i G ( n, i ) and G = ∪ n G n . Then G n = ∪ i G ( n, i ) has a∆-order by the ∆-order of H n . And then G = ∪ n G n has a ∆-order by the ∆-orderof G n , in the same way as the ∆-order of H n by the ∆-order of H ′ n .For 1 ≤ k ≤ n , take a projection π k : X → Q k = Q and let J nk = { [ q i k , q i k a ) : [ q i k , q i k a ) = π k [ g ( n, i, l )] for each g ( n, i, l ) ∈ G n } . Note D1: Let g ( n, i, l ) = g ( n, x ) = g ( n, i, x ) for x ∈ H ( n, i, l ) ⊂ H ( n, i ). Then G ( n, i ) = { g ( n, i, l ) : l ∈ N } = { g ( n, i, x ) : x ∈ H ( n, i ) } = { g ( n, x ) : x ∈ H ( n, i ) } . It is easy to see that g : N × X → ∪ n G n is a function. We prove that g : N × X →∪ n G n is a g -function of some stratifiable space in section 7. Note D2: Let x, y ∈ H ( n, i, l ). Then g ( n, i, l ) = g ( n, x ) = g ( n, y ). Let H ( n, i, l ) = H ( n, i, x ) = H ( n, x ) = H ( n, i, y ) = H ( n, y ) and n < n < ... < n k with H ( n k , x k ) ⊂ ... ⊂ H ( n , x ) ⊂ H ( n , x ) . Then H ( n i , x k ) = H ( n i , x i ) since x k ∈ H ( n i , x i ) for 1 ≤ i ≤ k . Note D3: G ( ∗ , ∗ , ∗ ) = { g ( n i , x i ) ∈ G : g ( n i , x i ) satisfies P } if and only if H ( ∗ , ∗ , ∗ ) = { H ( n i , x i ) ∈ H : g ( n i , x i ) satisfies P } . So we give only one of defini-tions G ( ∗ , ∗ , ∗ ) or H ( ∗ , ∗ , ∗ ) if we definite families. NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 9 Note D4: Let H ( n, i ) ∈ H ′ n with H ( n, i ) = { q i } × ... × { q i n } × X n . We have a = n + i , g ( n, i, l ) = [ q i , q i a ) × ... × [ q i n , q i n a ) × [ q l , q l a ) × ... × [ q l i , q l i a ) × X n + i and H ( n, i, l ) = { q i } × ... × { q i n } × [ q l , q l a ) × ... × [ q l i , q l i a ) × X n + i . Let g ( n, x ) = g ( n, i, l ) and H ( n, x ) = H ( n, i, l ) for arbitrary x ∈ H ( n, i, l ). Then g ( n, x ) ∩ H ( n, i ) = H ( n, x ) . Proposition 3.2. Let q i k , q j k ∈ Q = { q i : i ∈ ω } with ∆ -order and i k , j k ∈ N .Then: 1. If q j k ∈ ( q i k , q i k a ) , then j k > i k , j k > a and [ q j k , q j k b ) ⊂ ( q i k , q i k a ) .2. If q j k / ∈ [ q i k , q i k a ) with j k > i k , then [ q i k , q i k a ) ∩ [ q j k , q j k b ) = ∅ .3. If [ q j k , q j k b ) , [ q i k , q i k a ) ∈ J nk with b > a , then [ q i k , q i k a ) ∩ [ q j k , q j k b ) = ∅ or [ q j k , q j k b ) ⊂ [ q i k , q i k a ) .Proof. To prove 1 let q j k ∈ ( q i k , q i k ), q j k ∈ S j ∗ k and q i k ∈ S i ∗ k . Note, by B) inConstruction 1, ( q i k , q i k ) ∩ ( ∪ j ≤ i ∗ k S j ) = ∅ . Then j ∗ k > i ∗ k . Let j ∗ k = i ∗ k + h . Becauseof q j k ∈ ( q i k , q i k ), take S h ( q i k ) = { q i k l ∈ S h : l ∈ N } with h = i ∗ k + 1. Thenthere exists a q i k l ∈ S h ( q i k ) such that q j k ∈ [ q i k l , q i k l − ), q i k l = q k and 1 k > i k in Q by 3 of Proposition 2.1.If q j k = q i k l , then j k = 1 k > i k . If q j k = q i k l , then q j k ∈ ( q k , q k ). Take S h ( q k ) = { q k l ∈ S h : l ∈ N } with h = h + 1. Then there exists a q k l ∈ S h ( q k ) such that q j k ∈ [ q k l , q k l − ), q k l = q k and 2 k > k in Q by 3 ofProposition 2.1.Note i ∗ k < i ∗ k + 1 = h < ... < i ∗ k + h = j ∗ k . Then, by finite induction, thereis an m ≤ h , a q ℓ k l m ∈ S h m ( q ℓ k ) = { q ℓ k l ∈ S h m : l ∈ N } ( ℓ = m − 1) such that q j k ∈ [ q ℓ k l m , q ℓ k l m − ), q j k = q ℓ k l m = q m k and m k > ℓ k in Q by 3 of Proposition 2.1.Then j k = m k > ℓ k > ... > k > k > i k .On the other hand, q j k ∈ ( q i k , q i k a ) implies that there exists an l > a with q j k ∈ [ q i k l , q i k l − ). Let q i k l = q l k in Q . Then, by the above proof, we have j k ≥ l k in Q . Note that q i k l is the l th in S h ( q i k ) ⊂ Q , and q i k l = q l k is the l k th in Q .So l k > l by 1 of Proposition 2.1. So j k ≥ l k > l > a . And then, it is easy tosee [ q j k , q j k b ) ⊂ [ q i k , q i k a ) since q j k ∈ ( q i k , q i k a ) implies [ q j k , q j k b ) ⊂ [ q j k , q j k ) ⊂ ( q i k , q i k a ). This implies 1. The proof of 2. q j k / ∈ [ q i k , q i k a ) implies q j k < q i k or q i k a < q j k . Case 1, q i k a < q j k implies [ q i k , q i k a ) ∩ [ q j k , q j k b ) = ∅ . Case 2, q j k < q i k . Suppose q i k ∈ ( q j k , q j k ). Then i k > j k by 1 of Proposition 3.2, a contradiction to j k > i k . So q i k / ∈ ( q j k , q j k ).Then q j k < q i k implies [ q j k , q j k ) ∩ [ q i k , q i k a ) = ∅ . This implies 2. The proof of 3. Note b > a . Let [ q j k , q j k b ) ∩ [ q i k , q i k a ) = ∅ . If q i k = q j k ,then [ q j k , q j k b ) ⊂ [ q i k , q i k a ). If q i k = q j k , then q i k ∈ ( q j k , q j k b ) or q j k ∈ ( q i k , q i k a ). q i k ∈ ( q j k , q j k b ) implies i k > j k and i k > b by 1 of Proposition 3.2. i k > b implies i k > b = n + j > a = n + i > i , a contradiction to i ≥ i k by 5 of Proposition 3.1.So q j k ∈ ( q i k , q i k a ) implies [ q j k , q j k b ) ⊂ [ q i k , q i k a ). (cid:3) Proposition 3.3. Function G satisfies the following conditions:1 ∩ n g ( n, i, y ) = { y } .4 y ∈ g ( n, i, x ) implies g ( n, j, y ) ⊂ g ( n, i, x ) for some j .5 g ( n + 1 , j, x ) ⊂ g ( n, i, x ) .6 j > k implies H ( n, k ) ∩ ( ∪ G ( n, j )) = ∅ . ∗ AND BOSEN WANG ∗∗∗ ′ H ( n, k, h ) ∩ g ( n, j, l ) = ∅ and H ( n, k, h ) = H ( n, j, l ) imply k > j .7 Every G ( n, i ) is a c.o.D family.8 If g ( n, i, l ) , g ( n, j, ℓ ) ∈ G n with j > i , then g ( n, i, l ) ∩ g ( n, j, ℓ ) = ∅ or g ( n, j, ℓ ) ⊂ g ( n, i, l ) .Proof. We prove 8 at firstly. Let g ( n, i, l ) , g ( n, j, ℓ ) ∈ G n with j > i and g ( n, i, l ) ∩ g ( n, j, ℓ ) = ∅ . Then [ q j k , q j k b ) ⊂ [ q i k , q i k a ) by 3 of Proposition 3.2 since j > i implies b = n + j > n + i = a . Then [ q l k , q l k a ) ∩ [ q ℓ k , q ℓ k b ) = ∅ implies [ q ℓ k , q ℓ k b ) ⊂ [ q l k , q l k a )by 5 of Proposition 2.2. The proof of 1: Let x ∈ H ( n, i n , l n ) ⊂ g ( n, i n , x ) for n ∈ N . Then ∩ n H ( n, i n , l n ) = { x } . Then ∩ n g ( n, i n , x ) = { x } . The proof of 7: G ( n, i ) is a c.o.D family in ( X, ρ ) by C of Construction 3.2. The proof of 6: Take an H ( n, i, l ) and a g ( n, j, ℓ ) with j > i . Then there exists k ≤ n with q i k = q j k . If q i k < q j k , then q i k / ∈ [ q j k , q j k b ). So H ( n, i, l ) ∩ g ( n, j, ℓ ) = ∅ .If q i k ∈ ( q j k , q j k b ), then i k > j k and i k > b by 1 of Proposition 3.2. Note i k > b implies i k > b = n + j > n + i > i , a contradiction to i ≥ i k by 5 of Proposition3.1. So q j k b < q i k . Then q i k / ∈ [ q j k , q j k b ). Then H ( n, i, l ) ∩ g ( n, j, ℓ ) = ∅ . So H ( n, i, l ) ∩ [ ∪ G ( n, j )] = ∅ if j > i . The proof of 6 ′ : Suppose j ≥ k . Case 1, j = k . Note that h = l implies H ( n, k, h ) = H ( n, j, l ), a contradiction to H ( n, k, h ) = H ( n, j, l ). So H ( n, k, h ) ∩ H ( n, j, l ) = ∅ implies h = l . Then g ( n, k, h ) ∩ g ( n, j, l ) = ∅ by the above 7, acontradiction to H ( n, k, h ) ∩ g ( n, j, l ) = ∅ .Case 2, j > k . Then, by the above 6, H ( n, k, h ) ∩ g ( n, j, l ) = ∅ , a contradictionto H ( n, k, h ) ∩ g ( n, j, l ) = ∅ . The proof of 4: In fact, pick a y ∈ g ( n, i, x ) = g ( n, i, l ). If y ∈ H ( n, i, l ),then g ( n, i, y ) = g ( n, i, x ). If y / ∈ H ( n, i, l ), then y ∈ H ( n, j, ℓ ) with H ( n, j, ℓ ) ∩ g ( n, i, x ) = ∅ and H ( n, j, ℓ ) = H ( n, i, l ). Then j > i by 6 ′ of Proposition 3.3. Then b = n + j > n + i = a . Then [ q j k , q j k b ) ⊂ [ q i k , q i k a ) by 3 of Proposition 3.2. Then[ q ℓ k , q ℓ k b ) ⊂ [ q l k , q l k a ) by 5 of Proposition 2.2. This implies g ( n, j, y ) ⊂ g ( n, i, x ). The proof of 5: Let π k be a projection from X to Q k and let H ( n, i, l ) ⊂ H ( n − , j, ℓ ). Then i ≥ j by 4 of Proposition 3.1. Then b = n + i > ( n − 1) + j = a .So π k ( x ) = q i k ∈ [ q j k , q j k a ) implies [ q i k , q i k b ) ⊂ [ q j k , q j k a ) by 3 of Proposition3.2 for 1 ≤ k ≤ n − 1. On the other hand, H ( n, i, l ) ⊂ H ( n − , j, ℓ ) implies[ q l k , q l k b ) ⊂ [ q ℓ k , q ℓ k a ) by 5 of Proposition 2.2 since [ q l k , q l k b ) and [ q ℓ k , q ℓ k a ) in I for n < k ≤ n + i = b . And then π n ( x ) = q i n ∈ [ q ℓ , q ℓ a ) implies [ q i n , q i n b ) ⊂ [ q ℓ , q ℓ a )since b > a . So g ( n, x ) = g ( n, i, l ) ⊂ g ( n − , j, ℓ ) = g ( n − , x ). (cid:3) Proposition 3.4. G = ∪ n G n is a base of metric space ( X, ρ ) .Proof. Let B = B n × ... × B n n × X n ∈ B ′ . Then B = [ q n , q m ) × ... × [ q n n , q m n ) × X n .Let x = ( q i , ..., q i n , ... ) ∈ B . Then q i j ∈ [ q n j , q m j ) for j ≤ n .Case 1, q i j = q n j ∈ [ q n j , q m j ). Then [ q n j , q n j ℓ j ) ⊂ [ q n j , q m j ) for some ℓ j . Take ℓ j .Case 2, q i j ∈ ( q n j , q m j ). Let q i j ∈ S i and q m j ∈ S m . If i ≥ m , take ℓ j = 0.If i < m , then there exists [ q i j l j +1 , q i j l j ) with q m j ∈ [ q i j l j +1 , q i j l j ) and q i j l j +1 ∈ ( q n j , q m j ). Take ℓ j = l j + 1.Let k = max { ℓ j : j ≤ n } and l = max { k, n } . Note, for each H n = ∪ i H ( n, i ),there uniquely exists an H ( n, i n , l n ) ∈ H n with x ∈ H ( n, i n , l n ). Let H x = { H ( n, i n , l n ) ∈ H n : x ∈ H ( n, i n , l n ) } . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 11 Take H ( l, i, h ) = H ( l, i l , l l ) from H x . Let a = l + i l . Then g ( l, x ) = g ( l, i, h ) = [ q i , q i a ) × ... × [ q i l , q i l a ) × J ( l, i, h ) ⊂ B. (cid:3) In topological space ( X, ρ ), denote the closure of a set A by Cl ρ A , and letInt ρ A = X − Cl ρ ( X − A ) . Proposition 3.5. 1. Int ρ H ( n, x ′ ) = ∅ for arbitrary H ( n, x ′ ) ∈ H .2. Cl ρ [ H ( n, y ) − H ( n + 1 , y )] = H ( n, y ) for arbitrary H ( n, y ) ∈ H .3. Let H ( n i +1 , y i +1 ) ⊂ H ( n i , y i ) and n i < n i +1 . Then ∩ i H ( n i , y i ) = { y } .Proof. Note g ( ℓ, x ) − H ( l, y ) = ∅ for arbitrary g ( ℓ, x ) and arbitrary H ( l, y ). ThenInt ρ H ( n, x ′ ) = ∅ for arbitrary H ( n, x ′ ) ∈ H .To see 3 note H ( n i , y i ) = { q i } × ... × { q i n } × [ q l , q l a ) × ... × [ q l i , q l i a ) × X n i for n i = n by the definition of H ( n, i, l ). Note H ( n i +1 , y i +1 ) ⊂ H ( n i , y i ) for i ∈ N .Let n i +1 = m > n = n i . Then H ( n i +1 , y i +1 ) = { q i } × ... × { q i n } × ... × { q i m } × [ q h , q h b ) × ... × [ q h k , q h k b ) × X n i +1 . Then ∩ i H ( n i , y i ) = ( q i , q i , ..., q i n , ... ). (cid:3) To construct Ln covers sequences on G m We’ll use Ln covers and Ln covers sequences throughout this paper. To constructLn-covers sequences on g ( m, a ∗ , y ), let m = m + a ∗ , n ≥ m , G m = g ( m, a ∗ , y ) = I ( m, a ∗ , y ) × J ( m, a ∗ , y ) , H = H ( m, a ∗ ) ∩ g ( m, a ∗ , y ) , H ( m, 1) = { H ( n, ′ i , i l ) ∈ H n : H ( n, ′ i , i l ) ⊂ G m } and G ( m, 1) = { g ( n, ′ i , i l ) ∈ G n : H ( n, ′ i , i l ) ∈ H ( m, } . Then H = H ( m, a ∗ , y ) by Note D4. Then we have the following fact. Fact 4.0. H ( ℓ, h, k ) ∩ G m = ∅ implies g ( ℓ, h, k ) ⊂ G m if ℓ ≥ m . Construction 4. Condition. H ( m, 1) = { H ( n, ′ i , i l ) ∈ H n : H ( n, ′ i , i l ) ⊂ G m } with ∆-order.2. ∪ H ( m, 1) = ∪ G ( m, 1) = G m .Æ. We use Conditions of Construction 4 to construct a c.o.D family G ( n, m, X, ρ ) by induction. Operation O1. Conditions. H ( m, 1) with ∆-order.2. 1 ′ < ′ < ... < ′ i < ... .We use Conditions of Operation O1 to construct the first cover of G m by induc-tion.A. Pick the least number 1 ′ . Denote 1 ′ by 1 . Let H ∗ (1 , 1) = { H ( n, , l ) ∈ H ( n, ) : g ( n, , l ) ⊂ G m } . Then O ∗ (1 , 1) = { g ( n, , l ) : l ∈ N } is a c.o.D family. Let O m = ∪ O ∗ (1 , O ∗ (1 , k ), a c.D family H ∗ (1 , k ) and an O km for each k < h . Let G hm = G m − ∪ k 0) = ∪ k 0) = ∪ k 0) = ∪ h O ∗ (1 , h ) and H ( n, m, 0) = ∪ h H ∗ (1 , h ).We call the above induction Operation O1 on H ( m, and G ( m, .Claim 4.1. ∪ G ( n, m, 0) = G m and g ( n, h, k ) is a c.o set in ( X, ρ ) for every g ( n, h, k ) ∈ G ( n, m, .2. Every O ∗ (1 , h ) is c.o.D, and G ( n, m, 0) = ∪ h O ∗ (1 , h ) is c.o.D in ( G m , ρ ) .3. H ( n, m, is a c.D family and g [ n, H m ] = ∪{ g ( n, x ) : x ∈ H m } = G m for ∪ H ( n, m, 0) = H m .4. Let g ( ℓ, i ′′ , l ′′ ) ⊂ G m with ℓ ≥ n . Then there exists a g ( n, i ′ , l ′ ) ∈ G ( n, m, such that g ( ℓ, i ′′ , l ′′ ) ⊂ g ( n, i ′ , l ′ ) .5. Let g ( n, i, l ) be the first member in G ( n, m, with ∆ -order. Then H ( n, i, l ) = H ( n, , ) ⊂ ∪ H ∗ (1 , ⊂ H . Proof. To see 1 pick an x ∈ G m = ∪ H ( m, H ( n, ′ i , i h )with x ∈ H ( n, ′ i , i h ). Case 1, i = 1. Then H ( n, , e ) ∈ H ∗ (1 , 1) and x ∈ H ( n, , e ) ⊂ O m . Case 2, i > 1. Then 1 k < ′ i < k +1 . This implies x ∈ O km , and1 ′ i = 1 k +1 implies x ∈ O k +1 m . And then it is easy to see that g ( n, h, k ) is a c.o set in( X, ρ ). This implies 1. The proof of 2. Note that O km ∩ O k +1 m = ∅ and each O ∗ (1 , h ) is a c.o.D family.Then G ( n, m, 0) is a c.o.D family in ( X, ρ ) since ∪ G ( n, m, 0) = G m is a c.o set. The proof of 4. Let H ( n, i ′ , l ′ ) ∩ G m = ∅ . Then g ( n, i ′ , l ′ ) ⊂ G m by Fact 4.0.Then H ( n, i ′ , l ′ ) ∈ H ( m, i with 1 i ≤ i ′ < i +1 . If 1 i = i ′ ,then g ( n, i , i l ) = g ( n, i ′ , l ′ ) for some i l . If 1 i < i ′ , then i ′ < i +1 and H ( n, i ′ , l ′ ) ⊂ g ( n, i , i l ) for some i l by the definition of 1 i . Then g ( n, i ′ , l ′ ) ⊂ g ( n, i , i l ).Let ℓ > n . Then there exists an H ( n, i ′ , l ′ ) ∈ H ( m, 1) with H ( ℓ, i ′′ , l ′′ ) ⊂ H ( n, i ′ , l ′ ). Then there exists a g ( n, i ∗ , l ∗ ) ∈ G ( n, m, 0) such that H ( n, i ′ , l ′ ) ⊂ g ( n, i ∗ , l ∗ ). So g ( ℓ, i ′′ , l ′′ ) ⊂ g ( n, i ∗ , l ∗ ) by 4 and 5 of Proposition 3.3. The proof of 5. Let g ( n, i, l ) ∈ G ( n, m, 0) be the first member in G ( n, m, H ( n, i, l ) ⊂ G m = g ( m, a ∗ , y ). Then i ≥ since ∪ H ∗ (1 , ⊂ H and 1 isthe least number.Suppose i > . Then H ( n, i, l ) is in the i ’th line in the Figure II. It is a con-tradiction since H ( n, i, l ) is the first member in G ( n, m, 0) with ∆-order (followingthe direction of arrow in the Figure II).Then i = 1 and H ( n, i, l ) ⊂ ∪ H ∗ (1 , ⊂ H . Then, in the first line, we have H ( n, i, l ) = H ( n, , 1) = H ( n, , ). (cid:3) NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 13 Definition Ln. Call G ( n ) a least number cover ( Ln cover) on ( n, B ) (oron B ) if and only if G ( n ) and B satisfy 1, 2 and 4 of Claim 4.1.It is easy to check the following fact. Fact *. Let G ( n ) be a Ln cover on ( n, B ). Then G ′ is a Ln cover on ( n, ∪ G ′ ) if G ′ ⊂ G ( n ). Definition Ln ′ . Let H be a closed set in ( ∪ G ′ , ρ ) with H ∩ H ( n, i ′ , l ′ ) = ∅ foreach g ( n, i ′ , l ′ ) ∈ G ′ . Also call G ′ a Ln cover of ( n, H ) (or H ).Note Ln. Let H be a closed set in ( X, ρ ) and G ( n, H ) = { g ( n, x ) ∈ G n : x ∈ H } . Then there exists G ′ ( n, H ) ⊂ G ( n, H ) such that ∪ G ′ ( n, H ) = ∪ G ( n, H ) and G ′ ( n, H ) is a Ln cover of ( n, H ) by 3 of Claim 4.1. Call G ′ ( n, H ) a Ln subcover of( n, H ) also. Then ∪ G ′ ( n, H ) is c.o in ( X, ρ ).Œ. ( Construction 4 is continued. ) Assume that we have constructed a c.o.Dfamily G ( n a , m, 0) and a c.D family H ( n a , m, 0) in ( X, ρ ) such that Claim 4.1. Inthe following, we construct the next Ln cover by induction. To do it let b = a + 1and n b = n a + 1.Take a g ( n a , a j , j h ) ∈ G ( n a , m, H ( n, i, i l ) ∈ H ( m, H ( n, b ′ i , i l ) = H ( n, i, i l ) ∩ g ( n a , a j , j h ) if H ( n, i, i l ) ∩ g ( n a , a j , j h ) = ∅ , and let H ( n, a j , j h ) = { H ( n, b ′ i , i l ) = H ( n, i, i l ) ∩ g ( n a , a j , j h ) : H ( n, i, i l ) ∈ H ( m, } . Then b ′ < b ′ < ... < b ′ i < ... by ∆-order of H ( m, H ′ ( n b , a j , j h ) = { H ( n b , b i , i k ) ∈ H n b : H ( n b , b i , i k ) ⊂ g ( n a , a j , j h ) } . Then ∪ H ′ ( n b , a j , j h ) = ∪ G ′ ( n b , a j , j h ) = g ( n a , a j , j h ). Give H ′ ( n b , a j , j h ) a ∆-order as a subsequence of H n b . Then we have the following claim. Claim 4.2. Let H ( n b , b , e ) be the first member in H ′ ( n b , a j , j h ) with ∆ -order,and H ( n, b ′ , l ) be the first member in H ( n, a j , j h ) with ∆ -order. Then:1. H ( n b , b , e ) ⊂ H ( n a , a j , j h ) ⊂ H ( n, b ′ , l ) .2. b ′ is the least number in H ( n, a j , j h ) , and b is the least number in H ′ ( n b , a j , j h ) with b ′ < a j < b .Proof. At first, we prove H ( n a , a j , j h ) ⊂ H ( n, b ′ , ′ l ) for the first member H ( n, b ′ , ′ l ) ∈ H ( n, a j , j h ). To do it Suppose H ( n a , a j , j h ) ⊂ H ( n, b , l ) with b ′ = b .Then H ( n a , a j , j h ) ⊂ H ( n, b , l ) implies g ( n a , a j , j h ) ⊂ g ( n, b , l ) since n < n a .Then H ( n, b ′ , ′ l ) ∩ g ( n, b , l ) = ∅ . Then b ′ = b implies b < b ′ by 6 ′ of Proposition3.3. It is a contradiction to the definition of b ′ . Then b = b ′ .On the other hand, 1 ′ l = 1 l implies g ( n, b ′ , ′ l ) ∩ g ( n, b ′ , l ) = ∅ since G ( n, b ′ ) isdiscrete by 7 of Proposition 3.3. Then H ( n, b ′ , ′ l ) ∩ g ( n a , a j , j h ) = ∅ , a contradictionto H ( n, b ′ , ′ l ) ∈ H ( n, a j , j h ). This implies H ( n a , a j , j h ) ⊂ H ( n, b , l ) . We prove H ( n b , b , e ) ⊂ H ( n a , a j , j h ). To do it take the first member H ( n b , b , e )from H ′ ( n b , a j , j h ).Replace H with H ( n, b , l ), ∪ H ∗ (1 , 1) with H ( n a , a j , j h ), and G ( n, m, 0) with G ′ ( n b , a j , j h ) in 5 of Claim 4.1. Here H ′ ( n b , a j , j h ) = { g ( n b , x ) ∈ G ( n b , m, 0) : g ( n b , x ) ⊂ g ( n a , a j , j h ) } . Then, by 5 of Claim 4.1, we have H ( n b , b , e ) ⊂ H ( n a , a j , j h ) ⊂ H ( n, b ′ , l ). Then,by 4 of Proposition 3.1, b ′ < a j < b since n b > n a > n ≥ m . (cid:3) ∗ AND BOSEN WANG ∗∗∗ Construction 4 is continued. Take G ′ ( n b , a j , j h ) and H ′ ( n b , a j , j h ). Then H ′ ( n b , a j , j h ) = { H ( n b , b i , i k ) ∈ H n b : i, k ∈ N } , ∪ H ′ ( n b , a j , j h ) = ∪ G ′ ( n b , a j , j h ) = g ( n a , a j , j h ) and b < b < ... < b i < .... Replace n with n b , G m with g ( n a , a j , j h ), and H ( m, 1) with H ′ ( n b , a j , j h ) inoperation O1. Then Condition of Operation O1 is satisfied. Then we have