Monotone Normality and Nabla-Products
aa r X i v : . [ m a t h . GN ] J un Monotone Normality and Nabla-Products
H.A. Barriga-Acosta and P.M. Gartside
Abstract
Roitman’s combinatorial principle ∆ is equivalent to monotone nor-mality of the nabla product, ∇ ( ω + 1) ω . If { X n : n ∈ ω } is a familyof metrizable spaces and ∇ n X n is monotonically normal, then ∇ n X n is hereditarily paracompact. Hence, if ∆ holds then the box product (cid:3) ( ω + 1) ω is paracompact. Large fragments of ∆ hold in ZFC , yield-ing large subspaces of ∇ ( ω + 1) ω that are ‘really’ monotonically normal.Countable nabla products of metrizable spaces which are respectively:arbitrary, of size ≤ c , or separable, are monotonically normal under re-spectively: b = d , d = c or the Model Hypothesis.It is consistent and independent that ∇ A ( ω ) ω and ∇ ( ω + 1) ω arehereditarily normal (or hereditarily paracompact, or monotonically nor-mal). In ZFC neither ∇ A ( ω ) ω nor ∇ ( ω + 1) ω is hereditarily normal. Let { X i : i ∈ I } be a family of topological spaces. (All spaces in this articleare Tychonoff.) A box is a set Q i U i , where each U i is open in X i . The boxproduct , (cid:3) i X i , is the space with underlying set Q i X i and basis all boxes. Twoelements x and y of (cid:3) i X i are mod-finite equivalent , denoted x ∼ y , if the set { i ∈ I : x ( i ) = y ( i ) } is finite. The nabla product , ∇ i X i , is the quotient space, (cid:3) i X i / ∼ .It is unknown, in ZFC , whether the countable box product (cid:3) [0 , ω , or evenits closed subspace, (cid:3) ( ω + 1) ω , is normal. This question was asked (orally) forthe first time by Tietze sometime in the 1940’s. See [12] for a survey of the boxproduct problem. Central to almost all positive results on paracompactness,and hence normality, of box products, is a connection to the nabla productdue to Kunen [6]: let { X n : n ∈ ω } be a family of compact spaces, then, (cid:3) n X n is paracompact if and only if ∇ n X n is paracompact. In particular, itis now known that under certain set theoretic assumptions the nabla product ∇ ( ω + 1) ω is paracompact and so the box product (cid:3) ( ω + 1) ω is paracompact.These assumptions include the small cardinal conditions, b = d [2], and d = c [7, 8], and also the so called Model Hypothesis [9], which holds in any forcingextension by uncountably many Cohen reals.In an insightful analysis of the combinatorics behind these consistency re-sults, Roitman [9] extracted a combinatorial principle, ∆. She showed ∆ is aconsequence of each of the set theoretic axioms mentioned above, and further1laimed that ∆ implies the paracompactnes of ∇ ( ω + 1) ω . Unfortunately not allthe details for the latter deduction were presented, and the authors and Roit-man [11] are unclear how to fill the gap. See Section 4.1 for the definition of ∆,additional notation and more details on the gap.In Section 4 we close the gap by connecting ∆ to monotone normality of ∇ products. Indeed, (Theorem 20) the property ∆ holds if and only if ∇ ( ω + 1) ω is monotonically normal. Monotonically normal spaces are not automaticallyparacompact, but (Theorem 8) we show: if { X n : n ∈ ω } is a family of metriz-able spaces and ∇ n X n is monotonically normal, then it is hereditarily paracom-pact. It follows that if ∆ holds, then ∇ ( ω + 1) ω is monotonically normal, andso hereditarily paracompact, and hence (cid:3) ( ω + 1) ω is paracompact, as Roitmanoriginally claimed.Recall that a space X is monotonically normal if for every pair of disjointclosed sets A, B there is an open set H ( A, B ) such that (i) A ⊆ H ( A, B ) ⊆ H ( A, B ) ⊆ X \ B (so H ( · , · ) separates A from B , and thus witnesses normality)and (ii) if A ′ ⊆ A and B ⊆ B ′ then H ( A ′ , B ′ ) ⊆ H ( A, B ) (‘monotonicity’, theseparation respects set inclusion). An alternative characterization is that forevery point x in an open set U there is assigned an open set G ( x, U ) such that x ∈ G ( x, U ) ⊆ U , and if G ( x, U ) ∩ G ( y, V ) = ∅ , then x ∈ V or y ∈ U . Observe thatthe restriction of a monotone normality operator, G ( · , · ), to a subspace yields amonotone normality operator for the subspace, and so monotone normality ishereditary. It follows that monotone normality does not transfer from ∇ ( ω +1) ω to (cid:3) ( ω + 1) ω . Indeed, see [13], if { X i : i ∈ I } is a family of compact or firstcountable spaces, then (cid:3) i X i is not hereditarily normal.The authors do not know how to prove ∆ in ZFC , or to prove that its negationis consistent. In an effort to shed light on this conundrum we have attemptedto ‘parametrize’ the problem: either ‘from below’ in order to see how close wecan get to establishing ∆ in
ZFC , or ‘from above’ to determine when naturalstrengthenings of ∆ are false either consistently or in