thefollowing claim. Claim 4.3. G ( n b , a j , j h ) is a Ln cover on g ( n a , a j , j h ) for every g ( n a , a j , j h ) ∈ G ( n a , m, .2. H ( n b , a j , j h ) is a c.D family in ( G m , ρ ) .3. Let ∪ H ( n b , a j , j h ) = H . Then g [ n b , H ] = ∪ G ( n b , a j , j h ) = g ( n a , a j , j h ) .4. Let g ( ℓ, i ′ , l ′ ) ⊂ G m with ℓ ≥ n b . Then there exists a g ( n b , i ∗ , l ∗ ) ∈ G ( n b , m, such that g ( ℓ, i ′ , l ′ ) ⊂ g ( n b , i ∗ , l ∗ ) .5. G ( n b , m, 0) = ∪{ G ( n b , a j , j h ) : g ( n a , a j , j h ) ∈ G ( n a , m, } is a Ln cover on G m .Proof. Note 2 of Claim 4.1. Then G ( n b , a j , j h ) = ∪{ G ( n b , b k , k ) : k ∈ N } is a c.o.Dfamily in ( X, ρ ). Then ∪ G ( n b , a j , j h ) = g ( n a , a j , j h ) by 1 of Claim 4.1 if we replace G m with g ( n a , a j , j h ). Then G ( n b , a j , j h ) is a Ln cover on g ( n a , a j , j h ) by 1,2 and4 of Claim 4.1. Then 2 is a corollary of 1.To prove 3, replace ∪ H ( n, m, 0) = H m with ∪ H ( n b , a j , j h ) = H in 3 of Claim4.1. Then g [ n b , H ] = ∪ G ( n b , a j , j h ) = g ( n a , a j , j h ).We prove 4 by induction. Note g ( ℓ, i ′ , l ′ ) ⊂ G m for ℓ ≥ m . Assume that g ( ℓ, i ′ , l ′ ) ⊂ G m with ℓ ≥ n a implies g ( ℓ, i ′ , l ′ ) ⊂ g ( n a , a j , j h ) for some g ( n a , a j , j h ) ∈ G ( n a , m, g ( ℓ, i ′ , l ′ ) ⊂ G m with ℓ ≥ n b . Then there exists a g ( n a , a j , j h ) ∈ G ( n a , m, g ( ℓ, i ′ , l ′ ) ⊂ g ( n a , a j , j h ) by inductive assumption. Note g ( ℓ, i ′ , l ′ ) ⊂ g ( n a , a j , j h ) and G ( n b , a j , j h ) is a Ln cover on g ( n a , a j , j h ). Then there exists a g ( n b , i ∗ , l ∗ ) ∈ G ( n b , a j , j h ) ⊂ G ( n b , m, 0) such that g ( ℓ, i ′ , l ′ ) ⊂ g ( n b , i ∗ , l ∗ ) by thedefinition of Ln cover since G ( n b , a j , j h ) is a Ln cover of g ( n a , a j , j h ). This implies4. Note G ( n a , m, 0) is a c.o.D family by inductive assumption. Then G ( n b , m, 0) isa c.o.D family in ( G m , ρ ). So G ( n b , m, 0) is a Ln cover on ( n b , G m ). (cid:3) Note Æ and Œ. We have completed induction on b ∈ N . Let n b = n a + 1 and G ( n, m, L ) = { G ( n i , m, 0) : n i ≥ n } . Then we have proved the following claim. Claim 4.4. G ( n, m, L ) is a Ln covers sequence on G m .2. Every G ( n b , m, is a Ln cover on G m .3. Every H ( n b , m, is a c.D family in ( G m , ρ ) .4. For every g ( n a , i, j ) ∈ G ( n a , m, , there exists a family G ( n b , i, j ) ⊂ G ( n b , m, , ∪ G ( n b , i, j ) = g ( n a , i, j ) and [ ∪ ( G ( n b , m, − G ( n b , i, j ))] ∩ g ( n a , i, j ) = ∅ .5. Let H ( ℓ, i ′ , l ′ ) ∩ G m = ∅ with ℓ ≥ n b . Then there exists a g ( n b , i ∗ , l ∗ ) ∈ G ( n b , m, such that g ( ℓ, i ′ , l ′ ) ⊂ g ( n b , i ∗ , l ∗ ) . This completes Construction 4. NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 15 In the following, we describe a Ln cover by coordinates. To do it let G ( n, m, ∈G ( n, m, L ). Then ∪ G ( n, m, 0) = G m . Note, by Operation O1, H ( n, m, 0) = ∪ h H ∗ (1 , h ) and H ∗ (1 , h ) = { H ( n, h , h l ) ∈ H ( n, h ) : H ( n, h , h l ) ⊂ G hm } .A. Take H ∗ (1 , g ( n, i, l ). Let 1 be the least numberand a = n + 1 . Take g ( n, , l ) ∈ O ∗ (1 , ′ = 1 . Then g ( n, , l ) = [ q , q a ) × ... × [ q n , q n a ) × [ q l , q l a ) × ... × [ q l ′ , q l ′ a ) × X a . Then O ∗ (1 , 1) = { g ( n, , l ) : l ∈ N } .B. Assume O ∗ (1 , k ) = { g ( n, k , k l ) : l ∈ N } with a k = n + 1 k . Take H ∗ (1 , h ) for h = k + 1 in Operation O1. Let h ′ = n + 1 h , g ( n, h , h l ) ∈ O ∗ (1 , h ) and a h = n + 1 h .Then a h > a k > a and g ( n, h , h l ) = [ q h , q h a h ) × ... × [ q h n , q h n a h ) × [ q ℓ , q ℓ a h ) × ... × [ q ℓ h ′ , q ℓ h ′ a h ) × X a h . Then O ∗ (1 , h ) = { g ( n, h , h l ) : l ∈ N } is c.o.D in ( X, ρ ). Then, by induction on N ,we have O ∗ (1 , h ) for every h ∈ N .Take Q ∗ a h from Proposition 2.2. Note m = m + a ∗ for G m = g ( m, a ∗ , y ). Let j ≤ n and I ( j, G m ) = { [ q h j , q h j a h ) : q h j ∈ Q ∗ a h ∩ [ q i j , q i j m ) } . Then ∪ I ( n, G m ) = [ q i j , q i j m ) and I ( j, G m ) is c.o.D in ( Q, ρ ) for j < n by 3 ofProposition 2.2.Let g ( n, h , h l ) ∈ O ∗ (1 , h ), J ( n, h , h l ) = [ q ℓ , q ℓ a h ) × ... × [ q ℓ h ′ , q ℓ h ′ a h ) × X a h and J (1 h , G m ) = { J ( n, h , h l ) : l ∈ N } . Then, for every h ∈ N , we have ∪ J (1 h , G m ) = J ( m, a ∗ , y ) since G m = I ( m, a ∗ , y ) × J ( m, a ∗ , y ) . Then we have the following Proposition. Proposition 4.5. Let m = m + a ∗ and G ( n, m, 0) = ∪ h G ∗ (1 , h ) be a Ln cover on G m and G m = [ q i , q i m ) × ... × [ q i m , q i m m ) × [ q l , q l m ) × ... × [ q l ′ , q l ′ m ) × X m . 1. Let J (1 h , G m ) = { J ( n, h , h l ) : g ( n, h , h l ) ∈ G ∗ (1 , h ) } for every h ∈ N .Then ∪ J (1 h , G m ) = J ( m, a ∗ , y ) and J (1 h , G m ) is c.o.D in ( J ( m, a ∗ , y ) , ρ ) .2. Let j ≤ m and I ( j, G m ) = { [ q h j , q h j a h ) : g ( n, h , h l ) ∈ G ( n, m, } . Then ∪ I ( j, G m ) = [ q i j , q i j m ) for j ≤ m , and I ( j, G m ) is c.o.D in ( Q, ρ ) . Note 4.6. When g ( n a , i, j ) ∈ G ( n a , m, 0) with b > a , let G ( n b , i, j ) = { g ( n b , i ′ , j ′ ) ∈ G ( n b , m, 0) : g ( n b , i ′ , j ′ ) ⊂ g ( n a , i, j ) } and H ( n b , i, j ) = { H ( n b , i ′ , j ′ ) ∈ H ( n b , m, 0) : g ( n b , i ′ , j ′ ) ∈ G ( n b , i, j ) } . Then G ( n b , i, j ) is a Ln cover on ( n b , g ( n a , i, j )). Both G ( n b , i, j ) and H ( n b , i, j ) are used always throughout this paper for g ( n a , i, j ). Then we may replace G m with g ( n a , i, j ), and G ( n b , i, j ) = ∪ h G ( n b , h ) in Proposition 4.5. We have the followingcorollary. Corollary 4.6. Let G ( n b , i, j ) = ∪ h G ( n b , h ) be a Ln cover on g ( n a , i, j ) .1. Let J (1 h , n b ) = { J ( n b , h , h l ) : g ( n b , h , h l ) ∈ G ( n b , h ) } for every h ∈ N .Then ∪ J (1 h , n b ) = J ( n a , i, j ) and J (1 h , n b ) is c.o.D in ( J ( n a , i, j ) , ρ ) .2. Let l ≤ n a , a h = n b + 1 h , c = n a + i for g ( n a , i, j ) and I ( l, g ( n a , i, j )) = { [ q h l , q h l a h ) : g ( n b , h , h l ) ∈ G ( n b , h ) } . Then ∪ I ( l, g ( n a , i, j )) = [ q i l , q i l c ) for l ≤ n a , and I ( l, g ( n a , i, j )) is c.o.D in ( Q, ρ ) . ∗ AND BOSEN WANG ∗∗∗ Note 4.7. Note G ( n b , i, j ) = ∪ h G ( n b , h ) if g ( n a , i, j ) ∈ G ( n a , m, 0) with b > a by Corollary 4.6. Let G ( n b , i, j, H a ) = { g ( n b , h , h l ) ∈ G ( n b , i, j ) : H ( n b , h , h l ) ⊂ H ( n a , i, j ) } and H ( n b , i, j, H a ) = { H ( n b , h , h l ) ∈ H ( n b , i, j ) : g ( n b , h , h l ) ∈ G ( n b , i, j, H a ) } . Then G ( n b , i, j, H a ) is a Ln cover of H ( n a , i, j ). Let G ( n b , H a , h ) = { g ( n b , h , h l ) ∈ G ( n b , h ) : H ( n b , h , h l ) ⊂ H ( n a , i, j ) } and H ( n b , H a , h ) = { H ( n b , h , h l ) : g ( n b , h , h l ) ∈ G ( n b , H a , h ) } . Corollary 4.7. G ( n b , i, j, H a ) = ∪ h G ( n b , H a , h ) is a Ln cover of H ( n a , i, j ) .2. G ( n b , H a , h ) is a Ln cover of H ( n a , i, j ) ∩ H ( n b , h ) for every h ∈ N . We use Ln covers sequence G ( n, m, L ) = { G ( n i , m, 0) : n i ≥ n } on G m to provethe following proposition. Proposition 4.8. Let H ( n, i ∗ , l ∗ ) ⊂ G m . Then there exists an a ∈ N and a family G ( a, i ∗ , l ∗ ) ⊂ G ( a, m, such that ∪ H ( a, i ∗ , l ∗ ) ⊂ H ( n, i ∗ , l ∗ ) ⊂ ∪ G ( a, i ∗ , l ∗ ) .Proof. Let H ( n, i ∗ , l ∗ ) ∩ g ( n, h, k ) = ∅ , H ( n, i ∗ , l ∗ ) ∩ H ( n, h, k ) = ∅ and G ( ℓ, n, h, k ) = { g ( ℓ, h ′ , k ′ ) ∈ G ( ℓ, m, 0) : g ( ℓ, h ′ , k ′ ) ∩ H ( n, h, k ) = ∅} . We prove that there exists an a h with H ( n, i ∗ , l ∗ ) ∩ [ ∪ G ( ℓ, n, h, k )] = ∅ for each ℓ ≥ a h .At first, H ( n, i ∗ , l ∗ ) ∩ g ( n, h, k ) = ∅ and H ( n, i ∗ , l ∗ ) ∩ H ( n, h, k ) = ∅ implies i ∗ > h by 6 ′ of Proposition 3.3. Note that H ( n, i ∗ , l ∗ ) ∩ g ( n, h, k ) = ∅ implies g ( n, i ∗ , l ∗ ) ⊂ g ( n, h, k ) by 4 of Proposition 3.3. Let G ( n, f ) = { g ( n, h, k ) ∈ G n : H ( n, i ∗ , l ∗ ) ∩ g ( n, h, k ) = ∅ and H ( n, i ∗ , l ∗ ) ∩ H ( n, h, k ) = ∅} . Then G ( n, f ) = { g ( n, h i , k i ) : h i < i ∗ } is finite by 7 and 8 of Proposition 3.3.And then, π j [ H ( n, i ∗ , l ∗ )] = { q i ∗ j } and π j [ H ( n, h, k )] = { q h j } with q i ∗ j = q h j forsome j ≤ n . Here π j is a projection from X to Q j = Q .Suppose q i ∗ j = q h j for each j ≤ n . Then P ( n, i ∗ ) = P ( n, h ). Then H ( n, i ∗ , l ∗ ) ∩ H ( n, h, k ) = ∅ implies J ( n, i ∗ , l ∗ ) ∩ J ( n, h, k ) = ∅ . So g ( n, i ∗ , l ∗ ) ∩ g ( n, h, k ) = ∅ , acontradiction to H ( n, i ∗ , l ∗ ) ∩ g ( n, h, k ) = ∅ .Let q i ∗ j = q h j . Then q i ∗ j / ∈ [ q h j , q h j a i ) for some a i . Pick an ℓ ≥ a i . Let x ∈ H ( ℓ, h ′ , k ′ ) ⊂ H ( n, h, k ). Then H ( n, i ∗ , l ∗ ) ∩ g ( ℓ, x ) = H ( n, i ∗ , l ∗ ) ∩ g ( ℓ, h ′ , k ′ ) = ∅ since q i ∗ j / ∈ [ q h j , q h j a ′ ) = π j [ g ( ℓ, x )] with a ′ = ℓ + h ′ > a i . Then H ( n, i ∗ , l ∗ ) ∩ [ ∪ G ( ℓ, n, h, k )] = ∅ for each ℓ ≥ a i .Let a = max { a i : h i < i ∗ } . Take G ( a, n, i ∗ , l ∗ ). Then H ( n, i ∗ , l ∗ ) ⊂ ∪ G ( a, n, i ∗ , l ∗ )since G ( a, m, ∈ G ( n, m, L ). Take a g ( a, i, l ) ∈ G ( a, n, i ∗ , l ∗ ). Then g ( a, i, l ) ∩ H ( n, i ∗ , l ∗ ) = ∅ . Let H ( a, i, l ) ⊂ H ( n, i a , l a ). Then g ( n, i a , l a ) ∩ H ( n, i ∗ , l ∗ ) = ∅ .Then g ( n, i ∗ , l ∗ ) ⊂ g ( n, i a , l a ).Suppose H ( n, i ∗ , l ∗ ) ∩ H ( n, i a , l a ) = ∅ . Then g ( n, i a , l a ) = g ( n, h i , k i ) ∈ G ( n, f )for some h i < i ∗ . Then, for each ℓ ≥ a i , H ( n, i ∗ , l ∗ ) ∩ [ ∪ G ( ℓ, n, i a , l a )] = H ( n, i ∗ , l ∗ ) ∩ [ ∪ G ( ℓ, n, h i , k i )] = ∅ . Then H ( n, i ∗ , l ∗ ) ∩ [ ∪ G ( a, n, i a , l a )] = ∅ for a ≥ a i . Note H ( a, i, l ) ⊂ H ( n, i a , l a ) and g ( a, i, l ) ∈ G ( a, m, 0) since g ( a, i, l ) ∈ G ( a, n, i ∗ , l ∗ ). Then g ( a, i, l ) ∈ G ( a, n, i a , l a ).Then g ( a, i, l ) ∩ H ( n, i ∗ , l ∗ ) = ∅ , a contradiction to g ( a, i, l ) ∈ G ( a, n, i ∗ , l ∗ ) and g ( a, i, l ) ∩ H ( n, i ∗ , l ∗ ) = ∅ . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 17 So H ( n, i ∗ , l ∗ ) ∩ H ( n, i a , l a ) = ∅ . Then H ( n, i ∗ , l ∗ ) = H ( n, i a , l a ). Then, foreach g ( a, i, l ) ∈ G ( a, n, i ∗ , l ∗ ), we have H ( a, i, l ) ⊂ H ( n, i ∗ , l ∗ ). So ∪ H ( a, i ∗ , l ∗ ) ⊂ H ( n, i ∗ , l ∗ ) ⊂ ∪ G ( a, n, i ∗ , l ∗ ). Then G ( a, n, i ∗ , l ∗ ) = G ( a, i ∗ , l ∗ ) is desired. (cid:3) Corollary 4.9. Let H ( n ∗ , i ∗ , l ∗ ) ⊂ G m with n ∗ ≥ n . Then there exists a k ∈ N and a family G ( k, i ∗ , l ∗ ) ⊂ G ( k, m, such that ∪ H ( k, i ∗ , l ∗ ) ⊂ H ( n ∗ , i ∗ , l ∗ ) ⊂∪ G ( k, i ∗ , l ∗ ) .Proof. Let H ( n ∗ , i ∗ , l ∗ ) ⊂ G m . Then H ( n ∗ , i ∗ , l ∗ ) ⊂ ∪ G ( n ∗ , n ∗ , i ∗ , l ∗ ). Then G ( n ∗ , n ∗ , i ∗ , l ∗ ) = { g ( n ∗ , i ′ , l ′ ) } by 5 of Claim 4.4. If H ( n ∗ , i ′ , l ′ ) = H ( n ∗ , i ∗ , l ∗ ),then G ( n ∗ , n ∗ , i ∗ , l ∗ ) is desired. If H ( n ∗ , i ′ , l ′ ) = H ( n ∗ , i ∗ , l ∗ ), then H ( n ∗ , i ′ , l ′ ) ∩ H ( n ∗ , i ∗ , l ∗ ) = ∅ and H ( n ∗ , i ∗ , l ∗ ) ⊂ g ( n ∗ , i ′ , l ′ ). We consider Ln covers sequenceon G m ∩ g ( n ∗ , i ′ , l ′ ): G ( n ∗ , m, | g ( n ∗ , i ′ , l ′ ) , G ( n ∗ + 1 , m, | g ( n ∗ , i ′ , l ′ ) , G ( n ∗ + 2 , m, | g ( n ∗ , i ′ , l ′ ) , .... Then, by Proposition 4.8, there exists a k ∈ N and a family G ( k, n ∗ , i ∗ , l ∗ ) | g ( n ∗ , i ′ , l ′ ) ⊂ G ( k, m, | g ( n ∗ , i ′ , l ′ ) such that ∪ [ H ( k, n ∗ , i ∗ , l ∗ ) | g ( n ∗ , i ′ , l ′ )] ⊂ H ( n ∗ , i ∗ , l ∗ ) ⊂ ∪ [ G ( k, n ∗ , i ∗ , l ∗ ) | g ( n ∗ , i ′ , l ′ )] . Note G ( k, n ∗ , i ∗ , l ∗ ) | g ( n ∗ , i ′ , l ′ ) = G ( k, n ∗ , i ∗ , l ∗ ) ⊂ G ( k, m, G ( k, n ∗ , i ∗ , l ∗ ) | g ( n ∗ , i ′ , l ′ ) = G ( k, n ∗ , i ∗ , l ∗ ) = G ( k, i ∗ , l ∗ ) is desired. (cid:3) Professor Gary Gruenhage went through the primitive paper, and told us thatsome sections are too complicated and too hard to understand. So we are going torewriting the paper from Section 5.5. To construct a Ln cover on g c ( n, i, l ) − H In order to construct holes in neighborhoods, we have to construct a Ln coverson g ( n, i, l ) − H also.Let family H with ∆-order and H ′ ⊂ H . Call H ′ with ∆ -order if H ′ as asubsequence of H . We always use H ′ with ∆-order throughout this paper. Proposition 5.1. Fix an n ∈ N and an m < n . Then, for arbitrary g ( n, i, l ) ∈ G n , there exists c ∈ N such that k > c and H ( m, k, h ) ∩ g ( n, i, l ) = ∅ implies g ( m, k, h ) ⊂ g ( n, i, l ) .Proof. 1. Let g ( n, i, l ) = [ q i , q i a ) × ... × [ q i n , q i n a ) × [ q l , q l a ) × ... × [ q l i , q l i a ) × X n + i and c = a . Assume q i j ∈ S h ′ j for 1 ≤ j ≤ n , and q l j ∈ S h ′ n + j for 1 ≤ j ≤ i . Let h ′ = max { h ′ j : 1 ≤ j ≤ n + i } . Then q i j ∈ S h ′ j implies q i j = q ℓ j k j ∈ S h ′ j ( q ℓ j ) forsome q ℓ j ∈ S ℓ ∗ j with ℓ ∗ j ≤ h ′ − 1. So do q l j ∈ S h ′ n + j for 1 ≤ j ≤ i . If q i j = q = 0 forsome j , let k j = 0.2. Let c = max { k j : 1 ≤ j ≤ n + i } . Then, for each j , q ℓ j ∈ S l ∗ j implies q ℓ j = q m j k j ∈ S l ∗ j ( q m j ) for some q m j ∈ S m ∗ j with m ∗ j ≤ l ∗ j − ≤ h ′ − 2. If q ℓ j = q = 0 for some j , let k j = 0.3. Let c = max { k j : 1 ≤ j ≤ n + i } . Note h ′ is finite. Then, by finiteinduction at most h ′ times, we have c , c , ..., c h ′ . Let c = max { c i : i ≤ h ′ } . Then q i j , q ℓ j , ..., q l j in Q ∗ c for j ≤ n + i . Let I ( n, i, l ) be the family of all [ q ℓ j , q ℓ j c )’s,...,[ q i j , q i j c )’s and [ q l j , q l j c )’s for j ≤ n + i . Then I ( n, i, l ) ⊂ I c by 2 of Proposition2.2. ∗ AND BOSEN WANG ∗∗∗ Let H ( m, k, h ) ∈ H m with H ( m, k, h ) ∩ g ( n, i, l ) = ∅ , k > c , b = m + k and H ( m, k, h ) = { q k } × ... × { q k m } × [ q h , q h b ) × ... × [ q h k , q h k b ) × X b . Case 1. 1 ≤ j ≤ m . H ( m, k, h ) ∩ g ( n, i, l ) = ∅ implies q k j ∈ [ q i j , q i j a ). k > c ≥ c implies b = m + k > k > c ≥ a . If q k j = q i j , then [ q k j , q k j b ) ⊂ [ q i j , q i j c ) ⊂ [ q i j , q i j a ).If q k j = q i j , then q k j ∈ ( q i j , q i j a ). Then [ q k j , q k j b ) ⊂ [ q i j , q i j c ) ⊂ [ q i j , q i j a ) by 1 ofProposition 3.2. Case 2. m < j ≤ n . Then b = m + k > k > c ≥ a . Let [ q h j , q h j b ) ∩ [ q i j , q i j a ) = ∅ .Suppose q h j / ∈ [ q i j , q i j a ). Then q h j < q i j or q i j a < q h j . Subcase 2.1. q i j a < q h j . Then [ q i j , q i j a ) ∩ [ q h j , q h j b ) = ∅ . It is a contradictionto [ q h j , q h j b ) ∩ [ q i j , q i j a ) = ∅ . Subcase 2.2. q h j < q i j . (We want to know q h j b < q i j a or q i j a < q h j b .)(a). q i j = q ℓ j k j ∈ S h ′ j ( q ℓ j ) for some q ℓ j ∈ S ℓ ∗ j . Then q ℓ j < q i j .And then q ℓ j = q m j k j ∈ S l ∗ j ( q m j ) for some q m j ∈ S m ∗ j . Then q m j < q ℓ j .(b). Note that q i j ∈ S h ′ j and h ′ is finite. Let q h j ∈ S h ∗ j − .If h ∗ j − ≥ h ′ j , then q h j < q i j implies [ q h j , q h j , ) ∩ [ q i j , q i j , ) = ∅ by (1) and(2) of B in Construction 1. It is a contradiction to [ q h j , q h j b ) ∩ [ q i j , q i j a ) = ∅ . So h ∗ j − < h ′ j by Construction 1 since q i j ∈ S h ′ j and q h j / ∈ [ q i j , q i j a ). Then h ∗ j ≤ h ′ j and q h j b ∈ S h ∗ j ( q h j ). Then there is a p ∈ N such that p < h ′ and c ≥ c p ≥ k pj , andthere is a q m ′ j ∈ S h ∗ j − such that q m ′ j < q m ′ j k pj = q m j < ... < q ℓ j < q i j . Subcase (b1). q h j = q m ′ j . Then q h j ≤ q m ′ j by (2) of B in Construction 1since both q h j and q m ′ j are in S h ∗ j − with q h j < q i j . Then [ q h j , q h j b ) ∩ [ q i j , q i j a ) = ∅ since q h j ≤ q m ′ j < q m ′ j k pj = q m j < ... < q ℓ j < q i j . It is a contradiction to[ q h j , q h j b ) ∩ [ q i j , q i j a ) = ∅ . Subcase (b2). q h j = q m ′ j . Note b > c ≥ c p ≥ k pj . Then it is easy to see that q h j b < q h j k pj = q m ′ j k pj = q m j < ... < q ℓ j < q i j . Then [ q h j , q h j b ) ∩ [ q i j , q i j a ) = ∅ . It isa contradiction to [ q h j , q h j b ) ∩ [ q i j , q i j a ) = ∅ .Summarizing Subcase 2.1, Subcase (b1) and Subcase (b2), then q h j < q i j implies q h j b < q i j . It is a contradiction to [ q h j , q h j b ) ∩ [ q i j , q i j a ) = ∅ .Then we have q h j ∈ [ q i j , q i j a ). Then b > a implies [ q h j , q h j b ) ⊂ [ q i j , q i j a ). Thiscomplete a proof of Case 2.We use the above Case 2 throughout this paper.Case 3, n < j ≤ n + i . Note b = m + k > k > c > a . So [ q h j , q h j b ) ∩ [ q l j , q l j a ) = ∅ implies [ q h j , q h j b ) ⊂ [ q l j , q l j c ) ⊂ [ q l j , q l j a ) by 5 of Proposition 2.2.Case 4, a = n + i < j ≤ m + k . Then [ q h j , q h j b ) ⊂ [0 , (cid:3) Corollary 5.2. Fix an n ∈ N and an m < n . Then for arbitrary g ( n, i, l ) ∈ G n ,there exists at most finitely many k ’s such that H ( m, k, h ) ∩ g ( n, i, l ) = ∅ and g ( m, k, h ) − g ( n, i, l ) = ∅ . Note* . By Proposition 5.1 and 5.2, we can construct a Ln cover G ( m, i, l )on g ( n, i, l ) − ∪ i ≤ ℓ H ( m i , k i ) with m i < n in the following Construction 5. Let H ( n, i, l ) ⊂ ... ⊂ H (1 , i , l ) with H ( n, i, l ) ⊂ g ( n, i, l ). Then, by Proposition 5.1,there is a c ∈ N such that g (1 , k, h ) ⊂ g ( n, i, l ) if H (1 , k, h ) ∩ g ( n, i, l ) = ∅ with k > c . Note H ( m, k, h ) = { q k } × ... × { q k m } × [ q h , q h b ) × ... × [ q h k , q h k b ) × X b . Let π aa + b [ H ( m, k, h )] = π a [ H ( m, k, h )] × ... × π b [ H ( m, k, h )] . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 19 Then π m [ H ( m, k, h )] = { q k }× ... ×{ q k m } = { ( q k , ..., q k m ) } . Denote π m [ H ( m, k, h )]by ( q k , ..., q k m ) if π m [ H ( m, k, h )] is a point. Then we have π m + k [ H ( m, k, h )] = { q k } × ... × { q k m } × ... × [ q h , q h b ) × ... × [ q h k , q h k b ) and π m +1 m + k [ H ( m, k, h )] = [ q h , q h b ) × ... × [ q h k , q h k b ) . Let I c ( n, i, l ) = [ q i , q i c ) × ... × [ q i n , q i n c ), J ( n, i, l ) = [ q l , q l a ) × ... × [ q l i , q l i a ) × X a and g c ( n, i, l ) = I c ( n, i, l ) × J ( n, i, l ). Corollary 5.3. Let ℓ ≤ m ≤ n , g ( n, i, l ) ∈ G n and g c ( n, i, l ) = I c ( n, i, l ) × J ( n, i, l ) .1. If H ( ℓ, k, h ) ∩ g c ( n, i, l ) = ∅ with k > c , then g ( ℓ, k, h ) ⊂ g c ( n, i, l ) .2. If H ( m, k, h ) ∩ g c ( n, i, l ) = ∅ with H ( m, k, h ) ∩ H ( ℓ, i ℓ , l ℓ ) = ∅ and H ( n, i, l ) ⊂ H ( ℓ, i ℓ , l ℓ ) , then g ( m, k, h ) ⊂ g c ( n, i, l ) and g c ( n, i, l ) − H ( ℓ, i ℓ , l ℓ )= ∪ { H ( m, k, h ) : H ( m, k, h ) ∩ H ( ℓ, i ℓ , l ℓ ) = ∅ and H ( n, i, l ) ⊂ H ( ℓ, i ℓ , l ℓ ) } = ∪ { g ( m, k, h ) : H ( m, k, h ) ∩ H ( ℓ, i ℓ , l ℓ ) = ∅ and H ( n, i, l ) ⊂ H ( ℓ, i ℓ , l ℓ ) } . Proof. To check 1 let H ( ℓ, k, h ) ∩ g c ( n, i, l ) = ∅ with k > c for arbitrary ℓ ∈ N .Let j ≤ ℓ . Then π j [ H ( ℓ, k, h )] = q k j ∈ [ q i j , q i j c ). Then b = ℓ + k > k > c implies[ q k j , q k j b ) ⊂ [ q i j , q i j c ) by 3 of Proposition 3.2 for j ≤ ℓ . Then, in the same way asProposition 5.1, we can prove g ( ℓ, k, h ) ⊂ g c ( n, i, l ) by four Cases in the proof ofProposition 5.1. This implies 1. proof of 2. Let H ( m, k, h ) ∩ g c ( n, i, l ) = ∅ , ℓ ≤ m ≤ n and H ( m, k, h ) ⊂ H ( ℓ, k ℓ , h ℓ ) for some H ( ℓ, k ℓ , h ℓ ). Note g c ( n, i, l ) ⊂ g ( n, i, l ) ⊂ g ( ℓ, i ℓ , l ℓ ) and H ( ℓ, k ℓ , h ℓ ) ∩ H ( ℓ, i ℓ , l ℓ ) = ∅ . Then we have H ( ℓ, k ℓ , h ℓ ) ∩ g ( ℓ, i ℓ , l ℓ ) = ∅ . Then k ℓ > i ℓ by 6 ′ of Proposition 3.3. Then k ≥ k ℓ > i ℓ since H ( m, k, h ) ⊂ H ( ℓ, k ℓ , h ℓ )by 4 of Proposition 3.1. Then, by k ℓ > i ℓ , we have π ℓ [ H ( m, k, h )] = ( q k ′ , q k ′ , ..., q k ′ ℓ ) = ( q i ′ , q i ′ , ..., q i ′ ℓ ) = π ℓ [ H ( ℓ, i ℓ , l ℓ )] . Then there exists a j ≤ ℓ with q k ′ j = q i ′ j . Then, for j ≤ ℓ , π j [ H ( ℓ, k ℓ , h ℓ )] = q k ′ j = q i ′ j and π j [ H ( ℓ, k ℓ , h ℓ )] = q k ′ j ∈ ( q i ′ j , q i ′ j c ) since H ( ℓ, k ℓ , h ℓ ) ∩ g ( ℓ, i ℓ , l ℓ ) = ∅ .Then k ℓ ≥ k ′ j by 5 of Proposition 3.1, and k ′ j > c by 1 of Proposition 3.2. Then g ( ℓ, k ℓ , h ℓ ) ⊂ g c ( n, i, l ) since k ℓ ≥ k ′ j > c and H ( ℓ, k ℓ , h ℓ ) ∩ g c ( n, i, l ) = ∅ by theabove 1. Then g ( m, k, h ) ⊂ g ( ℓ, k ℓ , h ℓ ) ⊂ g c ( n, i, l ) by 5 of Proposition 3.3 since H ( m, k, h ) ⊂ H ( ℓ, k ℓ , h ℓ ). (cid:3) Call c on ( g ( n, i, l ) , m ) . To construct a Ln cover of g ( n, i, l ) − H , we take H ( n, i, l ) ⊂ g ( n, i, l ). Construction 5. Conditions. H ( n, i, l ) ⊂ ... ⊂ H ( m, k, h ) ⊂ ... ⊂ H (2 , i , l ) ⊂ H (1 , i , l ).2. H ( n, i, l ) ⊂ g ( n, i, l ).Let H ∗ ( m, n, 1) = { H ( m, k ′ , h ′ ) ∈ H m : H ( m, k ′ , h ′ ) ∩ g ( n, i, l ) = ∅ and g ( m, k ′ , h ′ ) − g ( n, i, l ) = ∅} . Then H ( m, k, h ) ∈ H ∗ ( m, n, 1) since H ( n, i, l ) ⊂ H ( m, k, h ). Let H ( m, k ′ i ) = { H ( m, k ′ i , i l ) ∈ H ∗ ( m, n, 1) : H ( m, k ′ i , i l ) ⊂ H ( m, k ′ i ) } ∗ AND BOSEN WANG ∗∗∗ and H ′ ( m, k ′ i ) = ∪ H ( m, k ′ i ). Then H ′ ( m, n, 1) = { H ′ ( m, k ′ j ) : j ≤ k ( m ) } is finiteby Corollary 5.2. Let H ′ ( m, k j , h j ) = H ′ ( m, k ′ j ) ∩ g ( n, i, l ) for 1 ≤ j ≤ k ( m ) , H ( m, n, f ) = { H ′ ( m, k j , h j ) : 1 ≤ j ≤ k ( m ) } ,O ( m, n, 1) = g ( n, i, l ) − ∪ H ( m, n, f ) , G ( m, n, 1) = { g ( m, ′ i , i h ) ∈ G m : g ( m, ′ i , i h ) ⊂ O ( m, n, } and H ( m, n, 1) = { H ( m, ′ i , i h ) ∈ H m : g ( m, ′ i , i h ) ∈ G ( m, n, } . Then ∪ G ( m, n, 1) = ∪ H ( m, n, 1) = O ( m, n, 1) and Cl ρ ( ∪ G ( m, n, g ( n, i, l )by Corollary 5.2. Fact 5.4. ∪ G ( m, n, 1) = ∪ H ( m, n, 1) = O ( m, n, .Proof. Pick an x ∈ g ( n, i, l ) − ∪ H ( m, n, f ). Then there exists an H ( m, k ∗ , l ∗ ) ∈ H m with x ∈ H ( m, k ∗ , l ∗ ). Then H ( m, k ∗ , l ∗ ) ∩ g ( n, i, l ) = ∅ . Then we have H ( m, k ∗ , l ∗ ) ⊂ g ( n, i, l ) ⊂ g ( m, k, h ) since H ( m, k ∗ , l ∗ ) ∩ [ ∪ H ( m, n, f )] = ∅ . Sup-pose g ( m, k ∗ , l ∗ ) ∩ H ( m, h j , k j ) = ∅ for some H ( m, h j , k j ) ⊂ H ′ ( m, h j , k j ) ∈ H ( m, n, f ). Then h j > k ∗ by 6 ′ of Proposition 3.3 and g ( m, h j , k j ) − g ( n, i, l ) = ∅ by the definition of H ( m, n, f ). Then g ( m, h j , k j ) ⊂ g ( m, k ∗ , l ∗ ) by 4 of Proposition3.3. Then g ( m, k ∗ , l ∗ ) − g ( n, i, l ) = ∅ , a contradiction to H ( m, k ∗ , l ∗ ) / ∈ H ( m, n, f ).So g ( m, k ∗ , l ∗ ) ∩ [ ∪ H ( m, n, f )] = ∅ . Then g ( m, k ∗ , l ∗ ) ⊂ O ( m, n, (cid:3) We construct a Ln cover G ( m, n, ∗ ) on ( m, O ( m, n, ∪ G ( m, n, ∗ ) = ∪ G ( m, n, 1) by induction. Note 1 ′ < ′ < ... < ′ i < ... .A. Take the least number 1 ′ and denote 1 ′ by 1 . Let G ( m, n, ) = { g ( m, , e ) : g ( m, , e ) ∈ G ( m, n, } = { g ( m, , e ) : e ∈ N } . Then G ( m, n, ) is c.o.D family in ( X, ρ ) by 7 of Proposition 3.3. Let H ( m, n, ) = { H ( m, , e ) : e ∈ N } . B. Assume we have had G ( m, n, i ) and H ( m, n, i ) for i < k . Let O k = ∪ i 1) = O ( m, n, 1) = g ( n, i, l ) − ∪ H ( m, n, f ). Let H = g ( n, i, l ) ∩ [ ∪ H ( m, n, f ). Then, by 1 of Proposition 3.5, H ⊂ Cl ρ [ ∪ G ( m, n, ∗ )]. NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 21 Summarizing the above A and B, we have proved the following claim. Claim 5.5. G ( m, n, ∗ ) = ∪ k G ( m, n, k ) is a c.o.D family in ( O ( m, n, , ρ ) and g ( m, h, k ) is a c.o set in ( X, ρ ) for each g ( m, h, k ) ∈ G ( m, n, ∗ ) .2. H ( m, n, ∗ ) = ∪ k H ( m, n, k ) is a c.D family in ( O ( m, n, , ρ ) .3. H ( m, n, f ) = { H ′ ( m, k j , h j ) : 1 ≤ j ≤ k ( m ) } is finite.4. ∪ G ( m, n, ∗ ) = ∪ G ( m, n, 1) = O ( m, n, 1) = g ( n, i, l ) − ∪ H ( m, n, f ) .5. Let g ( ℓ, h, k ) ⊂ O ( m, n, with ℓ ≥ m . Then there exists a g ( m, i ′ , l ′ ) ∈ G ( m, n, ∗ ) such that g ( ℓ, h, k ) ⊂ g ( m, i ′ , l ′ ) .6. Let H = g ( n, i, l ) ∩ [ ∪ H ( m, n, f )] . Then H ⊂ Cl ρ [ ∪ G ( m, n, ∗ )] . Call the above process Construction 5 on g ( n, i, l ).Call G ( m, n, ∗ ) a least number cover ( Ln cover) on ( m, O ( m, n, ( oron O ( m, n, ) if and only if G ( m, n, ∗ ) and O ( m, n, 1) satisfy 1 and 4 - 5 of Claim5.4. Corollary 5.6. There exists a Ln covers sequence G ( n, m, L ) on O ( m, n, suchthat:1. Every G ( ℓ, m, ∗ ) ∈ G ( n, m, L ) is a Ln cover on O ( m, n, .2. Every H ( ℓ, m, ∗ ) is a c.D family in ( O ( m, n, , ρ ) . Construction 5.1. Conditions. { G ( ℓ, n, , G ( ℓ + 1 , n, } is a Ln covers sequence on O ( m, n, ℓ th, take G ( ℓ, n, g ( ℓ, i h , h j ) ∈ G ( ℓ, n, ℓ = n a and d = ℓ + 1 = n b . Then we have H ′ ( d, i h , h j ) = { H ( d, i k , h k ) ∈ H ( d, n, 0) : H ( d, i k , h k ) ⊂ H ( ℓ, i h , h j ) } and G ′ ( d, i h , h j ) = { g ( d, i k , h k ) ∈ G ( d, n, 0) : H ( d, i k , h k ) ∈ H ′ ( d, i h , h j ) } . Then, by Corollary 4.7, we have the following claim. Claim 5.7. G ′ ( d, i h , h j ) = ∪ u G ( d, i h , e u ) is a Ln cover of H ( ℓ, i h , h j ) .2. G ( d, i h , e u ) is a Ln cover of H ( ℓ, i h , h j ) ∩ H ( d, k e ) for every e ∈ N .3. H ′ ( d, i h , h j ) is infinite and H ′ ( d, i h , h j ) has a ∆ -order as a subsequence of H d .Proof. n a = ℓ and n b = ℓ + 1 = d .To prove 3, note the definition of H ( n, i, l ). Then π b [ H ( ℓ, i h , h j )] = [0 , 1) if b = d + i k > ℓ + i h = a by 4 of Proposition 3.1 and π b [ H ( d, i k , h k )] = [ q h ′ b , q h ′ b b ). Let J db = { π b [ H ( d, i k , h k )] : H ( d, i k , h k ) ∈ H ′ ( d, i h , h j ) } . Then J db is a c.o.D familyin ([0 , , ρ ′ ) and ∪ J db = [0 , 1) by definitions of H ( d, i k , h k ) and H ( ℓ, i h , h j ). So J db is infinite. Then H ′ ( d, i h , h j ) is infinite. Give H ′ ( d, i h , h j ) a ∆-order as asubsequence of H d . Then, for d = ℓ + 1 and g ( ℓ, i h , h j ) ∈ G ( ℓ, n, H ′ ( d, i h , h j ) = { H ( ℓ , x i ) : i ∈ N } . (cid:3) ℜ Condition: { G ( ℓ, n, , G ( ℓ + 1 , n, } is a Ln covers sequence on G m and g ( ℓ, i h , h j ) ∈ G ( ℓ, n, H d ( d, i h , h j ) = { H ( ℓ , x i ) ∈ H ′ ( d, i h , h j ) : i = 1 } = { H ( ℓ , x ) } . ∗ AND BOSEN WANG ∗∗∗ Then ∪ H d ( d, i h , h j ) ⊂ H ( ℓ, i h , h j ) and G d ( d, i h , h j ) ⊂ G ( ℓ + 1 , n, 0) by 2 of Claim5.7. Let G d ( ℓ + 1 , n ) = ∪{ G d ( d, i h , h j ) : g ( ℓ, i h , h j ) ∈ G ( ℓ, n, } .Call G d ( ℓ + 1 , n ) .Recall A a mad family on N in page 115 of Van Douwen [7] if it is a maximalpairwise almost disjoint subfamily of [ N ] ω . α, β, δ and γ are used to denote membersin A . Take an α from A . Then α = { α (1) , α (2) , ... } ⊂ N . ℜ . Let m + j = m j and ℓ + 1 = m . Take H ( m , x ) = H ( ℓ , x ). Let G ( m , x )with ∆-order be a Ln cover of H ( m , x ). Then G ( m , x ) = { g ( m , x i ) : i ∈ N } . Let H ( m , x i ) = g ( m , x i ) ∩ H ( m , x ). Then ∪ i H ( m , x i ) = H ( m , x ). Let G ( m , x i ) with ∆-order be a Ln cover of H ( m , x i ). Then G ( m , x i ) = { g ( m , x ij ) : j ∈ N } Let H ( m , x ij ) = g ( m , x ij ) ∩ H ( m , x ). Then ∪ j H ( m , x ij ) = H ( m , x i ) and ∪ ij H ( m , x ij ) = H ( m , x ) . Take a g ( m , x ij ) ∈ G ( m , x i ). Let G ( m , x ij ) with ∆-order be a Ln cover of H ( m , x ij ). Then G ( m , x ij ) = { g ( m , x ijl ) : l ∈ N } . Let H ( m , x ijl ) = H ( m , x ) ∩ g ( m , x ijl ), H ( m , x i ) = { H ( m , x ijl ) : j, l ∈ N } and G ( m , x i ) = ∪{ G ( m , x ij ) : j ∈ N } . Then ∪ l H ( m , x ijl ) = H ( m , x ij ) and ∪ ijl H ( m , x ijl ) = H ( m , x ) . Fix an α = { α (1) , α (2) , ... } ⊂ N for α ∈ A . Let H d ( m , αx i ) = { H ( m , x ijl ) ∈ H ( m , x ij ) : l > α ( i ) for j ≤ α ( i ) } , H d ( m , αx i ) = { H ( m , x ijl ) ∈ H ( m , x ij ) : l ≤ α ( i ) for j > α ( i ) } and H d ( m , αx i ) = H d ( m , αx i ) ∪ H d ( m , αx i ) . Take G d ( m , αx i ) = { g ( m , x ijl ) ∈ G ( m , x ij ) : H ( m , x ijl ) ∈ H d ( m , αx i ) } . Let G d ( m , αx ) = ∪{ G d ( m , αx i ) : i ∈ N } and H C ( m , αx ) = ∪{ H d ( m , αx i ) : i ∈ N } . Then H C ( m , αx ) = ∪ H C ( m , αx ) ⊂ H ( m , x ).Call H C ( m , αx ) a characterization set of α in H ( m , x ) (or α ch-set). (cid:3) Let H − ( ℓ + 1 , n ) = H ( ℓ + 1 , n, − H d ( ℓ + 1 , n ) . Then both H d ( ℓ + 1 , n ) and H − ( ℓ + 1 , n ) are discrete families of closed sets in( O ( m, n, , ρ ). Take the relative G d ( ℓ + 1 , n ) and G − ( ℓ + 1 , n ). Then both G d ( ℓ + 1 , n ) and G − ( ℓ + 1 , n ) are c.o.D families in ( O ( m, n, , ρ ).Let H d = ∪ H d ( ℓ + 1 , n ) and H − = ∪ H − ( ℓ + 1 , n ). Then g [ ℓ + 1 , H d ] = ∪ G d ( ℓ + 1 , n ) , g [ ℓ + 1 , H − ] = ∪ G − ( ℓ + 1 , n ) and g [ ℓ, H − ] = g [ ℓ, H d ] = O ( m, n, . Summarizing the above Construction 5.1, we have proved the following claim. NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 23 Claim 5.8. A. If { G ( ℓ, n, , G ( ℓ + 1 , n, } is a Ln covers sequence on O ( m, n, ,then there uniquely exists ’th Ln family G d ( ℓ + 1 , n ) such that:1. Both G d ( ℓ + 1 , n ) and G − ( ℓ + 1 , n ) are infinite.2. G d ( ℓ + 1 , n ) is c.o.D, and H d ( ℓ + 1 , n ) is c.D in ( O ( m, n, , ρ ) .3. Let H = ∪ H ( m, n, f ) . Then H ⊂ Cl ρ [ ∪ G d ( ℓ + 1 , n )] .4. H ∩ O ( m, n, 1) = ∅ and H ∪ O ( m, n, 1) = g ( n, i, l ) .B. Let g ( ℓ, i h , h j ) ∈ G ( ℓ, n, .1. G ′ ( d, i h , h j ) ∩ G d ( ℓ + 1 , n ) = G d ( d, i , h ) = { g ( d, i , h ) } = { g ( ℓ , x ) } .2. ∪ H d ( ℓ + 1 , i h , h j ) ⊂ H ( ℓ, i h , h j ) .3. g [ ℓ + 1 , H d ] = ∪ G d ( ℓ + 1 , n ) , g [ ℓ + 1 , ∪ G d ( ℓ + 1 , n )] = ∪ G d ( ℓ + 1 , n ) , g [ ℓ + 1 , H − ] = ∪ G − ( ℓ + 1 , n ) and g [ ℓ, H − ] = g [ ℓ, H d ] = O ( m, n, . To construct holes in neighborhoods In this section, we use Operation ℓ th to construct holes families which makebases of neighborhoods are not C.P.Take H ( n, i ) with ∪ H ( n, i ) = H ( n, i ) in Construction 3.2. Note, before Con-struction 4, we have H = H ( m, a ∗ , y ) and G m = g ( m, a ∗ , y ). Let n ≥ m, H ( n, i ) ∩ H = ∅ , H ′ ( n, i ) = { H ( n, i , i l ) ∈ H ( n, i ) : H ( n, i , i l ) ⊂ H ( n, i ) ∩ G m } ,H ′ ( n, i ) = ∪ H ′ ( n, i ) , < < ... < i < ... and H ( n, G m ) = { H ′ ( n, i ) : 1 i ∈ N ( G m ) } . Then ∪ H ( n, G m ) = g ( m, a ∗ , y ) − H ( m, a ∗ , y ). Proposition 6.1. Let G ( n + 1 , ) be Ln cover of H ′ ( n, ) . Then there exists anopen set g ( n + 1 , ) such that:1. g ( n + 1 , ) ∩ [ ∪ G ( n + 1 , )] = ∅ .2. Let G ( n, ) be a Ln cover of H ′ ( n, ) . Then there exist infinitely many g ( n, x j ) ’s in G ( n, ) such that g ( n, x j ) ⊂ g ( n + 1 , ) .3. Let G ( n + 1 , ) = ∪ j G ( n + 1 , , c j ) be a Ln cover on H ′ ( n, ) . Then ∪ G ( n + 1 , , c j ) ⊂ g ( n + 1 , ) for infinitely many G ( n + 1 , , c j ) ’s.Proof. Take H ′ ( n, ). Then H ′ ( n, ) = H ( n, ) ∩ G m by the definitions of G m and H ( n, i ) in (B) of Construction 3.1. Let H ( n, ) = { q } × ... × { q n − } × { q n } × X n . Note m = m + a ∗ , n ≥ m , G m = g ( m, a ∗ , y ) = I ( m, a ∗ , y ) × J ( m, a ∗ , y ) and g ( n, , l ) = [ q , q b ) × ... × [ q n , q n b ) × [ q l , q l b ) × ... × [ q l ′ , q l ′ b ) × X b if H ( n, , l ) ∈ H ′ ( n, ) with ∪ H ′ ( n, ) = H ′ ( n, ), 1 ′ = 1 and b = n + 1 .Let J n ( m, a ∗ , y ) = [ q l , q l m ) × [ q l , q l m ) × ... × X m . Note G m = [ q i , q i m ) × ... × [ q i n , q i n m ) × J n ( m, a ∗ , y ). Then H ′ ( n, ) = G m ∩ H ( n, ) = { q } × ... × { q n } × J n ( m, a ∗ , y ) if m ≤ n < m and H ′ ( n, ) = G m ∩ H ( n, ) = { q } × ... × { q n } × X n if n ≥ m . Let G ( n, ) = { g ( n, x j ) : H ( n, x j ) ⊂ H ′ ( n, ) } be a Ln cover of H ′ ( n, ) and H ( n, x j ) ∈ H ( n, ). Then H ( n, x j ) = H ( n, , j ) ⊂ H ′ ( n, ) = H ( n, ) ∩ G m . ∗ AND BOSEN WANG ∗∗∗ Then 1 > a ∗ since n ≥ m , H ′ ( n, ) ⊂ G m = g ( m, a ∗ , y ) and H ′ ( n, ) ∩ H = ∅ .Then b = n + 1 > m = m + a ∗ . Let π n : X → Q n be a project map. Then π n [ H ′ ( n, )] = q n ∈ [0 , 1) = π n [ G m ] for n > m , or π n [ H ′ ( n, )] = q n ∈ [ q i n , q i n m ) = π n [ G m ] for m < n ≤ m . Then [ q n , q n b ) ⊂ [ q i n , q i n m ) by b > m .A. Let G ( n + 1 , ) be a Ln cover of H ′ ( n, ). Then, by 1 of Corollary 4.7,we have G ( n + 1 , ) = ∪{ G ( n + 1 , ′ j ) : j ∈ N } . Take a G ( n + 1 , ′ j ). Let g ( n + 1 , x ij ) ∈ G ( n + 1 , ′ j ), d = n + 1 and e j = ( n + 1) + 1 ′ j . Then e j > b = n + 1 , g ( n +1 , x ij ) = [ q , q e j ) × ... × [ q n , q n e j ) × [ q j d , q j d e j ) × [ q j i , q j i e j ) × ... × [ q j il , q j il e j ) × X e j ,I ( n + 1 , x ij ) = [ q , q e j ) × ... × [ q n , q n e j ) × [ q j d , q j d e j ) ,I n ( n + 1 , x ij ) = [ q , q e j ) × ... × [ q n , q n e j ) and J ( n + 1 , x ij ) = [ q j i , q j i e j ) × ... × [ q j il , q j il e j ) × X e j . Then, for every g ( n + 1 , x hj ) ∈ G ( n + 1 , ′ j ), we have I n ( n + 1 , x hj ) = I n ( n + 1 , x ij ) = I n ( n + 1 , e j ) = [ q , q e j ) × ... × [ q n , q n e j ) and I ( n + 1 , x hj ) = I ( n + 1 , x ij ) = I ( n + 1 , e j ) = [ q , q e j ) × ... × [ q n , q n e j ) × [ q j d , q j d e j ) . Let I ( n, ) = { I n ( n + 1 , x ij ) : g ( n + 1 , x ij ) ∈ G ( n + 1 , ) } . Then I ( n, ) = { I n ( n + 1 , e j ) : j ∈ N } . Note the definition of e j . Then ∩ j I n ( n + 1 , e j ) = p ( n, ) = { q } × ... × { q n } . Let p ( n, ) = { q } × ... × { q n } and H ( n, ) = p ( n, ) × X n . Then H ′ ( n, ) = H ( n, ) ∩ G m = p ( n, ) × J n ( m, a ∗ , y ) . Then p ( n, ) = p ( n, ). Then q h = q h for some h ≤ n .B. Note b = n + 1 . Let g ( n, , b ) = I ( n, , b ) × J ( n, , b ). Here I ( n, , b ) = [ q , q b ) × ... × [ q n , q n b ) and J ( n, , b ) = [ q l , q l b ) × ... × [ q ′ b , q ′ b b ) × X b . Let c = n + 1 . Then c > b = n + 1 . Let g ( n, , c ) = I ( n, , c ) × J ( n, , c ). Here I ( n, , c ) = [ q , q c ) × ... × [ q n , q n c ) and J ( n, , c ) = [ q l ′ , q l ′ c ) × ... × [ q l ′ c , q l ′ c c ) × X c . Case 1. There exists an h ≤ n with [ q h , q h b ) ∩ [ q h , q h c ) = ∅ . Then we have I ( n, , b ) ∩ I ( n, , c ) = ∅ . Then it is easy to see I n ( n + 1 , x ij ) ∩ I ( n, , c ) = ∅ if g ( n + 1 , x ij ) ∈ G ( n + 1 , ) with π h [ g ( n + 1 , x ij )] = [ q h , q h e j ) ⊂ [ q h , q h b ). Then g ( n + 1 , x ij ) ∩ g ( n, , c ) = ∅ .Case 2. [ q h , q h b ) ∩ [ q h , q h c ) = ∅ for each h ≤ n . Note c > b > m . Then, by 3of Proposition 3.2, [ q h , q h b ) ⊃ [ q h , q h c ) for each h ≤ n . Note p ( n, ) = p ( n, ).Then there exists an h ≤ n with q h ∈ ( q h , q h b ). Let I ( n, , ∈ ) = { I n ( n + 1 , e j ) : q h ∈ ( q h , q h e j ) = π h [ I n ( n + 1 , x ij )] − { q h }} and I ( n, , / ∈ ) = { I n ( n + 1 , e j ) : q h / ∈ ( q h , q h e j ) = π h [ I n ( n + 1 , x ij )] − { q h }} . Then I ( n, , ∈ ) ∩ I ( n, , / ∈ ) = ∅ and I ( n, , ∈ ) ∪ I ( n, , / ∈ ) = I ( n, ). Then I ( n, , ∈ ) is finite since q h / ∈ ∩ j π h [ I n ( n + 1 , e j )] = { q h } . Then we have I n ( n +1 , x ij ) ∩ I ( n, , c ) = ∅ for every I ( n + 1 , x ij ) ∈ I ( n, , / ∈ ). Let I ( n + 1 , , ∈ ) = { I ( n + 1 , e j ) : I n ( n + 1 , e j ) ∈ I ( n, , ∈ ) } , G ( n + 1 , , ∈ ) = { g ( n + 1 , x ij ) ∈ G ( n + 1 , ) : I ( n + 1 , e j ) ∈ I ( n + 1 , , ∈ ) } , NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 25 I ( n + 1 , , / ∈ ) = { I ( n + 1 , e j ) : I n ( n + 1 , e j ) ∈ I ( n, , / ∈ ) } and G ( n + 1 , , / ∈ ) = { g ( n + 1 , x ij ) ∈ G ( n + 1 , ) : I ( n + 1 , e j ) ∈ I ( n + 1 , , / ∈ ) } . Then g ( n + 1 , x ij ) ∩ g ( n, , c ) = ∅ for every g ( n + 1 , x ij ) ∈ G ( n + 1 , , / ∈ ). Then wehave the following claim: Claim 6.2. 1. Both I ( n, , ∈ ) and I ( n + 1 , , ∈ ) are finite.2. g ( n + 1 , x ij ) ∩ g ( n, , c ) = ∅ for every g ( n + 1 , x ij ) ∈ G ( n + 1 , , / ∈ ) . C. Note c > b > m . Take I b from 1 of Proposition 2.2. Let I bm = I b | [ q l , q l m ). Then I bm is infinite. Let F bm (1) = { [ q i d , q i d b ) ∈ I bm : there exits an I ( n + 1 , e j ) ∈ I ( n + 1 , , ∈ )with π d [ I ( n + 1 , e j )] = [ q j d , q j d e j ) ⊂ [ q i d , q i d b ) } . Claim 6.3. F bm (1) is finite.2. I bm − F bm (1) is infinite.3. There exists a q l p ≥ max [ ∪ F bm (1)] with p > a = m = m + a ∗ such that [ q l p , q l a ) ∩ [ ∪ F bm (1)] = ∅ . D. Note J n ( m, a ∗ , y ) = [ q l , q l m ) × [ q l , q l m ) × ... × X m . Let O ( b, m ) = ∪ ( I bm − F bm (1)) , J n +1 ( m, a ∗ , y ) = [ q l , q l m ) × ... × X m and g ( n + 1 , ) = I ( n, , c ) × O ( b, m ) × J n +1 ( m, a ∗ , y ) . Then g ( n + 1 , ) ⊂ G m , and H ′ ( n, ) ∩ g ( n + 1 , ) = { q } × ... × { q n } × O ( b, m ) × J n +1 ( m, a ∗ , y ) . Let J ( n + 1 , ) = O ( b, m ) × J n +1 ( m, a ∗ , y ). Then g ( n + 1 , ) = I ( n, , c ) × J ( n + 1 , ) . Let g ( n + 1 , x ) ∈ G ( n + 1 , , ∈ ). Note d = n + 1. Then π d [ g ( n + 1 , x )] = [ q j d , q j d e j ) ⊂ [ q j ′ d , q j ′ d b ) ∈ F bm (1) . Then J ( n +1 , x ) ∩ J ( n +1 , ) = ∅ . Then g ( n +1 , x ) ∩ g ( n +1 , ) = ∅ . Let g ( n +1 , x ) ∈ G ( n + 1 , , / ∈ ). Then I ( n + 1 , x ) = I ( n + 1 , e j ) and I n ( n + 1 , e j ) ∩ I ( n, , c ) = ∅ .Then g ( n + 1 , x ) ∩ g ( n + 1 , ) = ∅ . Claim 6.4. g ( n + 1 , ) ∩ [ ∪ G ( n + 1 , )] = ∅ . E. Let G ( n + 1 , ) be a Ln cover of H ′ ( n, ). Then, by 1 of Corollary 4.7, G ( n + 1 , ) = ∪ j G ( n + 1 , , c j ). Note d = n + 1. Then O ( b, m ) ⊂ [ q l , q l m ) = π d [ H ′ ( n, )] ⊂ π d [ ∪ G ( n + 1 , )] . Let [ q h , q hb ) ∈ I bm − F bm (1). Then q h ∈ [ q h , q hb ) ⊂ O ( b, m ) ⊂ π d [ ∪ G ( n +1 , )].Then there exists a g ( n + 1 , j , y h ) ∈ G ( n + 1 , , c j ) with g ( n + 1 , j , y h ) = [ q , q c j ) × ... × [ q n , q n c j ) × [ q l ′ , q l ′ c j ) × ... × [ q l ′ ′ , q l ′ ′ c j ) × X c j and q h ∈ π d [ g ( n + 1 , j , y h )] = [ q l ′ , q l ′ c j ). Here q l ′ h ∈ [ q l h , q l h m ) for 1 ≤ h ≤ a ∗ ,2 ′ = 2 j and c j = ( n + 1) + 2 j > n + 1 = c > b = n + 1 by 4 of Proposition 3.1since H ( n + 1 , j , y h ) ⊂ H ′ ( n, ). ∗ AND BOSEN WANG ∗∗∗ Note c j > b implies Q ∗ b ⊂ Q ∗ c j by 2 of Proposition 2.2, and q i ∈ Q ∗ b with[ q i , q ib ) ∈ I b implies q i ∈ Q ∗ c j with [ q i , q ic j ) ∈ I c j and [ q i , q ic j ) ⊂ [ q i , q ib ). Then q h ∈ [ q h , q hb ) ∈ I bm ⊂ I b implies q h ∈ Q ∗ b ⊂ Q ∗ c j . Then, by 3 of Proposition 3.2, q h = q l ′ ∈ [ q l ′ , q l ′ c j ) = [ q h , q hc j ) ⊂ [ q h , q hb ) . Let I ( n + 1 , j , y h ) = [ q , q c j ) × ... × [ q n , q n c j ) × [ q l ′ , q l ′ c j ) and J ( n + 1 , j , y h ) = [ q l ′ , q l ′ c j ) × ... × [ q l ′ ′ , q l ′ ′ c j ) × X c j . And then let I ( c j , m , l h ) = I c j | [ q l h , q l h m ) for 2 ≤ h ≤ j = 2 ′ . Then, by 3 ofProposition 2.2, ∪ I ( c j , m , l h ) = [ q l h , q l h m ). Let J ( n + 1 , c j ) = I ( c j , m , l ′ ) × ... × I ( c j , m , l ′ ′ ) × { X c j } . Then J ( n + 1 , j , y h ) ∈ J ( n + 1 , c j ) by the definition of g ( n, i, l ) in A-C of Con-struction 3.2. Let G ∗ ( n + 1 , , c j ) = { I ( n + 1 , j , y h ) × J ( n + 1 , j , j l ) : J ( n + 1 , j , j l ) ∈ J ( n + 1 , c j ) } . Then g ( n + 1 , j , y h ) ∈ G ∗ ( n + 1 , , c j ) since J ( n + 1 , j , y h ) ∈ J ( n + 1 , c j ). Then G ∗ ( n + 1 , , c j ) = G ( n + 1 , , c j ) by the definition of G ( n, i ) in C of Construction3.2 and 2 of Corollary 4.7. Then there exist infinitely many G ( n + 1 , , c j )’s suchthat ∪ G ( n + 1 , , c j ) ⊂ g ( n + 1 , ).Note Claim 6.4 and the definition of g ( n + 1 , ). It is easy to see 1.Note the definitions of I ( n, , c ) and g ( n + 1 , ), and 2 of Claim 6.3. Then itis easy to see 2. (cid:3) Proposition 6.5. Let G ( n + 1 , i ) be Ln cover of H ′ ( n, i ) for i < j . There existsan open set g ( n + 1 , j ) such that:1. g ( n + 1 , j ) ∩ [ ∪ G ( n + 1 , i )] = ∅ for every i < j .2. Let G ( n, j ) be a Ln cover of H ′ ( n, j ) . Then there exist infinitely many g ( n, x i ) ’s in G ( n, j ) such that g ( n, x i ) ⊂ g ( n + 1 , j ) .3. Let G ( n + 1 , j ) = ∪ b G ( n + 1 , j , c b ) be a Ln cover of H ′ ( n, j ) . Then ∪ G ( n + 1 , j , c b ) ⊂ g ( n + 1 , j ) for infinitely many G ( n + 1 , j , c b ) ’s.Proof. Let G ( n +1 , j ) = ∪ b G ( n +1 , j , c b ) be a Ln cover of H ′ ( n, j ) with ∆-order.Fix an H ′ ( n, i ) for i < j . Then, by Claim 6.2, there exists I ( n + 1 , i , ∈ ), anda set g ( n, j , c ) with c = n + 1 j such that I ( n + 1 , i , ∈ ) is finite, and g ( n + 1 , y ) ∩ g ( n, j , c ) = ∅ for every g ( n +1 , y ) ∈ G ( n +1 , i , / ∈ ). Note b = n +1 > m + a ∗ = m .Let G ( n + 1 , i ) be Ln cover of H ′ ( n, i ). Then G ( n + 1 , i ) = ∪ h G ( n + 1 , i , i h )by 1 of Corollary 4.7. Let g ( n + 1 , i h , x ) = g ( n + 1 , x ) ∈ G ( n + 1 , i , i h ). Then e ih = ( n + 1) + i h > n + 1 i > n + 1 = b . Take F bm = I b | [ q l , q l m ) from 2 ofClaim 6.3. Let F bm ( i ) = { [ q j d , q j d b ) ∈ I bm : there exits a [ q j ′ d , q j ′ d e ih ) ∈ I ( n + 1 , i , ∈ )such that [ q j ′ d , q j ′ d e ih ) ⊂ [ q j d , q j d b ) } . In the same way, F bm ( i ) is finite for each i < j . Then ∪ i Note g ( m, a ∗ , y ) = G m . Let O ( m, y ) = g ( m, a ∗ , y ) − H ( m, a ∗ , y ) and G ( m, m, L ) = { G ( m i , m, 0) : m i ≥ m } be a Ln covers sequence on O ( m, y ) whichsatisfies 1 of Corollary 5.6.Take H ′ ( n, y ) = H ( n, G m ) for every n ≥ m before Proposition 6.1. Let H ′ ( m, m, H ) = { H ′ ( n, y ) : n ≥ m } . Take H ′ ( m, y ) ∈ H ′ ( m, m, H ) for n = m . Then H ′ ( m, y ) = { H ′ ( m, h ) : h ∈ N } and 1 < < ... < i < ... with ∆-order in Construction 3.1. Then we have the following Condition B ∗ . Condition B ∗ : O ( m, y ) = ∪ H ′ ( m, y ).2. H ′ ( m, y ) = { H ′ ( m, h ) : h ∈ N } with 1 < < ... < h < ... . B. Take H ′ ( m, j ) ∈ H ′ ( m, y ). Let m = m + 1 and G ( m , j ) be a Ln coverof H ′ ( m, j ). Take a g ( m , z ) ∈ G ( m , j ). Let m = m + 1, H ( m , jz ) = H ′ ( m, j ) ∩ g ( m , z ) and G ( m , z ) = { g ( m , z l ) ∈ G m : l ∈ N } with ∆-orderbe a Ln cover of H ( m , jz ). Take an g ( m , z n ) ∈ G ( m , z ). Let H ( m , jz n ) = H ( m , jz ) ∩ g ( m , z n ) = H ′ ( m, j ) ∩ g ( m , z n ) . Let m = m + 2 and G ( m , jz n ) be a Ln cover of H ( m , jz n ). Let G ( m , j , z ) = ∪{ G ( m , jz n ) : g ( m , z n ) ∈ G ( m , z ) } . Then G ( m , j , z ) is a Ln cover of H ( m , jz ). Take an H ( m , jz n ). Let G ( m , jz n ) = { g ( m , z nv ) : v ∈ N } with ∆-order ,H ( m , jz nv ) = H ( m , jz n ) ∩ g ( m , z nv ) = H ′ ( m, j ) ∩ g ( m , z nv ) and H ( m , jz n ) = { H ( m , jz nv ) : v ∈ N } . Claim 6.6. ∪ v H ( m , jz nv ) = H ( m , jz n ) and ∪ n H ( m , jz n ) = H ( m , jz ) .2. G ( m , j , z ) is a Ln cover of H ( m , jz ) .3. G ( m , jz n ) is a Ln cover of H ( m , jz n ) for every n ∈ N . B1. Take an H ( m , jz nv ). Let v α = α ( m + v ), v α = α ( m + v ) + 1 and G ( v α , jz nv ) = { G ( v α , jz nv ) , G ( v α , jz nv ) } be a covers sequence of H ( m , jz nv ). Then there exists 1’th Ln family H ∗ d ( v α , jz nv )with ∆-order by ℜ 1. Then G ∗ d ( v α , jz nv ) is c.o.D in ( X, ρ ) by 2 of A in Claim 5.8. B2. For the same H ( m , jz nv ) as B1, take H ( m , x v ) and G ( m , x v ) for v = i from ℜ 2. Let x = z n and x v = z nv . Then, by ℜ 2, there exists an α ch-set ∗ AND BOSEN WANG ∗∗∗ H C ( m , αz n ), G ( m , z nv ) and G d ( m , αz nv ). Note both G ( m , z nv ) and G ( v α , jz nv )are Ln covers of H ( m , jz nv ). Let G (5 , αz nv ) = ∪ G d ( m , αz nv ). Then g ( v α , t ) ∩ G (5 , αz nv ) = ∅ implies g ( v α , t ) ⊂ G (5 , αz nv ) if v α ≥ m . Let v ( v α ) be the leastnumber such that v ≥ v ( v α ) implies v α ≥ m . Let G d ( v α , jz nv ) = { g ( v α , t ) ∈ G ∗ d ( v α , jz nv ) : g ( v α , t ) ∩ G (5 , αz nv ) = ∅} if v ≥ v ( v α ) , G d ( v α , jz nv ) = G d ( m , αz nv ) if v < v ( v α ) , G − ( v α , jz nv ) = { g ( v α , t ) ∈ G ( v α , jz nv ) : g ( v α , t ) ∩ [ ∪ G d ( v α , jz nv )] = ∅} , G d ( nσ α , jz, pσ ) = ∪{ G d ( v α , jz nv ) : v ≥ p and g ( m , z nv ) ∈ G ( m , jz n ) } and G − ( nσ α , jz, pσ ) = ∪{ G − ( v α , jz nv ) : v ≥ p and g ( m , z nv ) ∈ G ( m , jz n ) } . Note that G ( m , j ) is a Ln cover of H ′ ( m, j ) for H ′ ( m, j ) ∈ H ′ ( m, y ). Let G d ( nσ α , j , pσ ) = ∪{ G d ( nσ α , jz, pσ ) : g ( m , z ) ∈ G ( m , j ) } , G − ( nσ α , j , pσ ) = ∪{ G − ( nσ α , jz, pσ ) : g ( m , z ) ∈ G ( m , j ) } , G d ( nσ α , pσ ) = ∪{ G d ( nj α , j , pσ ) : H ′ ( m, j ) ∈ H ′ ( m, y ) } and G − ( nσ α , pσ ) = ∪{ G − ( nj α , j , pσ ) : H ′ ( m, j ) ∈ H ′ ( m, y ) } . Claim 6.7. Let H ( nσ α , pσ ) = ∪ H d ( nσ α , pσ ) .1. H ( nσ α , qσ ) ⊂ H ( nσ α , pσ ) if q > p .2. ∩ p H ( nσ α , pσ ) = ∅ and H ( m , jz nu ) ∩ H ( m , jz nv ) = ∅ if u = v .3. ∪ H ( nσ α , jz, pσ ) ⊂ H ( m , jz n ) .4. Let x ∈ H ( m , jz n ) for some jz n . Then there uniquely exists u such that x ∈ g ( u α , t ) ∈ G d ( u α , jz nu ) ∪ G − ( u α , jz nu ) .5. G ( nσ α , j , pσ ) = G d ( nσ α , j , pσ ) ∪ G − ( nσ α , j , pσ ) is c.o.D in ( X, ρ ) . B3. Take H (1 σ α , pσ ) = ∪{ H (1 j α , j , pσ ) : H ′ ( m, j ) ∈ H ′ ( m, y ) } . Note ∪ H d ( v α , jz v ) ⊂ H ( m , jz v ) for every v, j ∈ N . We construct families by inductionon H ′ ( m, y ). B3.1. Let H ′ ( m, ) ∈ H ′ ( m, y ) and G ( m , ) be a Ln cover of H ′ ( m, ).For j = 1 and n = 1, take G d (1 σ α , , pσ ) and G − (1 σ α , , pσ ) from B2. Let H ( m, ) = H ′ ( m, ) − ∪ G d (1 σ α , , pσ ) . B3.2. For n = 1, assume that we have had G d (1 σ α , ℓ , pσ ), G − (1 σ α , ℓ , pσ ) and H ( m, ℓ ) for each ℓ < j . Take H ′ ( m, j ). Let H ∗ ( m, j ) = H ′ ( m, j ) − ∪ ℓ Let D (1 σ α , G ) = ∪ G (1 σ α , G ) and H ( m, G ) = ∪ H ( m, G ) . Then:1. G d (1 σ α , j , pσ ) is c.o.D in ( X, ρ ) for every j ∈ N .2. D (1 σ α , G ) ∪ H ( m, G ) = O ( m, y ) and D (1 σ α , G ) ∩ H ( m, G ) = ∅ .3. Cl ρ D (1 σ α , G ) = g ( m, y ) .4. H ( m, j ) = H ′ ( m, j ) − D (1 σ α , G ) = H ′ ( m, j ) − ∪ i ≤ j ∪ G (1 σ α , i , pσ ) forevery H ′ ( m, j ) ∈ H ′ ( m, y ) .5. g [ p, H (1 σ α , G )] = O ( m, y ) if p = m .6. Let t ∈ g [ p, D (1 σ α , G )] − D (1 σ α , G ) . Then, there exists a k such that H ( ℓ, t ) ∩ g [ q, D (1 σ α , G )] = ∅ if ℓ, q > k .Proof. It is easy to see 1, 2 and 4. 5 follows from 3 of B in Claim 5.8. Proof of 3. To do it pick an r ∈ H ( m, j ). Then, by the above definition of H ( m, j ) and 4 of Claim 6.7, there uniquely exists g ( v α , r ′ ) ∈ G ( v α , jz v ) such that H ( v α , r ) ⊂ H ( v α , jr ′ ) = g ( v α , r ′ ) ∩ H ( m, j ) ⊂ H ( m, j ) and g ( v α , r ′ ) ∩ [ ∪ i ≤ j ∪ G d (1 σ α , i , pσ )] = ∅ . Let p > v α and r ∈ H ( p, y ′ ) ⊂ H ( v α , jr ′ ). Then there exists a c with g ( m, ℓ , t ) ⊂ g c ( p, y ′ ) if 1 ℓ > c and H ( m, ℓ , t ) ∩ g c ( p, y ′ ) = ∅ by 1 of Corollary 5.3. Let 1 ℓ be theleast number for m < p . Then, for every t ∈ H ′ ( m, ℓ ) ∩ g c ( p, y ′ ), we have g ( m, ℓ , t ) ⊂ g c ( p, y ′ ) ⊂ g ( p, y ′ ) . And then we may prove the following Fact. Fact 6.9. Let G ( m , k ) be a Ln cover of H ′ ( m, k ) for H ′ ( m, k ) ∈ H ′ ( m, y ) , G ( m , ℓ ) = ∪{∪ G ( m , k ) : 1 j < k < ℓ } and H ( m , ℓ , t ) ⊂ g c ( p, y ′ ) . Then g ( m , ℓ , t ) ∩ G ( m , ℓ ) = ∅ .Proof. Note 1 j < ℓ . Let 1 j < k < ℓ for some 1 k . Let g ( m , k , s ) ∈ G ( m , k ), I c ( p, y ′ ) = [ q , q c ) × ... × [ q m , q m c ) , g c ( p, y ′ ) = I c ( p, y ′ ) × J ( p, y ′ ) ,I ( m , j , r ) = [ q , q c j ) × ... × [ q m , q m c j ) , g ( m , j , r ) = I ( m , j , r ) × J ( m , j , r ) ,I ( m , ℓ , t ) = [ q ′ , q ′ c ℓ ) × ... × [ q ′ m , q ′ m c ℓ ) , g ( m , ℓ , t ) = I ( m , ℓ , t ) × J ( m , ℓ , t ) ,I ( m , k , s ) = [ q ′′ , q ′′ c k ) × ... × [ q ′′ m , q ′′ m c k ) and g ( m , k , s ) = I ( m , k , s ) × J ( m , k , s ) . Note H ( m , ℓ , t ) ⊂ H ′ ( m , ′ ℓ ) ⊂ H ′ ( m, ℓ ), H ( m , k , s ) ⊂ H ′ ( m , ′ k ) ⊂ H ′ ( m, k )and H ( p, y ′ ) ⊂ H ( v α , r ) ⊂ H ( m , j , r ) ⊂ H ′ ( m , ′ j ) ⊂ H ′ ( m, j ) such that H ′ ( m , ′ k ) = [ { q ′′ } × ... × { q ′′ m } × X m ] ∩ g ( m, y ) and g c ( p, y ′ ) ⊂ g ( m, y ).Case 1. g ( m , k , s ) ∩ g c ( p, y ′ ) = ∅ . Then g ( m , k , s ) ∩ g ( m , ℓ , t ) = ∅ since g ( m, ℓ , t ) ⊂ g c ( p, y ′ ).Case 2. g ( m , k , s ) ∩ g c ( p, y ′ ) = ∅ . Note that 1 ℓ is the least number such that1 ℓ > c , 1 ℓ > j and H ′ ( m, ℓ ) ∩ g c ( p, y ′ ) = ∅ . Then 1 j < k < ℓ implies H ′ ( m, k ) ∩ g c ( p, y ′ ) = ∅ . ∗ AND BOSEN WANG ∗∗∗ Then, for every H ′ ( m , ′ k ) ⊂ H ′ ( m, k ), we have H ′ ( m , ′ k ) ∩ g c ( p, y ′ ) = ∅ . Then H ′ ( m , ′ k ) ∩ g c ( p, y ′ ) = ∅ implies q ′′ i / ∈ [ q i , q ic ) for some i ≤ m since g ( m , k , s ) ∈ H ( m , k ) and g ( m , k , s ) ∩ g c ( p, y ′ ) = ∅ .Note 1 j < k . This implies c j = m + 1 j < m + 1 k = c k . Note H ( p, y ′ ) ⊂ H ( v α , r ) ⊂ H ( m, j ) . Suppose g ( m , k , s ) ∩ g ( m , ℓ , t ) = ∅ for some 1 k < ℓ . Then g ( m , ℓ , t ) ⊂ g c ( p, y ′ ) ⊂ g ( m , j , r ) implies g ( m , k , s ) ∩ g ( m , j , r ) = ∅ . Then 1 j < k implies H ′ ( m, j ) ∩ g ( m , k , s ) = ∅ by 6 of Proposition of 3.3, and implies g ( m , k , s ) ⊂ g ( m , j , r ) by 8 of Proposition 3.3. Then [ q ′′ b , q ′′ bc k ) ⊂ [ q b , q bc j ) for each b ≤ m .Then q ′′ b ∈ [ q b , q bc j ) for each b ≤ m , and q ′′ i / ∈ [ q i , q ic ) for some i ≤ m . Then q ′′ i ∈ [ q i , q ic j ) and q ′′ i / ∈ [ q i , q ic ). Then q ic < q ′′ i . Then [ q i , q ic ) ∩ [ q ′′ i , q ′′ ic k ) = ∅ .Note g c ( p, y ′ ) ∩ g ( m , k , s ) = ∅ . Then [ q i , q ic ) ∩ [ q ′′ i , q ′′ ic k ) = ∅ for each i ≤ m ,a contradiction. Then g ( m , k , s ) ∩ g ( m , ℓ , t ) = ∅ for each 1 k < ℓ . Then G ( m , ℓ ) ∩ g ( m , ℓ , t ) = ∅ . (cid:3) Fact 6.10. Let H ( m + 1 , s ) ⊂ H ′ ( m, ℓ ) and g ( m + 1 , s ) ∩ g c ( p, y ′ ) = ∅ . Then H ( m + 1 , s ) ∩ g c ( p, y ′ ) = ∅ , g ( m + 1 , s ) ⊂ g c ( p, y ′ ) and g ( m, s ) ⊂ g c ( p, y ′ ) .Proof. Note g ( m, a ∗ , y ) = I ( m, a ∗ , y ) × J ( m, a ∗ , y ). Let J ( m, a ∗ , y ) = [ q l , q l m ) × [ q l , q l m ) × ... × X m . Let H ( m, ℓ ) = { q } × ... × { q m } × X m and P ( m, ℓ ) = { q } × ... × { q m } . Then H ′ ( m, ℓ ) = G m ∩ H ( m, ℓ ) = P ( m, ℓ ) × J ( m, a ∗ , y ) . Let g c ( p, y ′ ) = I m ( p, y ′ ) × J m ( p, y ′ ). Note that 1 ℓ is the least number such that H ′ ( m, ℓ ) ∩ g c ( p, y ′ ) = ∅ . Then P ( m, ℓ ) ∈ I m ( p, y ′ ) = [ q , q c ) × ... × [ q m , q mc ) . Let G ( m , ℓ ) be an Ln cover of H ′ ( m, ℓ ). Then there exists a g ( m , s ) ∈ G ( m , ℓ )such that H ( m , s ) ⊂ H ′ ( m, ℓ ) and g ( m , s ) ∩ g c ( p, y ′ ) = ∅ . Then we have H ( m , s ) ⊂ H ( m, s ) and g ( m, s ) ∩ g c ( p, y ′ ) = ∅ . Note g ( m, s ) = g ( m, ℓ , s ) = I ( m, ℓ , s ) × J ( m, ℓ , s ) . Then J ( m, ℓ , s ) ∩ J m ( p, y ′ ) = ∅ . Note H ( m, ℓ , s ) = P ( m, ℓ ) × J ( m, ℓ , s ) by thedefinition of H ( n, i, l ) in A of Construction 3.2. Then H ( m, ℓ , s ) ∩ g c ( p, y ′ ) = ∅ .Then g ( m, s ) ⊂ g c ( p, y ′ ) by the definition 1 ℓ . Then g ( m + 1 , s ) ⊂ g c ( p, y ′ ). (cid:3) The proof of 3 is continued. Let t ∗ ∈ H ′′ ( m , ℓ ) = H ′ ( m , ℓ ) ∩ g c ( p, y ′ ) with H ′′ ( m, ℓ ) ∩ H ( p, y ′ ) = ∅ , and G ( m , ℓ ) be an Ln cover of H ′ ( m, ℓ ). Then thereexists a g ( m , x ) ∈ G ( m , ℓ ) such that t ∗ ∈ H ′′ ( m, ℓ ) ∩ g ( m , x ) ⊂ g c ( p, y ′ ) ∩ g ( m , x ) . Then g ( m , x ) ∩ g c ( p, y ′ ) = ∅ . Then, by Fact 6.10, we have g ( m , x ) ⊂ g c ( p, y ′ ) and x ∈ H ′ ( m, ℓ ) ∩ g ( m , x ) ⊂ H ′′ ( m, ℓ ).Let G ∗∗ ( m , y ′ ℓ ) = { g ( m , x ) ∈ G ( m , ℓ ) : g c ( p, y ′ ) ∩ g ( m , x ) = ∅} and G ∗ ( m , y ′ ℓ ) = { g ( m , x ) ∈ G ( m , ℓ ) : g ( m , x ) ⊂ g c ( p, y ′ ) } . Then G ∗∗ ( m , y ′ ℓ ) = G ∗ ( m , y ′ ℓ ) = ∅ . Let G ( m, ℓ ) be a Ln cover of H ′ ( m , ℓ )and G ∗ ( m, y ′ ℓ ) = { g ( m, x ) ∈ G ( m, ℓ ) : g ( m, x ) ⊂ g c ( p, y ′ ) } . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 31 Then G ∗ ( m, y ′ ℓ ) = ∅ by Fact 6.10.A. Note G d (1 σ α , i , pσ ) = ∪{ G d (1 σ α , iz, pσ ) : g ( m , z ) ∈ G ∗ ( m , i ) } for i ∈ N .Let g ( m , z ) ∈ G ∗ ( m , y ′ ℓ ) Then g ( m , z ) ∩ [ ∪ i ≤ j ∪ G d (1 σ α , i , pσ )] = ∅ since g ( m , z ) ⊂ g c ( p, y ′ ) ⊂ g ( v α , r ′ ) and g ( v α , r ′ ) ∩ [ ∪ i ≤ j ∪ G d (1 σ α , i , pσ )] = ∅ . Note g ( m , z ) ∩ G ( m , ℓ ) = ∅ for j < k < ℓ by Fact 6.9. Let i > ℓ , G ( m , i ) be a Lncover of H ′ ( m , i ) and G ( m , i ) = ∪ G ( m , i ). Then, by 6 of Proposition 3.3, wehave G ( m , i ) ∩ [ ∪ H d (1 σ α , ℓ , pσ )] = ∅ . Note G ( m , ℓ ) is an Ln cover of H ′ ( m, ℓ )and G ∗ ( m , y ′ ℓ ) ⊂ G ( m , ℓ ). Then G d (1 σ α , ℓ , pσ ) | g ( m , z ) = G d (1 σ α , ℓz, pσ ) forevery g ( m , z ) ∈ G ∗ ( m , y ′ ℓ ) by the definition of H d (1 σ α , ℓz, pσ ) in B3.2.B. Let g ( m , z ) ∈ G ∗ ( m , y ′ ℓ ). Then ∪ G d (1 σ α , ℓz, pσ ) ⊂ g ( m , z ) ⊂ g c ( p, y ′ ).Let G d (1 σ α , ℓz, p ) = ∪ G d (1 σ α , ℓz, pσ ).Then we have the following fact: Fact 6.11. ∅ 6 = G ∗ ( m , y ′ ℓ ) ⊂ G ∗ ( m , ℓ ) and G ∗ ( m, y ′ ℓ ) = ∅ .2. H d (1 σ α , pσ ) | H ( m , z ) = H d (1 σ α , ℓz, pσ ) for every g ( m , z ) ∈ G ∗ ( m , y ′ ℓ ) .3. H ( m , z ) − G d (1 σ α , ℓz, p ) ⊂ g c ( p, y ′ ) and G d (1 σ α , ℓz, p ) ⊂ g c ( p, y ′ ) for every g ( m , z ) ∈ G ∗ ( m , y ′ ℓ ) . Call H ( m , z ) full if g ( m , z ) ∈ G ∗ ( m , y ′ ℓ ) . Pick an arbitrary x dj ∈ H ( m , z ) − G d (1 σ α , ℓz, p ). Let A ∗ dℓ = { x dz : x dz ∈ H ( m , z ) − G d (1 σ α , ℓz, p ) and g ( m , z ) ∈ G ∗ ( m , ℓ ) } and A ∗ d = { A ∗ dℓ : H ′ ( m, ℓ ) ∈ H ′ ( m, y ) } . Then we have the following fact: Fact 6.12. Let H ( v α , t ) ∈ H − (1 σ α , pσ ) . Then:1. H ( v α , t ) ⊂ Cl ρ A ∗ d .2. H ( v α , t ) ⊂ Cl ρ [ ∪ ℓ>j H ( m, ℓ )] , H ( v α , t ) ⊂ Cl ρ [ ∪ i>j ∪ H d (1 σ α , i , pσ )] and H ( v α , t ) ⊂ Cl ρ [ ∪ i>j ∪ G d (1 σ α , i , pσ )] .The proof of 3 is continued. Then, by the above Fact 6.12, Cl ρ D (1 σ α , G ) = g ( m, y ). Proof of 6. Let t ∈ g [ p, D (1 σ α , G )] − D (1 σ α , G ). Then there exists an H ′ ( m, j ) ∈ H ′ ( m, y ) such that t ∈ H ′ ( m, j ) ∩ [ g [ p, D (1 σ α , G )] − D (1 σ α , G )] . A. Let D (1 , < ) = ∪ i Then g [ q, D (1 , < )] ∩ H ( v α , t ) = ∅ . Let q = q .B. Let H ( m, j +) = { H ′ ( m, l ) ∈ H ′ ( m, y ) : 1 l > j } and H ( m, j +) = ∪ H ( m, j +) . Then H ′ ( m, j ) ∩ g [ m , H ( m , j +)] = ∅ . Let q > m .C. Note G ( m , j ) is a Ln cover of H ′ ( m, j ) and G ( m , j , z ) is a Ln cover of H ( m , jz ) for every g ( m , z ) ∈ G ( m , j ) by 2 of Claim 6.6. Then there uniquely ex-ists an g ( m , z v ) ∈ G ( m , j , z ) such that t ∈ H ( m , jz v ) = g ( m , z v ) ∩ H ( m , jz ).Then t ∈ g [ p, ∪ G d ( v α , jz v )] − ∪ G d ( v α , jz v ) since G ( m , j , z ) is a Ln cover of H ( m , jz ). Note, by 3 of B of Claim 5.8, q ≥ v α implies g [ q, ∪ G d ( v α , jz v )] = ∪ G d ( v α , jz v ) . Then there exists an ℓ > v α such that H ( l, t ) ∩ g [ q, ∪ G d ( v α , jz v )] = ∅ if l, q > ℓ . Let q > ℓ .Let l, q > max { q , q , q } . Then H ( l, t ) ∩ g [ q, D (1 σ α , G )] = ∅ . (cid:3) To calculate g [ p, D (1 σ α , G )] for p ≥ m , we prove the following proposition. Proposition 6.13. Let p ≥ m , H ′ ( m, y ) = { H ′ ( m, h ) : h ∈ N } with O ( m, y ) = ∪ H ′ ( m, y ) . Then:1. g [ p, D (1 σ α , G )] = D ( p, m α ) = ∪ h D ( p, m α , h ) .2. Cl ρ D ( p, m α ) = g ( m, y ) .3. D ( p, m α ) ∩ H ( p, m α ) = ∅ and D ( p, m α ) ∪ H ( p, m α ) = O ( m, y ) .4. Let G d (1 α, j ) = ∪ G d (1 σ α , j , qσ ) and g [ p, G d (1 α, j )] = D ( p, m α , j ) . Then D ( p, m α , j ) is a c.o set in ( X, ρ ) for every j ∈ N , D ( p, m α ) = ∪ j D ( p, m α , j ) and H ( m, j ) = H ′ ( m, j ) − D ( p, m α ) = H ′ ( m, l j ) − [ ∪ i ≤ j D ( p, m α , i )] .5. There exists a c.o.D family G d ( p, m α , j ) with D ( p, m α , j ) = ∪ G ( p, m α , j ) .6. Let H ( l, t ) ⊂ H ( p, m α ) = ∪ j H ( m, j ) . Then H ( l, t ) ⊂ Cl ρ [ ∪ j>h H ( m, j )] and H ( l, t ) ⊂ Cl ρ [ ∪ j>h D ( p, m α , j )] for arbitrary h ∈ N .Proof. A. Take H d (1 σ α , j , pσ ) = ∪{ H d (1 σ α , jz, pσ ) : g ( m , z ) ∈ G ∗ ( m , j ) } fromB3.2. Take H d ( v α , jz v ) ⊂ H d (1 σ α , j , pσ ). Let v α > q . Then D ( q, v ) = g [ q, ∪ H d ( v α , jz v )] = g [ q, ∪ G d ( v α , jz v )]is a c.o set in ( X, ρ ) by the definition of G ( v α , jz nv ) and 3 of B of Claim 5.8. Then D ( q, v ) ⊂ g ( m , z v ). Let D ( q + , z, j ) = { D ( q, v ) : v α > q } . Then D ( q + , z, j )is c.o.D in ( X, ρ ). Let v α ≤ q . Then g [ q, g ( v α , t )] = g ( v α , t ). Let D ( q − , z, j ) = ∪{ G d ( v α , jz v ) : v α ≤ q } . Then D ( q − , z, j ) is c.o.D in ( X, ρ ). Let D ( q, z, j ) = D ( q − , z, j ) ∪ D ( q + , z, j ) D d ( q, m α , j ) = ∪{ D ( q, z, j ) : g ( m , z ) ∈ G ∗ ( m , j ) } . Then D d ( q, m α , j ) is c.o.D in ( X, ρ ). Let G d (1 α, j ) = ∪ G d (1 σ α , j , pσ ). Then g [ q, G d (1 α, j )] = ∪ D d ( q, m α , j ) = D ( q, m α , j ) is a c.o set for every j ∈ N .B. I. Let k = 1. Note H ( m, ) = H ′ ( m, ) − D ( q, m α , ). Then we have H ( m, ) = H ′ ( m, ) − g [ q, G d (1 α, )]. Then H ( m, ) ∩ g [ q, ∪ j> G d (1 α, j )] = ∅ .Then H ( m, ) ∩ ∪ j> D ( q, m α , j ) = ∅ by the definition of D ( q, m α , j ). Then H ( m, ) = H ′ ( m, ) − g [ q, ∪ j G d (1 α, j )] = H ′ ( m, ) − D ( q, m α , ) . II. Assume g [ q, ∪ j ≤ n G d (1 α, j )] = ∪ j ≤ n D ( q, m α , j ). Let k = n + 1. Take H ′ ( m, k ). Then, by 1 and 2 Proposition 6.5, there exist infinitely many g ( m , z )such that H ( m , z ) ⊂ H ′ ( m, k ) and g ( m , z ) ∩ g [ q, ∪ n 1) = { H ( m, k ) : k ∈ N } and D ( q, m α , 1) = { D ( q, m α , k ) : k ∈ N } .This implies 1, 3, 4 and 5. Let G ′′ (1 , α ) = { G ′′ (1 α, k ) : k ∈ N } .Let A ∗ dj = { x dz : x dz ∈ H ( m , z ) − [ ∪ D ( q, z, j )] and g ( m , z ) ∈ G ′′ (1 α, j ) } and A ∗ d = { A ∗ dj : H ′ ( m , j ) ∈ H ′ ( m, y ) } . Then, in the same way as the proof of 3 of Claim 6.8, we have the following fact: Fact 6.14. Let H ( l, t ) ⊂ H ( q, m α ) = ∪ H ′ ( q, m α , . Then:1. H ( l, t ) ⊂ Cl ρ A ∗ d .2. H ( l, t ) ⊂ Cl ρ [ ∪ ℓ>j H ( m, ℓ )] , H ( l, t ) ⊂ Cl ρ [ ∪ i>j ∪ G (1 α, i )] and H ( l, t ) ⊂ Cl ρ [ ∪ i>j ∪ H (1 α, i )] . This implies 2 and 6. Then we complete the proof of proposition. (cid:3) Let p > m . Let D = D (1 σ α , G ) = ∪ G (1 σ α , G ). Then D is an open set in( X, ρ ). Let ∂D = [ Cl ρ D ] − D . Then D ∩ ∂D = ∅ . Take g [ p, D (1 σ α , G )] from 1 ofProposition 6.13. Let max { p, m + 1 } = q . Take H d (2 σ α , q σ ) in B2. Let H (2 σ α , q σ ) = [ ∪ H d (2 σ α , q σ )] − g [ p, D (1 σ α , G )] . Note Face 6.12 and Fact 6.14. Then we have the following denotation. Denotation 1. Denote g [ p, D (1 σ α , G )] ∪ H (2 σ α , q σ ) by Cl τ g [ p, D (1 σ α , G )]for p > m . Claim 6.15. Let p > m . Then:1. ∩ p Cl τ g [ p, D (1 σ α , G )] = D (1 σ α , G ) .2. Let t ∈ Cl τ g [ p, D (1 σ α , G )] − D (1 σ α , G ) . Then there exists a k > p suchthat H ( ℓ, t ) ∩ Cl τ g [ q, D (1 σ α , G )] = ∅ if ℓ, q > k .3. Cl τ g [ p, D (1 σ α , G )] = g [ p, D (1 σ α , G )] ∪ H (2 σ α , q σ ) . ∗ AND BOSEN WANG ∗∗∗ Proof. To see 2 let E ( α , p ) = Cl τ g [ p, D (1 σ α , G )]. Then, for q = max { p, m + 1 } , E ( α , p ) = g [ p, ∪ G (1 σ α , G )] ∪ H (2 σ α , q σ ) . Let t ∈ E ( α , p ) − D (1 σ α , G ). Then t ∈ g [ p, ∪ G (1 σ α , G )] or t ∈ H (2 σ α , q σ ).Case 1, t ∈ g [ p, ∪ G (1 σ α , G )]. Then t ∈ g [ p, ∪ G (1 σ α , G )] − ∪ G (1 σ α , G ). Then,by 6 of Claim 6.8, there exists a q > p such that H ( ℓ, t ) ∩ g [ q, D ] = ∅ if ℓ > q . Here D = D (1 σ α , G ). Then we have the following Fact. Fact 6.16. Let t ∈ g [ p, ∪ G (1 σ α , G )] − ∪ G (1 σ α , G ) . Then there exists a k ∈ N such that H ( ℓ, t ) ∩ g [ q, D (1 σ α , G )] = ∅ if ℓ, q > k . Case 2. t ∈ H (2 σ α , q σ ). Then t / ∈ D since H (2 σ α , q σ ) ∩ g [ p, D (1 σ α , G )] = ∅ .Then H ( ℓ, t ) ∩ g [ q, D ] = ∅ by Fact 6.16.On the other hand, by the definition of H (2 σ α , q σ ), there exists an H ( u α , t ′ ) ∈ H (2 σ α , q σ ) such that x ∈ H ( u α , t ′ ) = H ( u α , t ). Note the definition of H (2 σ α , q σ ).Let v p = u α and q > v p . Then H ( u α , t ) / ∈ H (2 σ α , qσ ) and H ( u α , t ) ∩ [ ∪ H (2 σ α , qσ )] = ∅ . Then we have the following Fact. Fact 6.17. Let t ∈ H (2 σ α , q σ ) . Then there exists a v p ∈ N such that H ( ℓ, t ) ∩ H (2 σ α , qσ ) = ∅ if q, ℓ > v p . Let q > n ( t ) = max { k , v p } . Then, by Denotation 1, we have g [ q, D (1 σ α , G )] ∪ H (2 σ α , qσ ) = Cl τ g [ q, D (1 σ α , G )] . Then H ( ℓ, t ) ∩ Cl τ g [ q, D (1 σ α , G )] = ∅ if ℓ > n ( t ). The proof of 2 is continued . Then, by Case 1 and Case 2, we have proved 2.Then it is easy to see ∩ p Cl τ g [ p, D (1 σ α , G )] ⊂ D (1 σ α , G ). This implies 1. (cid:3) Construction 6.1 is continued.Q. Assume, for n = h +1, and p n ≥ p h ≥ ... ≥ p ≥ p > m , we have constructed E ( αn, p n ) = Cl τ g [ p n , E ( αh, p h )] = g [ p n , E ( αh, p h )] ∪ H ( nσ α , q n σ ) . Here q n = max { p n , q h } and H ( nσ α , q n σ ) = [ ∪ H d ( nσ α , q n σ )] − g [ p n , E ( αh, p h )]. Byinduction in the same as method of B1 and B2, we construct E ( αk, p k ) = Cl τ g [ p k , E ( αn, p n )] = g [ p k , E ( αn, p n )] ∪ H ( kσ α , q k σ ) such that E ( αk, p k ) = D ( p , m α ) ∪ g [ p , H (2 σ α , q σ )] ∪ ... ∪∪ g [ p k , H ( nσ α , q n σ )] ∪ H ( kσ α , q k σ ) and H ( kσ α , q k σ ) = [ ∪ H d ( kσ α , q k σ )] − g [ p k , E ( αn, p n )]. Here k = n +1, q k = max { p k , q n } and p k ≥ p n ≥ ... ≥ p > m .By Proposition 6.13, assume we have had H ( p n , m α , n ) = { H ( m, n i ) : i ∈ N } , D ( p n , m α , n ) = { D ( p n , m α , n i ) : i ∈ N } and G ′′ ( n, α ) = { G ′′ ( nα, i ) : i ∈ N } with H ( m, n j ) = H ′ ( m, j ) − D ( αn, p n ) = H ′ ( m, j ) − [ ∪ i ≤ j D ( p, m α , n i )] . Here D ( αn, p n ) = ∪ D ( p n , m α , n ). Q1. Take H ′ ( m, ) ∈ H ′ ( m, y ) and D ( p n , m α , n ) ∈ D ( p n , m α , n ). Note k = n + 1. Let G ( m , k ) be a Ln cover of H ′ ( m, ), H ∗ ( m, k ) = H ′ ( m, ) − D ( p n , m α , n ) and G ∗ ( m , k ) = { g ( m , z ) ∈ G ( m , k ) : g ( m , z ) ∩ H ∗ ( m , k ) = ∅} . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 35 Let g ( m , z ) ∈ G ∗ ( m , k ) and H ( m , z ) = g ( m , z ) ∩ H ′ ( m, ). Then we have H ( m , z ) ∩ E ( αn, p n ) ⊂ ∪ i ≤ n H ( m , z i ) since ∪ H ( iσ α , z, pσ ) ⊂ H ( m , z i ) by1 of Claim 6.6 and 3 of Claim 6.7, and p i > m for i ≤ n . Let p k ≥ p n and q k = max { p k , q n } . Take G d ( kσ α , z, q k σ ) and G − ( kσ α , z, q k σ ) from B2 for every H ( m , z ) ∈ G ∗ ( m , k ). Let G d ( kσ α , , q k σ ) = ∪{ G d ( kσ α , z, q k σ ) : g ( m , z ) ∈ G ∗ ( m , k ) } ,H ( kσ α , , q k σ ) = ∪ H d ( kσ α , , q k σ ) , G − ( kσ α , , q k σ ) = ∪{ G − ( kσ α , z, q k σ ) : g ( m , z ) ∈ G ∗ ( m , k ) } and D ( p k , m α , k ) = D ( p n , m α , n ) ∪ g [ p k , H ( kσ α , , q k σ )] . Q2. Assume we have had H ( kσ α , i , q k σ ) and D ( p k , m α , k i ) for i < j . Take H ′ ( m, j ) ∈ H ′ ( m, y ) and D ( p n , m α , n j ) ∈ D ( q n , m α , n ). Let G ( m , k j ) be a Lncover of H ′ ( m, j ), H ∗ ( m, k j ) = H ′ ( m, j ) − ∪ i Let p ′ k > p k . Then Cl τ g [ p ′ k , E ( αn, p n )] ⊂ Cl τ g [ p k , E ( αn, p n )] . ∗ AND BOSEN WANG ∗∗∗ Proof. By 1 of Claim 6.7, we have H (1 σ α , q ′ σ ) ⊂ H (1 σ α , q σ ) if q ′ > q .A. Let p ′ > p . Then g [ p ′ , D (1 σ α , G )] ⊂ g [ p , D (1 σ α , G )]. Let q ′ = max { p ′ , q } and q = max { p , q } . Then q ′ ≥ q . Then, by B2 in Construction 6.1, we have H d (2 σ α , q ′ σ ) ⊂ H d (2 σ α , q σ ). Then Cl τ g [ p ′ , D (1 σ α , G )] = g [ p ′ , D (1 σ α , G )] ∪ H (2 σ α , q ′ σ )= g [ p ′ , D (1 σ α , G )] ∪ [ ∪ H d (2 σ α , q ′ σ ) − g [ p ′ , D (1 σ α , G )]]= g [ p ′ , D (1 σ α , G )] ∪ [ ∪ H d (2 σ α , q ′ σ )] ⊂ g [ p , D (1 σ α , G )] ∪ [ ∪ H d (2 σ α , q σ )]= g [ p , D (1 σ α , G )] ∪ [ ∪ H d (2 σ α , q σ ) − g [ p , D (1 σ α , G )]= g [ p , D (1 σ α , G )] ∪ H (2 σ α , q σ )= Cl τ g [ p , D (1 σ α , G )] . B. Assume that we have Cl τ g [ p ′ n , E ( αh, p h )] ⊂ Cl τ g [ p n , E ( αh, p h )] if p ′ n > p n for n = h + 1. Let p ′ k > p k . Then q ′ k = max { p ′ k , q n } ≥ q k = max { p k , q n } . Then g [ p ′ k , E ( αn, p n )] ⊂ g [ p k , E ( αn, p n )] and H d ( kσ α , q ′ k σ ) ⊂ H d ( kσ α , q k σ ). Then E ( αk, p ′ k ) = Cl τ g [ p ′ k , E ( αn, p n )] ⊂ Cl τ g [ p k , E ( αn, p n )] ⊂ E ( αk, p k )in the same way as the above A. (cid:3) Then we have E ( α , p ) = D (1 σ α , G ), E ( α , p ) = g [ p , G (1 σ α , G )] ∪ H (2 σ α , q σ ) with q = max { p , q } and E ( αk, p k ) = Cl τ g [ p k , E ( αn, p n )] = g [ p k , E ( αn, p n )] ∪ H ( kσ α , q k σ )with q k = max { p k , q n } . To construct a stratifiable space ( Y, τ ) Construction 7. We define a topological space ( Y, τ ). To do it take g ( m, a ∗ , y ). Let Y = G m = g ( m, a ∗ , y ) and H = g ( m, a ∗ , y ) ∩ H ′ ( m, n ) . Then H = H ( m, a ∗ , y ). For every x ∈ Y , let H x = { H ( l, x ) ∈ H : l ≥ m } and G x = { g ( l, x ) ∈ G : l ≥ m } . Then { x } = ∩ l ≥ m H ( l, x ) = ∩ l ≥ m g ( l, x ) . Let H Y = ∪{ H x : x ∈ Y } and G Y = ∪{ G x : x ∈ Y } . To define topological space ( Y, τ ), take G (1 σ α , G ) . Let D (1 σ α , G ) = ∪ G (1 σ α , G )for α ∈ A . Let k = n + 1, q k = max { p k , q n } for p k ≥ p n ≥ ... ≥ p > m ,H ( kσ α , q k σ ) = [ ∪ H d ( kσ α , q k σ )] − g [ p k , E ( αn, p n )] ,E ( αk, p k ) = g [ p , D (1 σ α , G )] ∪ g [ p , H (2 σ α , q σ )] ∪ ... ∪∪ g [ p k , H ( nσ α , q n σ )] ∪ H ( kσ α , q k σ ) ,D ( αk, p k ) = g [ p , D (1 σ α , G )] ∪ g [ p , H (2 σ α , q σ )] ∪ ... ∪ g [ p k , H ( nσ α , q n σ )] . Then E ( αk, p k ) = D ( αk, p k ) ∪ H ( kσ α , q k σ ) and E ( αk, p k ) ∩ H = ∅ . Definition A . 1. Let g ( l, x ) be closed and open if g ( l, x ) ∈ G Y . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 37 2. Let Cl τ g [ p , D (1 σ α , G )] = G m if p ≤ m , and let Cl τ g [ p , D (1 σ α , G )] = g [ p , D (1 σ α , G )] ∪ H (2 σ α , q σ ) = E ( α , p )be closed and open if p > m .3. Let U ( αℓ , p x ) = g ( ℓ, x ) − E ( α , p )be a closed and open neighborhood of x if x ∈ Y − E ( α , p ) and ℓ > m . Definition B . Let k = n + 1 > Cl τ D ( αk, p k ) = G m if p k ≤ m , and let Cl τ D ( αk, p k ) = D ( αk, p k ) ∪ H ( kσ α , q k σ ) = E ( αk, p k )be closed and open if p k > m .2. Let U ( αℓk, p k x ) = g ( ℓ, x ) − E ( αk, p k )be a closed and open neighborhood of x if x ∈ Y − E ( αk, p k ) and ℓ > m . Definition C . Let p h > p k and g [ p h , U ( αℓk, p k x )]be a closed and open if U ( αℓk, p k x ) = g ( ℓ, x ) − E ( αk, p k ). (cid:3) Proposition 7.1. Definition C is reasonable.Proof. Let y ′ ∈ H ( v α , t ) ∈ H d ( v α , jz nv ). Then H ( v α , t ) ⊂ H ( nσ α , p n σ ) ⊂ E ( αn, p n ).Then H ( v α , t ) ∩ [ Y − E ( αn, p n )] = ∅ . Let H ( v α , t ) ⊂ H ′ ( m, l j ), 1 j = l j and H ( mv α , − t ) = [ H ′ ( m, l j ) ∩ g ( v α , t )] − H ( v α , t ) . Then H ( mv α , − t ) ⊂ Y − E ( αn, p n ) by the definition of E ( αn, p n ). Take a g ( ℓ, s )such that g ( v α , t ) ⊂ g ( ℓ, s ). Then we have H ( mv α , − t ) ⊂ U s = U ( αℓn, p n s ) = g ( ℓ, s ) − E ( αn, p n ) with H ( v α , t ) ∩ U s = ∅ . I. g c ( p, y ′ ) ∩ h Y − [ g [ ℓ, U s ] ∪ H ( v α , t )] i = ∅ if H ( v α , t ) ∩ g [ ℓ, H ( mv α , − t )] = ∅ forevery p, ℓ ≥ max { v α , p , p , ..., p n , q } .In fact, y ′ ∈ H ( v α , t ) ⊂ H ′ ( m, l j ). Let G ( m , l j ) be a Ln cover of H ′ ( m, l j ), H ∗ ( m, n j ) = H ′ ( m, l j ) − ∪ i Note H ( v α , t ) ∩ g [ ℓ, H ( mv α , − t )] = ∅ for every p, ℓ ≥ max { v α , p , p , ..., p n , q } .Then H ( v α , t z ) ∩ g [ ℓ, U s ] = ∅ if ℓ ≥ max { v α , p , p , ..., p n , q } . Then it is easy to see H ( v α , t z ) ⊂ Y − g [ ℓ, U s ] and H ( v α , t z ) ⊂ g c ( p, y ′ ). Then g c ( p, y ′ ) ∩ h Y − [ g [ ℓ, U s ] ∪ H ( v α , t )] i = ∅ . II. We prove that: if g c ( p, y ′ ) ∩ h Y − [ g [ ℓ, U s ] ∪ H ( v α , t )] i = ∅ for every p, ℓ ≥ max { v α , p , p , ..., p n , q } , then Definition C is reasonable.In fact, y ′ ∈ H ( v α , t ) ∈ H d ( v α , jz nv ) and H ( v α , t ) ∩ [ Y − E ( αn, p n )] = ∅ . Suppose g c ( p, y ′ ) ∩ h Y − [ g [ ℓ, U s ] ∪ H ( v α , t )] i = ∅ for some p, ℓ ≥ max { v α , p , p , ..., p n , q } .Then we have g c ( p, y ′ ) − H ( v α , t ) ⊂ g [ ℓ, U s ]. Note that H ( v α , t ) ∩ [ g [ ℓ, U s ] = ∅ ,and g [ ℓ, U s ] is closed by Definition C in Section 7. Then we have H ( v α , t ) ∩ g c ( p, y ′ ) = g c ( p, y ′ ) − g [ ℓ, U s ] is open. Then H ( v α , t ) is open since y ′ is arbitraryin H ( v α , t ). Then, by the definition of D ( αn, p n ) in Q2 of Construction 6.1, thisimplies H ( nσ α , q n σ ) = ∪{ H ( v α , t ) ∈ H d ( nσ α , q n σ ) : H ( v α , t ) ∩ D ( αn, p n ) = ∅} .Then H ( nσ α , q n σ ) is open. On the other hand, by 1 of Definition B in Section 7, Cl τ D ( αn, p n ) − D ( αn, p n ) = H ( nσ α , q n σ ) . This is a contradiction. III. So, in the following we must prove that H ( v α , t ) ∩ g [ ℓ, H ( mv α , − t )] = ∅ forevery ℓ ≥ max { v α , p , p , ..., p n , q } .To do it let k = v α . Then H ( v α , t ) = H ( k, t ) and g ( v α , t ) = g ( k, t ). Let H ( k, t ) = H ( k, , t ) ⊂ H ′ ( m, l j ), a = k + 1 = k + d , H ( mv α , t ) = H ′ ( m, l j ) ∩ g ( v α , t ) ,g ( k, t ) = [ q , q a ) × ... × [ q k , q k a ) × [ q h , q h a ) × ... × [ q h d , q h d a ) × X k +1 and H ( k, t ) = { q } × ... × { q m } × ... × { q k } × [ q h , q h a ) × ... × [ q h d , q h d a ) × X k +1 . Note H ( k, t ) ⊂ H ′ ( m, l j ). Then H ′ ( m, l j ) = [ { q } × ... × { q m } × X m ] ∩ G m and P ( m, l j ) = { q } × ... × { q m } = ( q , ..., q m ). Let b = m + 1. Then H ( mv α , t ) = H ′ ( m, l j ) ∩ g ( k, t ) = P ( m, l j ) × [ q b , q b a ) × ... × [ q k , q k a ) × J ( k, , t ) . Let I ( mv α , t ) = ( q , ..., q m ) × [ q b , q b a ) × ... × [ q k , q k a ). Take a Ln cover of H ( mv α , − t ) for k = n + 1. To do it note Figure I in (B)of Construction 3.1. Let P ( k, i ) = P ( n, l ) × { q j } in Figure I satisfy P ( k, i ) = ( q , ..., q m ) × { q i b } × ... × { q i k } . Then P ( k, i ) × X k = H ( k, i ) satisfies ∪ i H ( k, i ) = H ( mv α , t ) . Then top m coor-dinate of H ( k, i ) is ( q , ..., q m ) for every i . Let H ( k, t ) = H ( k, , t ), g ( k, , t ) = I ( k, , t ) × J ( k, , t ) and H ( k, ) = P ( n, × { q } × X k = P ( k, ) × X k . Then H ( k, ) ∩ g ( k, , t ) = H ( k, , t ). Note we take a Ln cover of H ( mv α , − t ) for k = n + 1. Then we can not take P ( n, × { q } in Figure I in (B) of Construction3.1 since H ( k, t ) = H ( k, ) ∩ g ( k, , t ) and P ( n, × { q } = P ( k, ) = ( q , ..., q k ) ∈ I ( k, , t ) . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 39 We must take P ( n, × { q } = P ( k, ) since 1 is the least number suchthat H ( k, ) = P ( n, × { q } × X k = P ( k, ) × X k , H ( k, ) ∩ g ( k, t ) ⊂ H ( mv α , − t ) and H ( k, ) ∩ H ( k, ) = ∅ . Let a = k + 1 , P ( k, ) = ( q , ..., q k ) and I ( k, ) = [ q , q a ) × ... × [ q k , q k a ).Then P ( k, ) = ( q , ..., q m , q b , ..., q k ) by the definition of P ( k, i ) in Figure I.Then I ( k, ) = [ q , q a ) × ... × [ q m , q ma ) × [ q b , q b a ) × ... × [ q k , q k a ) . Take I a from 1 of Proposition 2.2. Note a = k + 1 > k + 1 = a . Let I a j = I a | [ q h j , q h j a ) for 1 ≤ j ≤ k, and I ( k, a , h ) = [ q h ′ , q h ′ a ) × ... × [ q h ′ k , q h ′ k a ) with [ q h ′ j , q h ′ j a ) ∈ I a j for 1 ≤ j ≤ k and I ∗ ( k, a ) be the family of all I ( k, a , h )’s. Then I ∗ ( k, a ) = { I ( k, a , h ) : h ∈ N } . Then ∪ I ∗ ( k, a ) = I ( k, , t ) by the definition of I a j for 1 ≤ j ≤ k . Let I + ( k, a ) = { I ( k, a , h ) ∈ I ∗ ( k, a ) : I ( k, a , h ) ∩ I ( mv α , t ) = ∅} . Then I + ( k, a ) = { I ( k, a , h ) : h ∈ N } with ∆-order. Then P ( k, ) ∈ I ( k, a , 1) and P ( k, ) ∈ I ( k, a , 2) = I ( k, ) . Then I ( k, a , ∩ I ( k, a , 2) = ∅ since I a j is pairwise disjoint for 1 ≤ j ≤ k . Then I ( k, a , 1) = [ q , q a ) × ... × [ q m , q m a ) × [ q b , q b a ) × ... × [ q k , q k a ) . Let I a j = I a | [ q j , q j a ) for k < j ≤ a = k + ,J ( k, a , h ) = [ q h ′ k +1 , q h ′ k +1 a ) × ... × [ q h ′ k + , q h ′ k + a ) × X a with [ q h ′ j , q h ′ j a ) ∈ I a j for k < j ≤ a = k +, and J ( k, a ) be the family of all J ( k, a , h )’s. Then J ( k, a ) = { J ( k, a , h ) : h ∈ N } . Then ∪ J ( k, a ) = J ( k, , t ) by the definition of I a j . Let G ( k, ) = { I ( k, ) × J ( k, a , l ) : J ( k, a , l ) ∈ J ( k, a ) } . Then G ( k, ) is a Ln cover of H ( k, ) ∩ g ( k, t ) and H ( k, ) ∩ g ( k, t ) ⊂ H ( mv α , − t ). Assume we have had G ( k, i ) for i < j . Following ∆-order (direction of thearrow) in Figure I, take the first point P ( n, i ) × { q j ′ } with P ( n, i ′ ) × { q j ′ } = P ( k, j ) ∈ I ( k, , t ) − ∪ i 1) and G ( k, j ) for j ∈ N . Let I ( k, − t ) = { I ( k, j ) : j ∈ N } , I ( k, a σ , t ) = { I ( k, a j , 1) : j ∈ N } and G ( k, − t ) = ∪ j G ( k, j ) . Then G ( k, − t ) is a Ln cover of H ( mv α , − t ).In fact, pick an x ∈ H ( mv α , − t ). Let I ( mv α , − t ) = ( q , ..., q m ) × [ q b , q b a ) × ... × [ q k , q k a ) − P ( k, ) ,x = P ( k, j ) × ( q ′ k +1 , q ′ k +2 , ... ) and P ( k, j ) ∈ I ( mv α , − t ). If P ( k, j ) ∈ ∪ i 1) = [ q , q a j ) × ... × [ q m , q m a j ) × [ q b , q b a j ) × ... × [ q k , q k a j ) . Let I bk ( k, a j , 1) = [ q b , q b a j ) × ... × [ q k , q k a j ) and P bk ( k, ) = ( q b , ..., q k ). Then P bk ( k, ) = ( q b , ..., q k ) ∈ I bk ( k, a j , . Note H ( mp, y ′ , c ) ⊂ ∪ G ( mp, y ′ , c ). Take a g ( m , l ′ ℓ , z ) ∈ G ( mp, y ′ , c ) with P bk ( k, ) ∈ I bk ( m , l ′ ℓ , z ). Then I bk ( m , l ′ ℓ , z ) = [ q b , q b e ′ ) × ... × [ q k , q k e ′ ) since q i ∈ Q ∗ a j ⊂ Q ∗ e ′ for 1 b ≤ i ≤ k by e ′ > a j . Then I bk ( m , l ′ ℓ , z ) ⊂ I bk ( k, a j , e ′ > a j . Let G ∗ ( mp, y ′ , c ) = { g ( m , z ) ∈ G ( mp, y ′ , c ) : P bk ( k, ) ∈ I bk ( m , l ′ ℓ , z ) } . Then we have G ∗ ( mp, y ′ , c ) = ∅ since g ( m , l ′ ℓ , z ) ∈ G ∗ ( mp, y ′ , c ). Take a g ( m , z )from G ∗ ( mp, y ′ , c ). It is easy to see g ( m , z ) ⊂ g c ( p, y ′ ) and H ( m , z ) ⊂ H ′ ( m, l ℓ )by the definition of G ( mp, y ′ , c ). And then we have the following claim.3D1. g ( m , z ) ∩ [ ∪ G ( k, j )] = ∅ if j < n .In fact, note I ( k, a j , ∩ I ( k, j ) = ∅ . Then we have I bk ( m , l ′ ℓ , z ) ∩ I bk ( k, j ) = ∅ for each j < n since ( q , ..., q m ) ∈ I m ( k, a j , ∩ I m ( k, j ) by 1 of Claim 1C, and I bk ( m , l ′ ℓ , z ) ⊂ I bk ( k, a j , I k ( m , l ′ ℓ , z ) ∩ I ( k, j ) = ∅ for every j < n .Then g ( m , z ) ∩ [ ∪ G ( k, j )] = ∅ if j < n .3D2. g ( m , z ) ∩ [ ∪ G ( k, j )] = ∅ if j ≥ n . ∗ AND BOSEN WANG ∗∗∗ In fact, j ≥ n implies q ′ h / ∈ [ q h , q h a j ) = p h [ I m ( k, j )] for some h ≤ m by theabove Then q h a j < q ′ h since q ′ h ∈ [ q h , q h a i ) = p h [ I m ( k, i )] if i < n . Then wehave [ q h , q h a j ) ∩ [ q ′ h , q ′ he ) = ∅ . Then g ( m , z ) ∩ [ ∪ G ( k, j )] = ∅ if j ≥ n . Take a g ( k, x ) with x ∈ H ( mv α , − t ). Then there exists a g ( k, x ′ ) ∈ G ( k, − t )with g ( k, x ) ∩ g ( k, x ′ ) = ∅ since G ( k, − t ) is a Ln cover of H ( mv α , − t ). Then g ( k, x ) ⊂ g ( k, x ′ ) by the definition of Ln covers. Then, for every g ( m , z ) ∈ G ∗ ( mp, y ′ , c ), we have g ( m , z ) ∩ g [ k, H ( mv α , − t )] = ∅ . Note ℓ > k implies g [ ℓ, H ( mv α , − t )] ⊂ g [ k, H ( mv α , − t )]. Then g ( m , z ) ⊂ g c ( p, y ′ )and [ g ( m , z ) ∩ E ( αk, p k )] − g [ ℓ, U s ] = ∅ . (cid:3) 1. Let x / ∈ E ( αk, p k ). Then U ( αℓk, p k x ) = g ( ℓ, x ) − E ( αk, p k ) is open by thedefinition A and B.2. Let ℓ > m , x ∈ E ( αk, p k ) and U ( αℓk, p k x ) = g ( ℓ, x ) ∩ E ( αk, p k ) . Then U ( αℓk, p k x ) = g ( ℓ, x ) ∩ E ( αk, p k ) is open by the definition A and B.3. Let x ∈ Y and U ( x ) = { U ( αℓk, p k x ) : α ∈ A , ℓ > m , k ≥ p k > m } . And then let U ′ ⊂ U ( x ) be finite, U ( αℓk, p k x ) ∈ U ′ , U [ α − ℓk ′ , p ′ k x ] = ∩ U ′ with α − = { α i : U ( α i ℓ i k i , p ik x ) ∈ U ′ } , U x = { U [ α − ℓk ′ , p ′ k x ] : U ′ ⊂ U ( x ) is finite } and U = ∪{ U x : x ∈ Y } . Then U is a family of open sets by the definition A and B.A base U x of neighborhoods of x is called an outer base of x in Definition 1.3 of[22] also. Proposition 7.2. U x is a base of neighborhoods of x in some topological space ( Y, τ ) for every x ∈ Y .2. H Y = ∪{ H x : x ∈ Y } is a countable network of ( Y, τ ) .3. U is a base of some topological space ( Y, τ ) .Proof. Let U ( αℓk, p k x ) , U ( βℓ ′ n, p n x ) ∈ U ( x ) and t ∈ U ( αℓk, p k x ) ∩ U ( βℓ ′ n, p n x ).Let H ( u α , t α ) ⊂ U ( αℓk, p k x ) and H ( v β , t β ) ⊂ U ( βℓ ′ n, p n x ). Pick a point t suchthat t ∈ H ( u α , t α ) ∩ H ( v β , t β ).Case 1, U ( βℓ ′ n, p n x ) = g ( ℓ ′ , x ) − E ( βn, p n ) and U ( αℓk, p k x ) = g ( ℓ, x ) − E ( αk, p k ).Then H ( u α , t α ) ∩ E ( αk, p k ) = ∅ and H ( v β , t β ) ∩ E ( βn, p n ) = ∅ . Let l > max { u α , v β , ℓ ′ , ℓ } . Then g ( l, t ) ∩ U ( αℓk, p k x ) = g ( l, t ) ∩ [ g ( ℓ, x ) − E ( αk, p k )]. Then U ( αlk, p k t ) = g ( l, t ) ∩ U ( αℓk, p k x ) = g ( l, t ) − E ( αk, p k ) ∈ U ( t ) . Then, in the similar way, we have g ( l, t ) ∩ U ( βℓ ′ n, p n x ) = g ( l, t ) − E ( βn, p n ) = U ( βln, p n t ) ∈ U ( t ) . Then, by the definitions of H ( u α , t α ) and H ( v β , t β ), we have H ( l, t ) ⊂ U [ α − lk, p k t ] = U ( αlk, p k t ) ∩ U ( βln, p n t ) ∈ U t . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 43 Case 2, U ( βℓ ′ n, p n x ) = g ( ℓ ′ , x ) − E ( βn, p n ) and U ( αℓk, p k x ) = g ( ℓ, x ) ∩ E ( αk, p k ).Then H ( v β , t β ) ∩ E ( βn, p n ) = ∅ and H ( a, t α ) ⊂ g ( ℓ, x ) ∩ E ( αk, p k ). Let l > max { a, v β , ℓ ′ , ℓ } . Then g ( l, t ) ∩ U ( αℓk, p k x ) = g ( l, t ) ∩ [ g ( ℓ, x ) ∩ E ( αk, p k )]. Then U ( αlk, p k t ) = g ( l, t ) ∩ U ( αℓk, p k x ) = g ( l, t ) ∩ E ( αk, p k ) ∈ U ( t ) . Then, in the similar way, we have g ( l, t ) ∩ U ( βℓ ′ n, p n x ) = g ( l, t ) − E ( βn, p n ) = U ( βln, p n t ) ∈ U ( t ) . Then, by the definitions of H ( u α , t α ) and H ( v β , t β ), we have H ( l, t ) ⊂ U [ α − lk, p k t ] = U ( αlk, p k t ) ∩ U ( βln, p n t ) ∈ U t . Case 3, U ( βℓ ′ n, p n x ) = g ( ℓ ′ , x ) ∩ E ( βn, p n ) and U ( αℓk, p k x ) = g ( ℓ, x ) ∩ E ( αk, p k ).Then H ( b, t β ) ⊂ g ( ℓ ′ , x ) ∩ E ( βn, p n ) and H ( a, t α ) ⊂ g ( ℓ, x ) ∩ E ( αk, p k ). Let l > max { a, b, ℓ ′ , ℓ } . Then we have g ( l, t ) ∩ U ( αℓk, p k x ) = g ( l, t ) ∩ E ( αk, p k ) = U ( αlk, p k t ) ∈ U ( t ) and g ( l, t ) ∩ U ( βℓ ′ n, p n x ) = U ( βln, p n t ) ∈ U ( t ) . Then, by the definitions of H ( u α , t α )and H ( v β , t β ), we have H ( l, t ) ⊂ U [ α − lk ′ , p ′ k t ] = U ( αlk, p k t ) ∩ U ( βln, p n t ) ∈ U t . Let U [ α − ℓk ′ , p ′ k x ] , U [ β − ℓ ′ n ′ , p ′ n y ] ∈ U and t ∈ U [ α − ℓk ′ , p ′ k x ] ∩ U [ β − ℓ ′ n ′ , p ′ n y ]in general. Then, in the same way as the above proof, there exists a g ( l, t ) and a U [ γ − lk ′ , p ′ k t ] ∈ U t such that H ( l, t ) ⊂ U [ γ − lh ′ , p ′ h t ] ⊂ U [ α − ℓk ′ , p ′ k x ] ∩ U [ β − ℓ ′ n ′ , p ′ n y ] . Then, by Proposition 1.2.3 in [5], U is a base for some topological space ( Y, τ ). (cid:3) Let B be a defined set in Definitions A-C. Then B is c.o in ( Y, τ ) because both B and Y − B are in U . Denote the closure of B in ( Y, τ ) by Cl ∗ B , and inDefinitions of A-C by Cl τ B . Then Cl ∗ B = B = Cl τ B . Take E ( αk, p k ) = D ( αk, p k ) ∪ H ( kσ α , q k σ ) from Definition A-B. Note ∁ . Cl ∗ D ( αk, p k ) = Cl τ D ( αk, p k ) for every α ∈ A , k ≥ p k ∈ N .In fact, let t / ∈ Cl τ D ( αk, p k ) = E ( αk, p k ). Then U ( αlk, p k t ) = g ( l, t ) − E ( αk, p k )satisfies U ( αlk, p k t ) ∩ E ( αk, p k ) = ∅ . Note Y − E ( αk, p k ) ∈ U . Then D ( αk, p k ) ⊂ E ( αk, p k ) implies Cl ∗ D ( αk, p k ) ⊂ E ( αk, p k ). Then t / ∈ Cl ∗ D ( αk, p k ). Then Cl ∗ D ( αk, p k ) ⊂ Cl τ D ( αk, p k ) . Pick an x ∈ H ( kσ α , q k σ ). Then there exists an H ( v α , x ) ∈ H d ( kσ α , q k σ ) with H ( v α , x ) ∩ D ( αk, p k ) = ∅ . Take a g c ( ℓ, x ) for ℓ > m . Then, by Fact 6.10-6.11and 3 of Proposition 6.5, there exists a full H ( m , z ) with H ( m , z ) ⊂ g c ( ℓ, x ) and g ( m , z ) ∈ G ∗ ( m , y ′ ℓ ). Then, for every E ( βh, p h ), we have H ( m , z ) ∩ [ g c ( ℓ, x ) ∩ E ( βh, p h )] = ∅ and H ( m , z ) ∩ [ g c ( ℓ, x ) − E ( βh, p h )] = ∅ . Then x ∈ Cl ∗ D ( αk, p k ). Then Cl τ D ( αk, p k ) ⊂ Cl ∗ D ( αk, p k ). (cid:3) ∗ AND BOSEN WANG ∗∗∗ Then the closure operation Cl τ in definitions A-C is the same as the closureoperation Cl ∗ in ( Y, τ ). Denote the topological space by ( Y, τ ) . So, in topological space ( Y, τ ), we can denote the closure of a set A by Cl τ A byNote ∁ . Let Int τ A = Y − Cl τ ( Y − A ) . Proposition 7.3. ∩ p Cl τ g [ p, E ( αk, p k )] = E ( αk, p k ) .2. Let t ∈ Cl τ g [ p, E ( αk, p k )] − E ( αk, p k ) . Then there exists an ℓ such that: if l, q > ℓ , H ( l, t ) ∩ Cl τ g [ q, E ( αk, q k )] = ∅ .Proof. Let p > p k . Then g [ p, E ( αk, p k )] = D ( αk, p k ) ∪ g [ p, H ( kσ α , q k σ )]. Then, by1 of Definition B, Cl τ g [ p, E ( αk, p k )] = g [ p, E ( αk, p k )] ∪ H ( hσ α , q h σ ) . Here h = k + 1 and q h = max { p k , q n } .Let t ∈ H ( m , jz ) = H ′ ( m, j ) ∩ g ( m , z ) for some g ( m , z ) ∈ G ∗ ( m , j ), H ( t, j − ) = { H ′ ( m, i ) ∈ H ′ ( m, y ) : i < j } , H − = ∪ H ( m, j − ) , H ( m, j +) = { H ′ ( m, i ) ∈ H ′ ( m, y ) : i > j } and H + = ∪ H ( m, j +) . t / ∈ g [ p, H + ] if p > m = ℓ by 6 ′ of Proposition 3.3.2. t / ∈ g [ p, H − ∩ E ( αk, p k )] if p > ℓ for some ℓ since H ( t , j − ) is finite.3. Note p > p k and t ∈ h H ( m , jz ) ∩ Cl τ g [ p, E ( αk, p k )] i − E ( αk, p k ). Then t / ∈ D ( αk, p k ) since g [ p, D ( αk, p k )] = D ( αk, p k ) ⊂ E ( αk, p k ) if p > p k . Then thereexists an ℓ such that l > ℓ implies H ( l, t ) ∩ D ( αk, p k ) = ∅ by Fact 6.16.4. t ∈ h g [ p, H ( kσ α , q k σ )] ∩ H ( m , jz ) i − E ( αk, p k ). Then, by the definition of H ( kσ α , q k σ ), there exists an H d ( v α , jz lv ) ⊂ H ( kσ α , jz, q k σ ) such that t ∈ g [ p, ∪ H d ( v α , jz lv )] − E ( αk, p k ) . Then there exists an H ( v α , s ′ ) ∈ H d ( v α , jz lv ) such that t ∈ g [ p, H ( v α , s ′ )] − E ( αk, p k ). Note H ( v α , s ′ ) is closed in ( Y, ρ ). Then there exists ℓ > v α suchthat H ( h, t ) ∩ g [ q, H ( v α , s ′ )] = ∅ if h, q > ℓ by 1 of Claim 6.8. Note p > p k > m , t ∈ g [ p, ∪ H d ( v α , jz lv )] ∩ H ( m , jz ) ⊂ H ( m , jz lv ) and H ( m , jz lv ) ∩ H ( m , jz nu ) = ∅ if l = n or v = u . Then H ( h, t ) ∩ g [ q, H ( kσ α , q k σ )] = ∅ . t ∈ H ( hσ α , q h σ ). Note H ( hσ α , q h σ ) ∩ g [ p, H ( kσ α , q k σ )] = ∅ . Then, inthe same way as the proof of Fact 6.17, there exists an ℓ such that H ( l, t ) ∩ H ( hσ α , q ∗ h σ ) = ∅ if q ∗ h , l > ℓ .Let ℓ, q > max { ℓ i : i ≤ } . Then, by 1 of Definition B, we have Cl τ g [ q, E ( αk, p k )] = g [ q, E ( αk, p k )] ∪ H ( hσ α , q h σ ) . Here q h = max { q, q k } and h = k + 1. Then H ( ℓ, t ) ∩ Cl τ g [ q, E ( αk, p k )] = ∅ . (cid:3) Proposition 7.4. Let U ( αℓk, p k x ) = g ( ℓ, x ) − E ( αk, p k ) . Then:1. ∩ p Cl τ g [ p, U ( αℓk, p k x )] = U ( αℓk, p k x ) .2. Let t ∈ Cl τ g [ p, U ( αℓk, p k x )] − U ( αℓk, p k x ) . Then there exists an ℓ such that H ( l, t ) ∩ Cl τ g [ q, U ( αℓk, p k x )] = ∅ if l, q > ℓ . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 45 Proof. Let U x = U ( αℓk, p k x ), p > p k and t ∈ Cl τ g [ p, U x ] − U x . Then t ∈ g [ p, U x ] − U x by Definition C.Let t ∈ H j = H ′ ( m, l j ) for some H ′ ( m, l j ) ∈ H ′ ( m, y ), H ( m, j +) = { H ( m, l i ) ∈ H ( m, y ) : i > j } , H + = ∪ H ( m, j +) , H ( m, j − ) = { H ( m, l j ) ∈ H ( m, y ) : i < j } and H − = ∪ H ( m, j − ) . Then H ( m, j − ) is finite. Let ℓ = m . Then g [ q, H + ] ∩ H j = ∅ if q > ℓ . Note ρ ( H − , H ′ ( m, l j )) = r > ℓ such that g [ q, H − ∩ U x ] ∩ H j = ∅ if q > ℓ . Note U x = U ( αℓk, p k x ) = g ( ℓ, x ) − E ( αk, p k ). Then H ′ ( m, l j ) ∩ U x = [ g ( ℓ, x ) ∩ H ′ ( m, l j )] − [ D ( αk, p k ) ∪ H ( kσ α , q k σ )] . Let G ( m , k j ) be a Ln cover of H ′ ( m, j ). From Q2 of Construction 6.1, take H ∗ ( m, k j ) = H ′ ( m, j ) − ∪ i Let g ( m , z ) ∈ G ∗ ( m , k j ) and s ∈ H ( m , jz n ) ∈ H ( m , jz ) with n < k . If s / ∈ ∪ i Let H ( m , jz n ) ∈ H ( m , jz ) with n < k . If p > p k ≥ p n , then g [ p, H ( m , jz n ) ∩ U x ] ⊂ U x . Proof. Note ∪ v H ( m , jz nv ) = H ( m , jz n ) by 1 of Claim 6.6. Take an H ( m , jz nv ).Case 1, v α < p n . Take H d ( v α , jz nv ) and H ( v α , t ) ∈ H d ( v α , jz nv ). Then we have ∅ 6 = H ( v α , t ) − g [ p n , H ( v α , t )] = H ( v α , − v α t ) ⊂ U x and g [ p n , H ( v α , − v α t )] ∩ g [ p n , H ( v α , t )] = ∅ by 8 of Proposition 3.3. ∗ AND BOSEN WANG ∗∗∗ Note G ( v α , jz nv ) is a Ln cover of H ( m , jz nv ). Let H d ( v α , jz nv ) = ∪ H d ( v α , jz nv ) and H ( m , − jz n< ) = ∪{ H ( m , jz nv ) − g [ p n , H d ( v α , jz nv )] : v α < p n } . Then H ( m , − jz n< ) ⊂ U x and g [ p n , H ( m , − jz n< )] ∩ g [ p n , H d ( v α , jz nv )] = ∅ for every v α < p n .Case 2, v α = p n . Then g [ p n , H ( v α , t )] = g ( v α , t ) for every H ( v α , t ) ∈ H d ( v α , jz nv ).Then g [ p n , ∪ H d ( v α , jz nv )] = [ ∪ G ( v α , jz nv )] ∩ G (5 , αz nv ). Then g [ p n , ∪ H d ( v α , jz nv )] ∩ H ( m , jz nv ) = H ( m , jz nv ) ∩ G (5 , αz nv ) . Let H ( m , − jz nv ) = H ( m , jz nv ) − G (5 , αz nv ). Note p n > m . Then g [ p n , H ( m , − jz nv )] ∩ g [ p n , G (5 , αz nv )] = ∅ . Case 3, v α > p n > m . Then we have H ( m , jz nv )] ∩ g [ p n , H d ( v α , jz nv )] = H ( m , jz nv )] ∩ G (5 , αz nv ) . Let H ( m , − jz nv ) = H ( m , jz nv ) − G (5 , αz nv ). Note p n > m . Then we have g [ p n , H ( m , − jz nv )] ∩ G (5 , αz nv ) = ∅ . Then g [ p n , H ( m , − jz nv )] ∩ g [ p n , G (5 , αz nv )] = ∅ . Let H ( m , − jz n ≥ ) = ∪{ H ( m , jz nv ) − g [ p n , H d ( v α , jz nv )] : v α ≥ p n } . Then H ( m , − jz n ≥ ) ⊂ U x and g [ p n , H ( m , − jz n ≥ )] ∩ g [ p n , H d ( v α , jz nv )] = ∅ for every v α ≥ p n .Let H ( m , − jz nσ ) = H ( m , − jz n< ) ∪ H ( m , − jz n ≥ ) . Then, for every v ∈ N , g [ p n , ∪ H d ( v α , jz nv )] ∩ g [ p n , H ( m , − jz nσ )] = ∅ . Then, for every s ∈ U x ∩ H ( m , jz n ) with n < k , we have g [ p, { s } ] ∩ H ( m , jz n ) ⊂ U x if p > p n . (cid:3) The proof of Proposition 7.4 is continued. Let G ( m , k j ) be a Ln cover of H ′ ( m, j ). Take H ∗ ( m, k j ) = H ′ ( m, j ) − ∪ i Let U x = U ( αℓk, p k x ) = g ( ℓ, x ) − E ( αk, p k ) . Then:1. Cl τ g [ l, g [ p, U x ]] = g [ p, U x ] if l ≥ p .2. ∩ l Cl τ g [ l, g [ p, U x ]] = g [ p, U x ] .Proof. Note g [ p, U x ] is a c.o set and g [ l, g [ p, U x ]] = g [ p, U x ] if l ≥ p . Then t / ∈ g [ p, U x ]implies H ( l, t ) ∩ g [ p, U x ] = ∅ for some l . Then we have Cl τ g [ l, g [ p, U x ]] = g [ p, U x ]by the definition C. (cid:3) Recall Proposition 3.3. Function G satisfies the following conditions:1 ∩ ℓ g ( ℓ, i, y ) = { y } .4 y ∈ g ( ℓ, i, x ) implies g ( ℓ, j, y ) ⊂ g ( ℓ, i, x ) for some j .5 g ( ℓ + 1 , j, x ) ⊂ g ( ℓ, i, x ).6 j > k implies H ( ℓ, k ) ∩ ( ∪ G ( ℓ, j )) = ∅ .6 ′ H ( ℓ, k, h ) ∩ g ( ℓ, j, l ) = ∅ and H ( ℓ, k, h ) = H ( ℓ, j, l ) imply k > j .7 Each G ( ℓ, i ) is c.o.D family.8 If g ( ℓ, i, l ) , g ( ℓ, j, e ) ∈ G ℓ with j > i , then g ( ℓ, i, l ) ∩ g ( ℓ, j, e ) = ∅ or g ( ℓ, j, e ) ⊂ g ( ℓ, i, l ). Theorem 1. a. ( Y, τ ) is a stratifiable space and G Y is a g -function of ( Y, τ ). Proof. To prove Theorem 1 we show the function G Y such that:2, x ∈ g ( ℓ, x ℓ ) ∈ G Y implies x ℓ → x and3, if H is closed and x / ∈ H , then x / ∈ Cl τ ( ∪{ g ( ℓ, x ′ ) ∈ G Y : x ′ ∈ H } ) for some ℓ . proof of 2. Let x ∈ Y . Let g ( ℓ, x ℓ ) = g ( ℓ, i ℓ , l ℓ ), x ℓ ∈ H ( ℓ, i ℓ , l ℓ ), x ∈ H ( ℓ, i ′ ℓ , l ′ ℓ ) = { q l } × ... × { q l ℓ } × J ( ℓ, i ′ ℓ , l ′ ℓ ) and S = { x ℓ : ℓ ∈ N } . Suppose that there exists an U [ α − lk, p k x ] with S − U [ α − lk, p k , x ] = S = { x n i : i ∈ N } . ∗ AND BOSEN WANG ∗∗∗ Let x ℓ i ∈ H ( ℓ i , i ℓ i , l ℓ i ) ⊂ H ( ℓ, j i , k i ) ⊂ g ( ℓ, j i , k i ) for i ∈ N , and let N = { j i : x ℓ i ∈ H ( ℓ i , i ℓ i , l ℓ i ) ⊂ H ( ℓ, j i , k i ) for i ∈ N } . Suppose that there exists a strictly increasing infinite subsequence of N , to say j < j < ... < j i < ... . Then x ℓ i ∈ g ( ℓ, j i , k i ) implies g ( ℓ i , x ℓ i ) ⊂ g ( ℓ, j i , k i ) by4 of Proposition 3.3. Note x ∈ H ( ℓ, i ′ ℓ , l ′ ℓ ). Then j i → + ∞ implies j i > i ′ ℓ when i > n for some n . Then, by 6 of Proposition 3.3, g ( ℓ, j i , k i ) ∩ H ( ℓ, i ′ ℓ , l ′ ℓ ) = ∅ , acontradiction to x ∈ g ( ℓ i , x ℓ i ) ⊂ g ( ℓ, j i , k i ).So we may assume j = j = j = ... = j i = ... and g ( ℓ, j i , k i ) = g ( ℓ, j, k i ). Let G ( ℓ, j ) = { g ( ℓ, j, k i ) : i ∈ N } . Then G ( ℓ, j ) is a c.o.D family in ( Y, ρ ) by 7 ofProposition 3.3. Note x ∈ g ( ℓ i , x ℓ i ) ⊂ g ( ℓ, j, k i ) for each i . Then it is easy to see k = k = ... = k i = ... . So H ( ℓ, j i , k i ) = H ( ℓ, j, k ) for each i .Suppose H ( ℓ, i ′ ℓ , l ′ ℓ ) = H ( ℓ, j, k ) = { q j } × ... × { q j ℓ } × J ( ℓ, j, k ). Note the defini-tion of H ( ℓ, i ′ ℓ , l ′ ℓ ). Then there exists an e ≤ ℓ with q j e = q l e .In fact, suppose q j e = q l e for each e ≤ ℓ . Then H ( ℓ, j, k ) = H ( ℓ, i ′ ℓ , l ′ ℓ ) implies J ( ℓ, j, k ) ∩ J ( ℓ, i ′ ℓ , l ′ ℓ ) = ∅ . Then g ( ℓ, j, k ) ∩ g ( ℓ, i ′ ℓ , l ′ ℓ ) = ∅ , a contradiction to x ∈ g ( ℓ i , x ℓ i ) ⊂ g ( ℓ, j, k ).So H ( ℓ, i ′ ℓ , l ′ ℓ ) = H ( ℓ, j, k ) implies q j ℓ = q l ℓ for some e ≤ ℓ . Note x ℓ i ∈ H ( ℓ, j, k ),and ∩ i g ( ℓ i , x ℓ i ) = { x } since x ∈ g ( ℓ i , x ℓ i ). Then x ℓ i converges to x in ( Y, ρ ),and x ∈ H ( ℓ, j, k ) since H ( ℓ, j, k ) is closed in ( Y, ρ ). Note x ∈ H ( ℓ, i ′ ℓ , l ′ ℓ ). Then H ( ℓ, i ′ ℓ , l ′ ℓ ) = H ( ℓ, j, k ) by 1 of Proposition 3.1, a contradiction to supposition H ( ℓ, i ′ ℓ , l ′ ℓ ) = H ( ℓ, j, k ).This implies H ( ℓ, j, k ) = H ( ℓ, i ′ ℓ , l ′ ℓ ). So S ⊂ H ( ℓ, j, k ) = H ( ℓ, i ′ ℓ , l ′ ℓ ), a con-tradiction to S ∩ U [ α − lk, p k x ] = ∅ and H ( ℓ, i ′ ℓ , l ′ ℓ ) ⊂ U [ α − lk, p k x ]. So S − U [ α − , l, p, h, x ] is finite for each U [ α − lk, p k x ]. So x ℓ → x . proof of 3. Let H be closed in ( Y, τ ). Then there exists an O ∈ τ and a family O = { U λ : λ ∈ Λ } ⊂ U such that O = ∪ O and H = Y − O = Y − ∪ O = Y − ∪{ U λ : λ ∈ Λ } . Note U λ ∈ U . Then U λ = ∩{ U iλ ∈ U ( x λ ) : i ≤ n ( λ ) } . Let U iλ = Y − H iλ . Then U λ = Y − ∪ i ≤ n ( λ ) H iλ . Then H = Y −∪{ U λ : λ ∈ Λ } = Y −∪{ Y −∪ i ≤ n ( λ ) H iλ : λ ∈ Λ } = ∩{∪ i ≤ n ( λ ) H iλ : λ ∈ Λ } . Let s / ∈ H . Then there exists an ∪ i ≤ n ( λ ) H iλ such that s / ∈ ∪ i ≤ n ( λ ) H iλ ⊃ H .Note s / ∈ ∪ i ≤ n ( λ ) H iλ . Then s / ∈ H iλ for each i ≤ n ( λ ). Then, by Proposition7.3-7.5, there exists an ℓ i such that H ( l i , s ) ∩ Cl τ g [ q i , H iλ ] = ∅ if l i , q i > ℓ i . Let ℓ > max { q i , l i : i ≤ n ( λ ) } . Then H ( l, s ) ∩ [ ∪ i ≤ n ( λ ) Cl τ g [ q, H iλ ]] = ∅ if l, q > ℓ . Note ∪ i ≤ n ( λ ) Cl τ g [ q, H iλ ] = Cl τ g [ q, ∪ i ≤ n ( λ ) H iλ ] . Then H ( l, s ) ∩ Cl τ g [ q, ∪ i ≤ n ( λ ) H iλ ] = ∅ if l, q > ℓ . Then H ( l, s ) ∩ Cl τ g [ q, H ] = ∅ if l, q > ℓ . (cid:3) Properties of neighborhoods of stratifiable space ( Y, τ ) Proposition 8.1. Let y ∈ H , y ′ ∈ U [ α − bk ′ , p ′ k y ] = ∩ i ≤ n U ( α i b i k i , p k i y ) ∈ U y , U [ α − bk ′ , p ′ k y ] = g ( b, y ) − E [ α − k ′ , p ′ k ] and U ( δbk, p k y ′ ) = g ( b, y ′ ) − E ( δk, p k ) with δ = α i for i ≤ n . Then there exists an H ( l, t ) ⊂ E ( δk, p k ) and H ( l, t ) ⊂ U [ α − bk ′ , p ′ k y ] . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 49 Proof. Let α − = { α i : i ≤ n } . Note E [ α − k ′ , p ′ k ] ∩ H = ∅ for α ∈ A . A. Suppose y = y ′ . Then there exists an l with g ( l, y ′ ) ∩ g ( l, y ) = ∅ . Then wemay assume y ′ = y ∈ H . B. Take g ( l, y ). Then there exists c such that H ( m, ℓ, h ) ∩ g c ( l, y ) = ∅ and ℓ > c imply g ( m, ℓ, h ) ⊂ g c ( l, y ) by 1 of Corollary 5.3. Let 1 j be the least number suchthat 1 j > c and H ′ ( m, j ) ∩ g c ( l, y ) = ∅ .Let G ( m , j ) be an Ln cover of H ′ ( m, j ) and G ∗ ( m , y j ) = { g ( m , x ) ∈ G ( m , j ) : g ( m , x ) ⊂ g c ( l, y ) } . Take G d ( nσ α , jz, p ) = ∪ G d ( nσ α , jz, pσ ) from Fact 6.11. Then, by Fact 6.11, wehave the following fact: Fact B**. ∅ 6 = G ∗ ( m , y j ) ⊂ G ∗ ( m , j ).2. H d ( nσ α , j , pσ ) | H ( m , z ) = H d ( nσ α , jz, pσ ) for every g ( m , z ) ∈ G ∗ ( m , y j ).3. H ( m , z ) − G d ( nσ α , jz, p ) ⊂ g c ( l, y ) and G d ( nσ α , jz, p ) ⊂ g c ( l, y ) for every g ( m , z ) ∈ G ∗ ( m , y j ). (cid:3) Let k > n ∗ = max { k i : i ≤ n } . If H ( l, t ) ⊂ H ( kσ δ , q k σ ) ∩ H ( m , jz k ) ⊂ H ( m , z )for g ( m , z ) ∈ G ∗ ( m , y j ), then H ( l, t ) ∩ E [ α − k ′ , p ′ k ] = ∅ since E [ α − k ′ , p ′ k ] ∩ H ( m , z ) ⊂ ∪ i ≤ n H ( m , jz i ) by 2 of Fact B ∗∗ for α − . So, without loss of generality,we may assume k ≤ min { k i : i ≤ n } . C. Take α − = { α i : i ≤ n } .1. Let e be the least number such that i > e implies that { α h ( i ) : h ≤ n } isdifferent each other. We may assume α ( i ) < ... < α b ( i ) = β ( i ) < δ ( i ) < γ ( i ) = α b +1 ( i ) < ... < α n ( i ) . 2. Take H d ( m , θt i ), H d ( m , θt i ) and H d ( m , θt i ) from ℜ 2. Let H − z ( m , θt i ) = { H ( m , t ijl ) ∈ H ( m , θt ij ) : l ≤ θ ( i ) for j ≤ θ ( i ) } and H − y ( m , θt i ) = { H ( m , t ijl ) ∈ H ( m , θt ij ) : l > θ ( i ) for j > θ ( i ) } . Note γ ( i ) = α b +1 ( i ) < ... < α n ( i ). Then H − z ( m , γt i ) ⊂ ... ⊂ H − z ( m , α n t i ) and H − z ( m , γt i ) = ∩{ H − z ( m , α j t i ) : b + 1 ≤ j ≤ n } . Note α ( i ) < ... < α b ( i ) = β ( i ). Then H − y ( m , α t i ) ⊃ ... ⊃ H − y ( m , βt i ) and H − y ( m , βt i ) = ∩{ H − y ( m , α j t i ) : 1 ≤ j ≤ b } . Take H d ( m , δt i ) from ℜ 2. Let H ( m δ, − βγ, t i ) = H d ( m , δt i ) ∩ [ H − y ( m , βt i ) ∩ H − z ( m , γt i )] . Then H ( m δ, − βγ, t i ) = { H ( m , t ijl ) : β ( i ) < l ≤ δ ( i ) if δ ( i ) < j ≤ γ ( i ) , or δ ( i ) < l ≤ γ ( i ) if β ( i ) < j ≤ δ ( i ) } . 3. We translate H ( m δ, − βγ, t i ) into H ( m δ, − βγ, z kv ).To do it take H ( m , x v ) and G ( m , x v ) for v = i from ℜ 2. Let x v = z kv andtake H ( m , jz kv ) from Claim 6.6 for v > e . Here e is in the above 1 of C. Let H ( m δ, − βγ, z kv ) = { H ( m , t vjl ) ∈ H ( m , x v ) : β ( v ) < l ≤ δ ( v ) if δ ( v ) < j ≤ γ ( v ) , or δ ( v ) < l ≤ γ ( v ) if β ( v ) < j ≤ δ ( v ) } . ∗ AND BOSEN WANG ∗∗∗ Note H ( m , t vjl ) = H ( m, j ′ z ) ∩ g ( m , t vjl ) for H ( m , t vjl ) ∈ H ( m δ, − βγ, z kv ). Let G ( m δ, − βγ, z kv ) = { g ( m , t vjl ) ∈ G ( m , x v ) : H ( m , t vjl ) ∈ H ( m δ, − βγ, z kv ) } . 4. Take H ( m , jz kv ) and G ∗ d ( v α , jz kv ) from B1 of Construction 6.1. Let H ( m δ, − βγ, q k z ) = ∪{ H ( m δ, − βγ, z kv ) : z kv ∈ H ( m , jz ) , v > e and v > q k } and H ( m δ, − βγ, q k z ) = ∪ H ( m δ, − βγ, q k z ) . Then, for j ≤ n , H ( m δ, − βγ, q k z ) ∩ E ( α j k j , p k j ) = ∅ . Then, by the definition of H ( m δ, − βγ, z kv ) , we have H ( m δ, − βγ, q k z ) ∩ [ ∪ H d ( v δ , jz kv )] ⊂ H ( kσ δ , q k σ ) ⊂ E ( δk, p k ) . Claim 8.2. Let G ( m δ, − βγ, q k ) = ∪{ G ( m δ, − βγ, q k z ) : g ( m , z ) ∈ G ∗ ( m , j ) and j ∈ N } and H ( m δ, − βγ, q k ) = ∪{ H ( m δ, − βγ, q k z ) : g ( m , z ) ∈ G ∗ ( m , j ) and j ∈ N } . Then: 1. G ( m δ, − βγ, q k z ) is a Ln family,2. For every H ( m , t vjl ) ∈ H ( m δ, − βγ, q k ) , there uniquely exists a g ( m , t vjl ) ∈ G ( m δ, − βγ, q k ) with H ( m , t vjl ) = g ( m , t vjl ) ∩ H ( m , j ′ z ) ,3. H ( m δ, − βγ, q k ) ∩ E ( α j k j , p k j ) = ∅ for each α j ∈ α − . Here H ( m δ, − βγ, q k ) = ∪ H ( m δ, − βγ, q k ) .4. H ( m δ, − βγ, q k ) ∩ [ ∪ H d ( v δ , jz kv )] ⊂ H ( kσ δ , q k σ ) ⊂ E ( δk, p k ) .The proof is continued. Let H ( v δ , t ) ∈ H d ( v δ , jz kv ) | H ( m δ, − βγ, q k ). Then H ( v δ , t ) ⊂ E ( δk, p k ), H ( v δ , t ) ⊂ g c ( l, y ) and H ( v δ , t ) ∩ E ( α j k j , p k j ) = ∅ . Then H ( v δ , t ) ⊂ g c ( l, y ) − E ( α j k j , p k j ) for j ≤ n . (cid:3) Let A be a mad family on N . Take an α ∈ A . Let k ≥ 1. Then there exists afamily E ( α, k ) = { E ( αk, p ) : p > m } by Definition B. Let E ( α ) = ∪ k E ( α, k ).Let V y = { V λ : λ ∈ Λ } be an arbitrary neighborhoods base of y ∈ H in ( Y, τ ).Then V λ is open in ( Y, τ ) for every V λ ∈ V y by the definition of neighborhoods basebefore Proposition 7.2. Let b > m , y ∈ H , p > m , c on ( g ( b, y ) , m ) and U ( y, , c ) = { U ( αb , p y ) = g c ( b, y ) − E ( α , p ) ∈ U ( y ) : α ∈ A } . Take U y from Proposition 7.2. Then, for arbitrary U ( αb , p y ) ∈ U ( y, , c ), thereexists a V α ∈ V y such that V α ⊂ U ( αb , p y ) by the definition of V y . Then thereexists an U [ α − bk ′ , p ′ k y ] ∈ U y such that U [ α − bk ′ , p ′ k y ] ⊂ V α by the definition of U y . Let U [ α − bk ′ , p ′ k y ] = ∩ i ≤ n U ( α i b i k i , p k i y ) ∈ U y , α − = { α i : i ≤ n } and E [ α − k ′ , p ′ k ] = ∪ i ≤ n E ( α i k i , p k i ).Take H d ( kσ α , q k σ ) from B2, and take H ( m α, − βγ, q k z ) from Claim 8.2 for g ( m , z ) ∈ G ∗ ( m , j ). Let H ( m σ α , − βγ, q k z ) = H d ( kσ α , q k σ ) | H ( m α, − βγ, q k z ) and H ( m σ α , − βγ, q k z ) = ∪ H ( m σ α , − βγ, q k z ) . Call V α fine if there exists an n ≤ k and a full g ( m , z ) ∈ G ∗ ( m , j ) such that H ( m α, − βγ, q n z ) − H ( m σ α , − βγ, q n z ) ⊂ U [ α − bn ′ , p ′ n s ] ⊂ V α ⊂ U ( αb , p y ) and H ( m σ α , − βγ, q n z ) ∩ Cl τ V α = ∅ for V α ∈ V y . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 51 Proposition 8.3. Let y ∈ H , U [ α − bk ′ , p ′ k y ] = ∩ i ≤ n U ( α i b i k i , p k i y ) ∈ U y , V α ∈ V y and U ( αb , p y ) ∈ U ( y, , c ) such that U [ α − bk ′ , p ′ k y ] ⊂ V α ⊂ U ( αb , p y ) ⊂ g c ( b, y ) . Then V α is fine.Proof. We prove the proposition by induction on k for U [ α − bk ′ , p ′ k y ]. To do itnote U [ α − bk ′ , p ′ k y ] = g c ( b, y ) − E [ α − k ′ , p ′ k ] ⊂ V α ⊂ U ( αb , p y ) since y ∈ H and H ∩ E ( βh, p h ) = ∅ for every β ∈ A and every h ≥ 2. Let α − = { α i : i ≤ n } and U [ α − bk ′ , p ′ k y ] = ∩ i ≤ n U ( α i b i k i , p k i y ). Then α ∈ α − by Proposition 8.1. Then α = α a and E ( αk, p k a ) ⊂ E [ α − k ′ , p ′ k ] = ∪ i ≤ n E ( α i k i , p k i ). Then we may assume E ( αk, p k ) ⊂ E [ α − k ′ , p ′ k ].In fact, if p k a ≤ p k , then E ( αk, p k ) ⊂ E ( αk, p k a ) ⊂ E [ α − k ′ , p ′ k ].Let p k < p k a . Then E ( αk, p k a ) ⊂ E ( αk, p k ) and U ( αbk, p k y ) ⊂ U ( αbk, p k a ).Then U [ α − bk ′ , p ′ k y ] ∩ U ( αbk, p k y ) ⊂ U [ α − bk ′ , p ′ k y ] ⊂ V α ⊂ U ( αb , p y ) . we de-note E [ α − k ′ , p ′ k ] ∪ E ( αk, p k ) by E [ α − k ′ , p ′ k ], and U [ α − bk ′ , p ′ k y ] ∩ U ( αbk, p k y ) by U [ α − bk ′ , p ′ k y ] still. Then E ( αk, p k ) ⊂ E [ α − k ′ , p ′ k ]. Then we have the followingfact. Fact 1. There exists an U [ α − bk ′ , p ′ k y ] ∈ U y such that E ( αk, p k ) ⊂ E [ α − k ′ , p ′ k ], α ∈ α − and U [ α − bk ′ , p ′ k y ] = g c ( b, y ) − E [ α − k ′ , p ′ k ] ⊂ V α ⊂ U ( αb , p y ) ∈ U ( y, , c ).A. k = 2 and U [ α − b ′ , p ′ y ] = g c ( b, y ) − E [ α − ′ , p ′ ] ⊂ V α ⊂ U ( αb , p y ).Then E ( α , p ) ⊂ E [ α − ′ , p ′ ] by Fact 1. Note V α ⊂ U ( αb , p y ). Then we have E ( α , p ) ∩ U ( αb , p y ) = ∅ . Then E ( α , p ) ∩ Cl τ V α = ∅ since E ( α , p ) is a c.oset by 1 of Definition B. Then, by 4 of Claim 8.2, H ( m σ α , − βγ, q z ) ∩ Cl τ V α = ∅ since H ( m σ α , − βγ, q z ) ⊂ E ( α , p ). Note H ( m , z ) − E ( α , p ) ⊂ U ( αb , p y ) bythe definition of U ( αb , p y ) for every H ′ ( m, j ) ∈ H ′ ( m, y ) and every g ( m , z ) ∈ G ∗ ( m , j ) with g ( m , z ) ⊂ g c ( b, y ). Then, by 3 and 4 of Claim 8.2, we have H = H ( m α, − βγ, q z ) − H ( m σ α , − βγ, q z ) = H ( m α, − βγ, q z ) − E ( α , p ) and H = H ( m α, − βγ, q z ) − E [ α − ′ , p ′ ] ⊂ U [ α − b ′ , p ′ y ] ⊂ V α . Then V α is fine if U [ α − b ′ , p ′ y ] ⊂ V α ⊂ U ( αb , p y ) with α ∈ α − .B. k = 3 and U [ α − b ′ , p ′ y ] = g c ( b, y ) − E [ α − ′ , p ′ ] ⊂ V α ⊂ U ( αb , p y ).Then α ∈ α − and E ( α , p ) ⊂ E [ α − ′ , p ′ ] by Fact 1.Case 1, there exists an H ′ ( m, j ) ∈ H ′ ( m, y ) and a full g ( m , z ) ∈ G ∗ ( m , j )such that H ( m σ α , − βγ, q z ) ∩ Cl τ V α = ∅ . Then we can prove that V α is fine.In fact, by 3 and 4 of Claim 8.2, we have H = H ( m α, − βγ, q z ) − H ( m σ α , − βγ, q z ) = H ( m α, − βγ, q z ) − E ( α , p ) and H = H ( m α, − βγ, q z ) − E [ α − ′ , p ′ ] ⊂ U [ α − b ′ , p ′ y ] ⊂ V α . Case 2, for every H ′ ( m, j ) ∈ H ′ ( m, y ) and every full g ( m , z ) ∈ G ∗ ( m , j ),we have H ( m σ α , − βγ, q z ) ∩ Cl τ V α = ∅ . Let δ ∈ α − with δ = α . Then H ( m σ α , − βγ, q z ) ∩ E ( δ , p ) = ∅ by 3 ofClaim 8.2 for every H ′ ( m, j ) ∈ H ′ ( m, y ) and every g ( m , z ) ∈ G ∗ ( m , j ). Then E ( α , p ) is the unique open set with H ( m σ α , − βγ, q z ) ⊂ E ( α , p ) in α − . Take D ( α , p ) = g [ p , D (1 σ α , G )] ∪ g [ p , H (2 σ α , q σ )]. Then, by the definition of ∗ AND BOSEN WANG ∗∗∗ E ( α , p ), g [ p , D (1 σ α , G )] ⊂ E ( α , p ). Note g [ p , H (2 σ α , q σ )] − E ( α , p ) isopen and dense in E ( α , p ) − E ( α , p ) by the definition of E ( α , p ). Then H ( m σ α , − βγ, q z ) ∩ Cl τ (cid:2) V α ∩ g [ p , H (2 σ α , q σ )] (cid:3) = ∅ . Let V α ( p ) = V α ∩ g [ p , H (2 σ α , q σ )] and x ∈ V α ( p ). Then there exists an openneighborhood U [ β − ℓk ′ , p ′ k x ] such that U [ β − ℓk ′ , p ′ k x ] ⊂ V α ( p ) because both V α and g [ p , H (2 σ α , q σ )] are open sets in ( Y, τ ). Then we have the following fact. Fact B.1. U [ β − ℓk ′ , p ′ k x ] ⊂ V α ( p ) ⊂ V α for some U [ β − ℓk ′ , p ′ k x ] ∈ U x .Note p > m and x ∈ V α ( p ) = V α ∩ g [ p , H (2 σ α , q σ )] ⊂ g [ p , H (2 σ α , q σ )].Then x ∈ E ( α , p ) or x ∈ g [ p , H (2 σ α , q σ )] − E ( α , p ).Note x ∈ V α ( p ) ⊂ V α ⊂ U ( αb , p y ) = g c ( b, y ) − E ( α , p ). Then x ∈ g [ p , H (2 σ α , q σ )] − E ( α , p ). Then x ∈ g ( ℓ, x ) − E ( α , p ) for some ℓ ∈ N with g ( ℓ, x ) ⊂ g c ( b, y ). Note 1 of Proposition 7.2. We may let E ∩ [ θh, p h ] = g ( ℓ, x ) ∩ \ i ≤ n E ( θ i h i , p ih ) , E [ β − ′ , p ′ ] = ∪ i ≤ n E ( β i i , p i ) and U [ β − ℓk ′ , p ′ k x ] = U [ β − ℓ ′ , p ′ x ] = E ∩ [ θh, p h ] ∩ g ( ℓ, x ) ∩ (cid:2) g c ( b, y ) − E [ β − ′ , p ′ ] (cid:3) . Note U [ β − ℓ ′ , p ′ x ] ⊂ U ( αb , p y ) = g c ( b, y ) − E ( α , p ). Then we have U [ β − ℓ ′ , p ′ x ] = U [ β − ℓ ′ , p ′ x ] ∩ U ( αb , p y ) and U [ β − ℓ ′ , p ′ x ] = E ∩ [ θh, p h ] ∩ g ( ℓ, x ) ∩ (cid:2) g c ( b, y ) − E [ β − ′ , p ′ ] (cid:3) ∩ [ g c ( b, y ) − E ( α , p )] . Then U [ β − ℓ ′ , p ′ x ] = E ∩ [ θh, p h ] ∩ g ( ℓ, x ) ∩ h g c ( b, y ) − (cid:2) E [ β − ′ , p ′ ] ∪ E ( α , p ) (cid:3)i .Let β − = { β i : i ≤ n } with α ∈ β − . Denote E [ β − ′ , p ′ ] ∪ E ( α , p ) by E [ β − ′ , p ′ ]still. Then we have proved the following fact. Fact B.2. α ∈ β − , U [ β − ℓ ′ , p ′ x ] = E ∩ [ θh, p h ] ∩ g ( ℓ, x ) ∩ (cid:2) g c ( b, y ) − E [ β − ′ , p ′ ] (cid:3) and E ( α , p ) ⊂ E [ β − ′ , p ′ ].In order to prove that V α is fine in case 2 if U [ β − ℓ ′ , p ′ x ] ⊂ V α ⊂ U ( αb , p y )and α ∈ β − , we give the following decompositions. Decomposition B.3. Decompose E ∩ [ θh, p h ].To do it note x ∈ U [ β − ℓ ′ , p ′ x ] ⊂ V α ( p ) = V α ∩ g [ p , H (2 σ α , q σ )].I. Note the definition of H ( kσ α , q k σ ). Then we have x / ∈ H (3 σ α , q σ ) since g [ p , H (2 σ α , q σ )] ∩ H (3 σ α , q σ ) = ∅ . Then x / ∈ H ( nσ α , q n σ ) since H ( nσ α , q n σ ) ∩ D ( nσ α , q n σ ) = ∅ and g [ p , H (2 σ α , q σ )] ⊂ D ( nσ α , q n σ ). Note x ∈ ∩ i ≤ n E ( θ i h i , p ih ).Then x ∈ H (2 σ θ i , q σ ) for n ′ < i ≤ n , and x ∈ D ( θ i h i , p ih ) for i ≤ n ′ ≤ n . Note x ∈ D ( θ i h i , p ih ) implies that there exists a g ( ℓ , x ) ⊂ D ( θ i h i , p ih ) ⊂ E ( θ i h i , p ih ) if ℓ = max { v θ i : i ≤ n ′ and x ∈ H ( m , j ′ z v ) } . Here H ( m , j ′ z v ) is defined in B2of Construction 6.1. Then we may assume x ∈ E ′∩ [ θh, p h ] = g ( ℓ , x ) ∩ \ n ′
Decompose U [ β − b ′ , p ′ y ] = g c ( b, y ) − E [ β − ′ , p ′ ].Note g ( ℓ s , s ) ⊂ E ∩ [ θh, p h ] and U [ β − ℓ ′ , p ′ x ] = E ∩ [ θh, p h ] ∩ h g ( ℓ, x ) − E [ β − ′ , p ′ ] i .Let U [ β − ℓ s ′ , p ′ s ] = g ( ℓ s , s ) − E [ β − ′ , p ′ ]. Then H ( ℓ s , s ) ⊂ U [ β − ℓ s ′ , p ′ s ] and g ( ℓ s , s ) ∩ U [ β − b ′ , p ′ y ] = g ( ℓ s , s ) ∩ U [ β − ℓ ′ , p ′ x ] = g ( ℓ s , s ) − E [ β − ′ , p ′ ] . Then U [ β − ℓ s ′ , p ′ s ] ⊂ U [ β − ℓ ′ , p ′ x ] ⊂ V α ( p ). (cid:3) The proof of Case 2 is continued. Note V α ⊂ U ( αb , p y ) = g c ( b, y ) − E ( α , p ).Then V α ∩ E ( α , p ) = ∅ . Then E ( α , p ) ∩ Cl τ V α = ∅ since E ( α , p ) is a c.o set.Let W β = U [ β − b ′ , p ′ y ]. Then E ( α , p ) ⊂ E [ β − ′ , p ′ ] and U [ β − b ′ , p ′ y ] ⊂ W β ⊂ U ( αb , p y ) . Take H ( m , z ) ∈ H ∗ ( m , s ′ ℓ ) since ∅ 6 = ∪ G ∗ ( m, s ′ ℓ ) ⊂ g c ( ℓ s , s ). Then, by 3 ofClaim 8.2, we have H ( m α, − βγ, q z ) ⊂ H ( m , z ), H = H ( m α, − βγ, q z ) − H ( m σ α , − βγ, q z ) = H ( m α, − βγ, q z ) − E [ β − ′ , p ′ ] . Note H ( m σ α , − βγ, q z ) ⊂ E ( α , p ) by 4 of Claim 8.2. Then H ( m σ α , − βγ, q z ) ∩ Cl τ V α = ∅ . Then H ( m , z ) ⊂ E ∩ [ θh, p h ] and H ( m α, − βγ, q z ) ⊂ H ( m , z ) ⊂ g c ( ℓ s , s ). Then H ( m α, − βγ, q z ) − E [ β − ′ , p ′ ] ⊂ g ( ℓ s , s ) − E [ β − ′ , p ′ ] = U [ β − ℓ s ′ , p ′ s ]. Then H ⊂ U [ β − ℓ s ′ , p ′ s ] = E ∩ [ θh, p h ] ∩ (cid:2) g ( ℓ s , s ) − E [ β − ′ , p ′ ] (cid:3) ⊂ U [ β − ℓ ′ , p ′ x ] ⊂ V α . This completes a proof of Case 2.Summarizing the above Case 1 and Case 2, we have the following Fact. Fact B.5. Let U [ α − b ′ , p ′ y ] = g c ( b, y ) − E [ α − ′ , p ′ ] ⊂ V α ⊂ U ( αb , p y ) with E ( α , p ) ⊂ E [ α − ′ , p ′ ] and α ∈ α − . Then the following 1 or 2 holds. ∗ AND BOSEN WANG ∗∗∗ 1. There exists an H ′ ( m, j ) ∈ H ′ ( m, y ) and a full g ( m , z ) ∈ G ∗ ( m , j ) suchthat H ( m σ α , − βγ, q z ) ⊂ H ( m , z ), H ( m σ α , − βγ, q z ) ∩ Cl τ V α = ∅ ,H = H ( m α, − βγ, q z ) − H ( m σ α , − βγ, q z ) = H ( m α, − βγ, q z ) − E ( α , p ) and H = H ( m α, − βγ, q z ) − E [ β − ′ i , p ′ ] ⊂ U [ α − b ′ , p ′ y ] ⊂ V α . 2. If there exists U [ β − ℓ ′ , p ′ x ] = E ∩ [ θh, p h ] ∩ g ( ℓ, x ) ∩ (cid:2) g c ( b, y ) − E [ β − ′ , p ′ ] (cid:3) with U [ β − ℓ ′ , p ′ x ] ⊂ V α ⊂ U ( αb , p y ) and E ( α , p ) = E ( α a a , p a ) ⊂ E [ β − ′ , p ′ ],then there exists U [ β − ℓ s ′ , p ′ s ] = g ( ℓ s , s ) − E [ β − ′ , p ′ ] ⊂ U [ β − ℓ ′ , p ′ x ] and exists H ( m , z ) ∈ H ∗ ( m , s ′ ℓ ) such that H = H ( m α, − βγ, q z ) − H ( m σ α , − βγ, q z ) = H ( m α, − βγ, q z ) − E [ β − ′ , p ′ ] ,H ( m σ α , − βγ, q z ) ∩ Cl τ V α = ∅ and H ⊂ U [ β − ℓ ′ , p ′ x ] ⊂ V α . The proof of B is continued. If k = 3 and U [ α − b ′ , p ′ y ] ⊂ V α ⊂ U ( αb , p y ),then V α is fine by Fact B.5.C. Assume, for k = h = l + 1, we have proved the following fact. Fact B.h. Let U [ α − bh ′ , p ′ h y ] = g c ( b, y ) − E [ α − h ′ , p ′ h ] ⊂ V α ⊂ U ( αb , p y ) with α ∈ α − and E ( αh, p h ) ⊂ E [ α − h ′ , p ′ h ]. Then the following 1 or 2 holds.1. There exists an H ′ ( m, j ) ∈ H ′ ( m, y ) and a full g ( m , z ) ∈ G ∗ ( m , j ) suchthat H ( m σ α , − βγ, q h z ) ⊂ H ( m , z ), H ( m σ α , − βγ, q h z ) ∩ Cl τ V α = ∅ ,H = H ( m α, − βγ, q h z ) − H ( m σ α , − βγ, q h z ) = H ( m α, − βγ, q h z ) − E ( αh, p h ) and H = H ( m α, − βγ, q h z ) − E [ α − h ′ , p ′ h ] ⊂ U [ α − bh ′ , p ′ h y ] ⊂ V α . 2. If there exists an l < h and exists an U [ β − ℓl ′ , p ′ l x ] such that U [ β − ℓl ′ , p ′ l x ] = E ∩ [ θh, p h ] ∩ g ( ℓ, x ) ∩ (cid:2) g c ( b, y ) − E [ β − l ′ , p ′ l ] (cid:3) ⊂ V α ⊂ U ( αb , p y )and E ( αl, p l ) ⊂ E [ β − l ′ , p ′ l ], then there exists an U s ⊂ U [ β − ℓl ′ , p ′ l x ] such that H ( ℓ s , s ) ⊂ U s = g ( ℓ s , s ) − E [ β − l ′ , p ′ l ], and exists an H ( m , z ) ∈ H ∗ ( m , s ′ ℓ )such that H = H ( m α, − βγ, q l z ) − H ( m σ α , − βγ, q l z ) = H ( m α, − βγ, q l z ) − E [ β − l ′ , p ′ l ] ,H ( m σ α , − βγ, q l z ) ∩ Cl τ V α = ∅ and H ⊂ U [ β − ℓl ′ , p ′ l x ] ⊂ V α . C*. Let k = h +1 and U [ α − bk ′ , p ′ k y ] = g c ( b, y ) − E [ α − k ′ , p ′ k ] ⊂ V α ⊂ U ( αb , p y ) . Then α ∈ α − and E ( αk, p k ) ⊂ E [ α − k ′ , p ′ k ] by Fact 1.Case 1, there exists an H ′ ( m, j ) ∈ H ′ ( m, y ) and a full g ( m , z ) ∈ G ∗ ( m , j )such that H ( m σ α , − βγ, q k z ) ∩ Cl τ V α = ∅ . Then, by 3 and 4 of Claim 8.2, we have H = H ( m α, − βγ, q k z ) − H ( m σ α , − βγ, q k z ) = H ( m α, − βγ, q k z ) − E ( αk, p k ) and H = H ( m α, − βγ, q k z ) − E [ α − k ′ , p ′ k ] ⊂ U [ α − bk ′ , p ′ k y ] ⊂ V α . Then V α is fine. Otherwise we have the following case.Case 2, for every H ′ ( m, j ) ∈ H ′ ( m, y ) and every full g ( m , z ) ∈ G ∗ ( m , j ), H ( m σ α , − βγ, q k z ) ∩ Cl τ V α = ∅ . NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 55 Let δ ∈ α − with δ = α . Then H ( m σ α , − βγ, q k z ) ∩ E ( δk, p k ) = ∅ by 3 of Claim8.2. Then we can prove that g [ p k , H ( hσ α , q h σ )] − E ( αh, p h ) is open and dense in (cid:16) D ( αk, p k ) − E ( αh, p h ) , τ (cid:17) for the unique α ∈ α − .In fact, note H ( kσ α , q k σ ) ∩ E ( αh, p h ) = ∅ . Then H ( kσ α , q k σ ) ⊂ Y − E ( αh, p h ).Note the definitions of D ( αk, p k ) and E ( αh, p h ). Then H ( kσ α , q k σ ) ⊂ Cl τ [ D ( αk, p k ) − E ( αh, p h )] and D ( αk, p k ) − E ( αh, p h ) = g [ p k , H ( hσ α , q h σ )] − E ( αh, p h ) . Then we have H ( kσ α , q k σ ) ⊂ Cl τ [ D ( αk, p k ) − E ( αh, p h )] = Cl τ (cid:2) g [ p k , H ( hσ α , q h σ )] − E ( αh, p h ) (cid:3) . Then H ( m σ α , − βγ, q k z ) ∩ Cl τ h V α ∩ (cid:2) g [ p k , H ( hσ α , q h σ )] − E ( αh, p h ) (cid:3)i = ∅ . Let V α ( p k ) = V α ∩ (cid:2) g [ p k , H ( hσ α , q h σ )] − E ( αh, p h ) (cid:3) and x ∈ V α ( p k ) . Then, in the same way as the proof of Fact B.1, we have the following fact. Fact C.1. There exists an open neighborhood U [ β − ℓh ′ , p ′ h x ] ∈ U x such that U [ β − ℓh ′ , p ′ h x ] ⊂ V α ( p k ) ⊂ V α ⊂ U ( αb , p y ).Then, by 1 of Proposition 7.2, we have E ∩ [ θh ′ , p ′ h ] = g ( ℓ, x ) ∩ \ i ≤ n E ( θ i h i , p ih ) , E [ β − e ′ , e ′ h ] = ∪ i ≤ n E ( β i e ′ i , e ih ) and U [ β − ℓh ′ , p ′ h x ] = E ∩ [ θh ′ , p ′ h ] ∩ g ( ℓ, x ) ∩ (cid:2) g c ( b, y ) − E [ β − e ′ , e ′ h ] (cid:3) ⊂ g c ( b, y ) . Note U [ β − ℓh ′ , p ′ h x ] ∩ E ( αh, p h ) = ∅ since U [ β − ℓh ′ , p ′ h x ] ⊂ V α ( p k ). Then we have U [ β − ℓh ′ , p ′ h x ] ⊂ g c ( b, y ) − E ( αh, p h ) = U ( αbh, p h y ). Then U [ β − ℓh ′ , p ′ h x ] = U [ β − ℓh ′ , p ′ h x ] ∩ U ( αbh, p h y ) and U [ β − ℓh ′ , p ′ h x ] = E ∩ [ θh ′ , p ′ h ] ∩ g ( ℓ, x ) ∩ h g c ( b, y ) − (cid:2) E [ β − e ′ , e ′ h ] ∪ E ( αh, p h ) (cid:3)i . Let E [ β − h ′ , e ′ h ] = E [ β − e ′ , e ′ h ] ∪ E ( αh, p h ) and β − = { β i : i ≤ n } .Then we have the following fact in the same way as the proof of Fact B.2. Fact C.2. α ∈ β − , U [ β − ℓh ′ , p ′ h x ] = E ∩ [ θh ′ , p ′ h ] ∩ g ( ℓ, x ) ∩ (cid:2) g c ( b, y ) − E [ β − h ′ , e ′ h ] (cid:3) and E ( αh, p h ) ⊂ E [ β − h ′ , e ′ h ].Then, in the same way as the proof of Decomposition B.3, there exists g ( ℓ s , s )and H ∗ ( m , s ′ ℓ ) such that g ( ℓ s , s ) ⊂ g ( ℓ, x ) and ∅ 6 = ∪ G ∗ ( m, s ′ ℓ ) ⊂ g c ( ℓ s , s ) forevery g ( m , z ) ∈ G ∗ ( m , s ′ ℓ ). Then we have the following fact. Fact C.3. There exist g c s ( ℓ s , s ) and H ∗ ( m , s ′ ℓ ) such that g ( ℓ s , s ) ⊂ E ∩ [ θh ′ , p ′ h ]and H ( ℓ s , s ) ⊂ U [ β − ℓ s h ′ , e ′ h s ] = g ( ℓ s , s ) − E [ β − h ′ , e ′ h ] ⊂ U [ β − ℓh ′ , p ′ h x ] ⊂ V α ( p k ). C**. Take G ∗ ( m , h σ ) = ∪ j G ∗ ( m , h j ) from Q2 of Construction 6.1. Let G ∗ ( m , s, h σ ) = { g ( m , z ) ∈ G ∗ ( m , h σ ) : g ( m , z ) ⊂ g c s ( ℓ s , s ) } . Then ∅ 6 = G ∗ ( m , s ′ ℓ ) ⊂ G ∗ ( m , s, h σ ).Case 1, there exists a full g ( m , z ) ∈ G ∗ ( m , s, h σ ) such that H ( m σ α , − βγ, q h z ) ∩ Cl τ V α = ∅ . Then, by 3 and 4 of Claim 8.2, we have H = H ( m α, − βγ, q h z ) − H ( m σ α , − βγ, q h z ) = H ( m α, − βγ, q h z ) − E ( αh, p h ) and ∗ AND BOSEN WANG ∗∗∗ H = H ( m α, − βγ, q h z ) − E [ α − h ′ , p ′ h ] ⊂ U [ β − ℓ s e, p e s ] ⊂ V α . Then V α is fine.If Case 1 is not true, then we have the following case.Case 2, for every g ( m , z ) ∈ G ∗ ( m , s, h σ ), H ( m σ α , − βγ, q h z ) ∩ Cl τ V α = ∅ . Let δ ∈ α − with δ = α . Then H ( m σ α , − βγ, q h z ) ∩ E ( δh, p h ) = ∅ by 3 of Claim8.2. Then E ( αh, p h ) is the unique open set with H ( m σ α , − βγ, q h z ) ⊂ E ( αh, p h )and α ∈ α − . Note h = l + 1 by inductive assumption C. Take D ( αh, p h ) = g [ p , D (1 σ α , G )] ∪ g [ p , H (2 σ α , q σ )] ∪ ... ∪ g [ p h , H ( lσ α , q l σ )] . Then we have g [ p h , H ( lσ α , q l σ )] ⊂ E ( αh, p h ) by the definition of E ( αh, p h ). Note g [ p h , H ( lσ α , q l σ )] − E ( αl, p l ) is open and dense in E ( αh, p h ) − E ( αl, p l ). Then H ( m σ α , − βγ, q h z ) ∩ Cl τ h V α ∩ (cid:2) g [ p h , H ( lσ α , q l σ )] − E ( αl, p l ) (cid:3)i = ∅ . Let V α ( p h ) = V α ∩ (cid:2) g [ p h , H ( lσ α , q l σ )] − E ( αl, p l ) (cid:3) and x ∈ V α ( p h ). Then thereexists an open neighborhood U [ γ − ℓl ′ , p ′ l x ] ∈ U x such that U [ γ − ℓl ′ , p ′ l x ] ⊂ V α ( p h ).Then U [ γ − ℓl ′ , p ′ l x ] ⊂ V α ( p h ) ⊂ V α . Then we have the following fact. Fact C.4. There exists an open neighborhood U [ γ − ℓl ′ , p ′ l x ] ∈ U x such that U [ γ − ℓl ′ , p ′ l x ] ⊂ V α ( p h ) ⊂ V α ⊂ U ( αb , p y ).Then, in the same way as the proof of Fact B.2 and C.2, we have the followingfact. Fact C.5. α ∈ γ − , U [ γ − ℓl ′ , p ′ l x ] = E ∩ [ θl ′ , p ′ l ] ∩ g ( ℓ, x ) ∩ (cid:2) g c ( b, y ) − E [ γ − l ′ , p ′ l ] (cid:3) and E ( αl, p l ) ⊂ E [ γ − l ′ , p ′ l ].Then Fact C.5 satisfies the assumption of 2 of Fact B.h. Then V α is fine by theinductive assumption Fact B.h.Then, by inductive proof A, B and C, we complete a proof of the proposition. (cid:3) Proposition 8.4. Let U [ α − bk ′ , p ′ k y ] ⊂ V α ⊂ U ( αb , p y ) and g ( m , z ) ∈ G ∗ ( m , ℓ ) such that H ( m α, − βγ, q k z ) − H ( m σ α , − βγ, q k z ) ⊂ U [ α − bk ′ , p ′ k y ] ⊂ V α and H ( m σ α , − βγ, q k z ) ∩ Cl τ V α = ∅ . Then, for every H ( v α , t ) ∈ H ( m σ α , − βγ, q k z ) and every H ( v α , t ) ∈ H − ( kσ α , jz, q k σ ) | H ( v α , t ) with H ( v α , t ) ⊂ H ( m α, − βγ, q k z ) − H ( m σ α , − βγ, q k z ) ,H ( v α , t ) and H ( v α , t ) satisfy H ( v α , t ) ∩ Cl τ V α = ∅ , H ( v α , t ) ⊂ V α and H ( v α , t ) ∪ H ( v α , t ) ⊂ H ( v α , t ) ∈ H ( v α , jz kv ) . Proof. Let U [ α − bk ′ , p ′ k y ] ⊂ V α ⊂ U ( αb , p y ). Then, by Proposition 8.3, V α is fine.Then there exists an H ( m σ α , − βγ, q k z ) with H ( m σ α , − βγ, q k z ) ∩ Cl τ V α = ∅ and H ( m α, − βγ, q k z ) − H ( m σ α , − βγ, q k z ) ⊂ U [ α − bk ′ , p ′ k y ] ⊂ V α . Take an H ( v α , t ) ∈ H ( m σ α , − βγ, q k z ), and an H ( v α , t ) ∈ H − ( v α , jz kv ) | with H ( v α , t ) ⊂ H ( m α, − βγ, q k z ) − H ( m σ α , − βγ, q k z )and H ( v α , t ) ∪ H ( v α , t ) ⊂ H ( v α , t ) by the definition of H d ( v α , jz kv ). Then H ( v α , t ) ∈ H ( v α , jz kv ). Then H ( v α , t ) ∩ Cl τ V α = ∅ and H ( v α , t ) ⊂ V α . (cid:3) NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 57 stratifiable space ( X, τ ) in not M -spaces Recall A a mad family on N . α, β, δ and γ are used to denote members in A . Proposition 9.1. Let A = ∪ i A i . Then there exists an A i such that A i is un-bounded.Proof. Suppose that each A i is bounded. Then A is bounded, a contradiction. (cid:3) Take a α ∈ A . Let k > 1. Then there exists a family E ( α, k ) = { E ( αk, p k ) : p k > m } by 1 of Definition B. Let E ( α ) = ∪ k E ( α, k ).Take V y = { V λ : λ ∈ Λ } and U ( y, , c ) from Proposition 8.3. Let b > m , y ∈ H and p > m . Note U ( y, , c ) = { U ( αb , p y ) = g c ( b, y ) − E ( α , p ) ∈ U ( y ) : α ∈ A } . Take U y . Then, for every U ( αb , p y ) ∈ U ( y, , c ), there exists an U [ α − b α k, p k y ] ∈ U y and a V α ∈ V y such that U [ α − b α k ′ , p ′ k y ] = ∩ i ≤ n U ( α i b i k i , p k i y ) ⊂ V α ⊂ U ( αb , p y ) ⊂ g c ( b α , y )with α ∈ α − = { α i : i ≤ n } by Proposition 8.1. Then, by Proposition 8.3, V α isfine. Then there exists a full g ( m , z ) such that H ( m α, − βγ, q k z ) − H ( m σ α , − βγ, q k z ) ⊂ U [ α − b α k ′ , p ′ k y ] ⊂ V α ⊂ g c ( b α , y ) and H ( m σ α , − βγ, q k z ) ∩ Cl τ V α = ∅ . Denote H ( m , z ) by H ( m , zα ). Let H ( y, A ) = { H ( m , zα ) : α ∈ A } and U [ y, A ] = { U [ α − b α k ′ , p ′ k y ] : U [ α − b α k ′ , p ′ k y ] ⊂ V α ⊂ U ( αb , p y ) and α ∈ A } . Note H ( m , zα ) ∈ H Y is countable. Then there exists an H ( m , zα ) = H ( m , z )and a unbounded subfamily A ⊂ A such that U [ y, A ] = { U [ α − b α k ′ , p ′ k y ] : H ( m , zα ) = H ( m , z ) and α ∈ A } . Take an U [ α − b α k ′ , p ′ k y ] = ∩ i ≤ n U ( α i b i k i , p k i y ) ∈ U [ y, A ]. Then α ∈ α − . Thenthere exists an unbounded subfamily A ⊂ A such that | α − | = | β − | = n ′ , k α = k β = k and p k α = p k β = p k if α, β ∈ A . Let U [ y, A ] = { U [ α − b α k ′ , p ′ k y ] : | α − | = n ′ , k α = k, p k α = p k and α ∈ A } . Take the relative H ( y, A ) = { H ( m , zα ) = H ( m , z ) : α ∈ A } . Then H ( y, A ) = { H ( m , z ) } .Call U [ y, A ] an idea family. Summing up to the above result we have the following proposition. Proposition 9.2. Let V α be fine for every α ∈ A . Then there exists an idea family U [ y, A ] with the same full set H ( m , z ) . Let [ n , n , ..., n h ] = { α ∈ A : α ( i ) = n i for i = 1 , , ..., h } . Then, for arbitrary h ∈ N , A = ∪{ [ n , n , ..., n h ] : n , n , ..., n h ∈ N with n < n < ... < n h } . Let A ′ ⊂ A . Fix an h ∈ N and let A ′ h = { [ n , n , ..., n h ] ′ : [ n , n , ..., n h ] ′ ⊂ A ′ } .Then A ′ = ∪A ′ h . Proposition 9.3. If there exists b ∈ N such that, for every h > b and every [ n , n , ..., n h ] ′ ∈ A ′ h , there exists a k ∈ N with α ( h + 1) < k for every α ∈ [ n , n , ..., n h ] ′ , then A ′ is bounded. ∗ AND BOSEN WANG ∗∗∗ Proof. 1. Note A ′ b is countable. Take a [1 i , i , ..., b i ] ′ from A ′ b . Then there existsa k such that α ( b + 1) = 2 ′ j < k for each α ∈ [1 i , ..., b i ] ′ . Let β i ( b + 1) = k .Let A ′ b +1 [1 i , ..., b i ] = { [1 i , ..., b i , j ] ′ ∈ A ′ b +1 : [1 i , ..., b i , j ] ′ ⊂ [1 i , ..., b i ] ′ } . Then A ′ b +1 [1 i , ..., b i ] is finite. Take a [1 i , ..., b i , j ] ′ from A ′ b +1 [1 i , ..., b i ]. Then there existsa k j such that α ( b + 2) < k j for each α ∈ [1 i , ..., b i , j ] ′ . Let β i ( b + 2) = k =max { k j : j ≤ k } .2. Assume that A ′ k [1 i , ..., k l ] is finite and β i ( b + k ) = k k . Denote k + 1 by n andlet A ′ n [1 i , ..., k l ] = { [1 i , ..., k l , n j ] ′ ∈ A ′ k : [1 i , ..., k l , n j ] ′ ⊂ [1 i , ..., k l ] ′ ∈ A ′ k [1 i , ..., k l ] } .Take a [1 i , ..., k l , n j ] ′ from A ′ n [1 i , ..., k l ]. Then there exists a k nj such that α ( n +1) Let A ′ be unbounded. Then, for each b ∈ N , there exists an h > b and a [ n , n , ..., n h ] ∈ A ′ such that for each k ∈ N , α ( h + 1) ≥ k for some α ∈ [ n , n , ..., n h ] .Proof. This is an inverse no proposition of Proposition 9.3. (cid:3) Proposition 9.5. V y = { V λ : λ ∈ Λ } is not closure preserving.Proof. Note v α = α ( m + v ) + 1 and H ( m , jz kv ) ⊂ H ( m , jz ) for α ∈ A with H ( m , z ) = H ( m , zα ) ⊂ H ( m , jz ). Then there exists a q k such that, for every α ∈ A , we have H ( m α, − βγ, q k z ) − H ( m σ α , − βγ, q k z ) ⊂ U [ α − bk ′ , p ′ k y ] ⊂ V α ⊂ g c ( b, y ) and H ( m σ α , − βγ, q k z ) ∩ Cl τ V α = ∅ . Let [ n , n , ..., n v ] ′ ⊂ A satisfy Corollary 9.4. Then, for arbitrary l ∈ N , thereexists an α h ∈ [ n , n , ..., n v ] ′ with α h ( v + 1) ≥ l . Let α ∗ = { α h : h ∈ N } . Then wemay assume α ( m + v ) < α ( m + v ) < ... < α h ( m + v ) < ... . A. Take α ∈ α ∗ for p = 1. Denote α by α . Take relative U [ α − b α k ′ , p ′ k y ], V α and U ( αb , p y ). Then U [ α − b α k ′ , p ′ k y ] = g ( b, y ) −∪ i ≤ n E ( α i k i , p k i ) ⊂ V α ⊂ U ( αb , p y ).Let α − = { β i : i ≤ n ′ } . Then α ∈ α − . Then, by Proposition 8.4, for every H ( v α , t ) ∈ H ( m σ α , − βγ, q k z ) and every H ( v α , t ) ∈ H − ( kσ α , jz, q k σ ) | H ( v α , t )with H ( v α , t ) ⊂ H ( m α, − βγ, q k z ) − H ( m σ α , − βγ, q ′ k z ) , H ( v α , t ) and H ( v α , t )satisfy H ( v α , t ) ∩ Cl τ V α = ∅ , H ( v α , t ) ⊂ V α and H ( v α , t ) ∪ H ( v α , t ) ⊂ H ( v α , t ) ∈ H ( v α , jz kv ) . Let H ( v, α − ) = H ( m σ α , − βγ, q k z ) | H ( m , jz kv ) and H ( v, α − ) = ∪ H ( v, α − ). B. Let p = n and δ = α n . Take H ( m δ, − βγ, z kv ) in 3 of proof of Proposition8.1. Let H ( m δ, − βγ, z kv ) = ∪ H ( m δ, − βγ, z kv ). Assume we have had H ( v, α − n ) = H ( m σ δ , − βγ, q k z ) | H ( m δ, − βγ, z kv )and H ( v, α − n ) = ∪ H ( v, α − n ). Then we have ∩ i ≤ n H ( v, α − i ) = ∅ by the definition of H ( m δ, − βγ, z kv ). Let H ( v, α − n , ∧ ) = ^ i ≤ n H ( v, α − i ) = {∩ i ≤ n H ( v α i , t i ) : H ( v α i , t i ) ∈ H ( v, α − i ) for i ≤ n } . Then H ( v, α − n , ∧ ) ⊂ H ( v, α − n ) and ∪ H ( v, α − n , ∧ ) = ∩ i ≤ n H ( v, α − i ). NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 59 Take an H ( v δ , t n ) ∈ H ( v, α − n , ∧ ). Then, by Corollary 4.9, there exists an ℓ ,exists a Ln cover G ( ℓ, jz kv ) of H ( m , jz kv ) and exists a G ( ℓ, t n ) ⊂ G ( ℓ, jz kv ) suchthat ∪ H ( ℓ, t n ) ⊂ H ( v δ , t n ) ⊂ ∪ G ( ℓ, t n ). Then there exists an α h ∈ α ∗ with α h ( m + v ) > ℓ . Denote α h by θ . Note that G ( v θ , jz kv ) = { G ( v θ , jz kv ) , G ( v θ , jz kv ) } is a covers sequence of H ( m , jz nv ). Then there exists an subfamily G ( v θ , t n ) ⊂ G ( v θ , jz kv ) such that ∪ H ( v θ , t n ) ⊂ H ( v δ , t n ) ⊂ ∪ G ( v θ , t n ) . On the other hand, note θ = α h ∈ α ∗ ⊂ [ n , n , ..., n v ] ′ ⊂ A . Then there exist U [ θ − bk ′ , p ′ k y ], V θ and U ( θb , p y ) such that U [ θ − bk ′ , p ′ k y ] ⊂ V θ ⊂ U ( θb , p y ) and H ( m , z ) ∈ H ( m , ℓ ) satisfying H ( m θ, − βγ, q ′ k z ) − H ( m σ θ , − βγ, q ′ k z ) ⊂ U [ θ − bk ′ , p ′ k y ] ⊂ V θ and H ( m σ θ , − βγ, q ′ k z ) ∩ Cl τ V θ = ∅ . Let h = n + 1. Then, by Proposition 8.4,for every H ( v θ , t h ) ∈ H ( m σ θ , − βγ, q ′ k z ) and every H ( v θ , t h ) ∈ H − ( kσ θ , jz, q k σ )with H ( v θ , t h ) ⊂ H ( v θ , t h ), we have H ( v θ , t h ) ⊂ H ( m θ, − βγ, q ′ k z ) − H ( m σ θ , − βγ, q ′ k z ) ,H ( v θ , t h ) ∩ Cl τ V θ = ∅ , H ( v θ , t h ) ⊂ V θ and H ( v θ , t h ) ∪ H ( v θ , t h ) ⊂ H ( v θ , t h ) . Then, by induction for every α h , we denote α h by θ . Then H ( v θ , t h ) ∩ Cl τ V θ = ∅ , H ( v θ , t h ) ⊂ V θ and H ( v θ , t h ) ∪ H ( v θ , t h ) ⊂ H ( v θ , t h ) . Let H ( v, α − h ) = H ( m σ θ , − βγ, q k z ) | H ( m θ, − βγ, z kv ) and H ( v, α − h , ∧ ) = ^ i ≤ h H ( v, α − i ) . C. Note v α = α ( m + v ) + 1, α ( m + v ) < α ( m + v ) < ... < α h ( m + v ) < ... and ∪ H ( v θ , t n ) ⊂ H ( v δ , t n ) ⊂ ∪ G ( v θ , t n ) for δ = α n and θ = α n +1 . Then, by3 of Proposition 3.5, we have ∩ h H ( v α h , t h ) = ∩ h H ( v α h , t h ) = { t } 6 = ∅ . Note t ∈ H ( v α h , t h ) and H ( v α h , t h ) ∩ Cl τ V α h = ∅ . Then t / ∈ Cl τ V α h for every h ∈ N .On the other hand, we have t ∈ H ( v α h , t h ) ⊂ g ( v α h , t h ). Then, by Note D2, t h ∈ H ( v α h , t h ) ⊂ H ( v α h , t h ) ⊂ g ( v α h , t h ) = g ( v α h , t h ) and H ( v α h , t h ) ⊂ V α h . Then t h → t if h → + ∞ by 2 in the proof of Theorem 1.a. Then we have t ∈ Cl τ [ ∪ h V α h ]. Then V y = { V λ : λ ∈ Λ } is not closure preserving. (cid:3) Theorem 1. b. ( Y, τ ) is not an M -space. Proof. Suppose that B = ∪ i B i is a σ -closure preserving base of ( Y, τ ). Let B ( i, y ) = { B ∩ g ( i, y ) : y ∈ B ∈ B i } . Then B ( y ) = ∪ i B ( i, y ) is a closurepreserving outer base of y in ( Y, τ ). It is a contradiction to Proposition 9.5. (cid:3) Problems. Theorem 1 suggest the following problems. Problem 1. Simplify space ( X, τ ) or look for a simpler counterexample.Note stratifiable µ -spaces have various dimension theoretical properties and nicepreservation properties under topological operations. But, for hereditarily M -spaces, we have only a simple definition. So we venture to hope the followingproblem having a positive answer after the counterexample. Problem 2. Are hereditarily M -spaces stratifiable µ -spaces? ∗ AND BOSEN WANG ∗∗∗ Liu [13] has given a characterization of M -spaces by g -functions. We don’t knowhow to construct a σ -closure preserving base for M -spaces. Problem 3. Construct a σ -closure preserving base for M -spaces.Recall a well-known theorem in [9]: A space X is stratifiable if and only if X issemi-stratifiable and monotonically normal. So we have the following problem 4. Problem 4. Character differences between stratifiable µ -spaces and stratifiablespaces.The following problem 4 is more difficult than Problem 4. Problem 5. Character differences between M -spaces and stratifiable spaces.Junnila [17] proved that every (first countable) topological space is the continu-ous image of a stratifiable (metrizable) σ -discrete T -space under an open mapping.Reference Lin [12]. The following problem should be difficult. Problem 6. Character open images of M -spaces.11. Acknowledgement. Professor Gary Gruenhage went through the primitive paper, and told us thatsome sections are too complicated and too hard to understand. So we are going torewriting the paper from Section 5. References [1] C.R. Borges, On stratifiable spaces, Pacific J. Math., 17(1966), 1-16.[2] D. Burke and D.J.Lutzer, Recent advances in the theory of generalized metric spaces, inTopology: Proc. 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Topology, budapest (Hungary), (1978), 689-703.[18] K. Kunen, Set Theory, A Introduction to Independence Proofs, Amsterdam: North-Holland,1980.[19] T.Mizokami, On a certain class on M -spaces, Proc. Amer. Math. Soc., 87(1983), 357-362.[20] T.Mizokami, On M -structures, Topology Appl., 17(1984), 63-89.[21] T.Mizokami, On closed subsets of M -spaces, Topology Appl., 141(2004), 197-206. NEGATIVE ANSWER TO THE PROBLEM: ARE STRATIFIABLE SPACES M ? 61 [22] K.Tamano, Generalized Metric Spaces II, Topics in General Topology, K.Morita andJ.Nagata, editors, 1989, 368-409. North-Holland, Amsterdam.[23] G. Wang, Edited by Daqian Li andJiping Zhang, Chinese Scienec Press, 2009, 211-212. ∗ School of Mathematics and computer science, Shaanxi SCI-TECH University, Hanzhong,723000, Shaanxi P. R. China E-mail address : [email protected] ∗∗∗ No.11 Jin Rong Avenue, Xi Cheng District, Beijing, P.R.China. 100033 E-mail address ::