ZFC .For example, pursuing an idea of Roitman, we characterize in Section 4.3when certain subspaces A of ∇ ( ω + 1) ω are monotonically normal in terms ofa combinatorial property ∆( A ), where ∆ is ∆( ∇ ( ω + 1) ω ). In particular, seeProposition 16, ∆( A ) is true in ZFC for A consisting of all finite disjoint unionsof increasing functions.In the other direction we have found combinatorial characterizations of whennabla products of certain spaces containing ω + 1 as a closed subspace, or other-wise naturally extending ω + 1, are monotonically normal. Specifically, observethat ω +1 is the one-point compactification of a countably infinite discrete space.Accordingly in Section 5, we characterize combinatorially monotone normalityof nabla products of spaces of the form, A ( κ ), the one point compactification ofa discrete space of size κ . We show that ∇ A ( ω ) ω is not hereditarily normal,and so not monotonically normal, in ZFC ; while ∇ A ( ω ) ω is consistently nothereditarily normal. A striking result of Williams [13], is that consistently anycountable nabla product of compact spaces of weight (minimal size of a base) nomore than ℵ is ( ω -metrizable and so) monotonically normal and hereditarilyparacompact. In particular, ∇ A ( ω ) ω is consistently monotonically normal and2ereditarily paracompact; and so each of the statements: ‘ ∇ A ( ω ) ω is monoton-ically normal’, ‘ ∇ A ( ω ) ω is hereditarily paracompact’ and ‘ ∇ A ( ω ) ω is heredi-tarily normal’ is consistent and independent. These results answer questions ofRoitman.Observing that ω + 1 can also be viewed as an ordinal with the order topol-ogy, in Section 6, we go on to characterize combinatorially monotone normalityof nabla products of ordinals. This yields a finer parametrization than lookingat one point compactifications. Indeed if ∇ ( ω + 1) ω is monotonically normal (inother words, ∆ holds) then for every n in ω , we also have ∇ ( ω.n + 1) ω mono-tonically normal. However, the combinatorial principle characterizing when ∇ ( ω.ω + 1) ω is monotonically normal is – at least formally – stronger than∆. By the result of Williams [13] mentioned above, ∇ ( ω + 1) ω is consistentlymonotonically normal and hereditarily paracompact. We show ∇ ( ω + 1) ω isconsistently not hereditarily normal. Thus each of the statements: ‘ ∇ ( ω +1) ω ismonotonically normal’, ‘ ∇ ( ω +1) ω is hereditarily paracompact’ and ‘ ∇ ( ω +1) ω is hereditarily normal’ is consistent and independent. In contrast we also showthat ∇ ( ω + 1) ω is not hereditarily normal, and so not monotonically normal,in ZFC . These results answer questions of Roitman.While in Section 7 – thinking of ω +1 as the simplest non-discrete metrizablespace – we investigate combinatorial characterizations of monotone normalityof nabla products of metrizable spaces. We conclude the paper with some openproblems and potential lines of research. Recall that b is the minimal size of an unbounded set in ω ω with the mod-finiteorder, ≤ ∗ , and d the minimal size of a cofinal (dominating) set. Further, b = d if and only if there is a dominating family { f α : α < b } ⊆ ω ω so that if α < β then f α ≤ ∗ f β (such a family is called a scale ). We record an additional usefulfact, a proof of which can be found in [9]. Lemma 1. If G ⊆ ω ω , A ⊆ P ( ω ) and |G| , |A| < d , then there is a function f ∈ ω ω so that for any g ∈ G and a ∈ A , |{ n ∈ a : f ( n ) > g ( n ) }| = ω . Definition 2 (Roitman [9]) . The
Model Hypothesis , abbreviated MH , is thefollowing statement: For some κ , H ( ω ) is the increasing union of H α ’s, for α < κ , where each H α is an elementary submodel of ( H ( ω ) , ∈ ) and each H α ∩ ω ω is not dominating. Here H ( κ ) is the collection of all sets whose transitive closures have sizeless than κ . In particular, both ω ω , P ( ω ) are contained in H ( ω ), and a spaceof countable weight can be coded as (hence is homeomorphic to) a subset of H ( ω ). 3 .2 Box and Nabla Products We will follow Roitman’s notation from [9]. For x ∈ (cid:3) n X n , we write x for itsmod-finite equivalence class, [ x ] ∼ , in ∇ n X n . If x ∈ (cid:3) n X n or x ∈ ∇ n X n , and U = h U n i n ∈ ω is a sequence of open sets, where x ( n ) ∈ U n ⊆ X n , define thebasic neighborhood around x by N ( x, U ) := (cid:3) n U n ⊆ (cid:3) n X n and N ( x, U ) := ∇ n U n ⊆ ∇ n X n . If the X n ’s are first countable, a basis of x or x is coded by ω ω as follows: if { U kn : k ∈ ω } is a base at x ( n ), we will write N ( x, f ) for (cid:3) n U f ( n ) n and N ( x, f ) for ∇ n U f ( n ) n , where f ∈ ω ω . Following Roitman, we donot distinguish between elements of (cid:3) n X n and ∇ n X n ( x versus x ) if there isno chance of confusion.A space X is said to be P κ if the intersection of strictly fewer than κ -manyopen sets is open. We recall: every nabla product, ∇ n X n , is a P ω -space; andif each X n is first countable, then ∇ n X n is a P b -space. Let A be a subspace of a space X . We say that A is monotonically normal in X if for every point x of A and set U open in X containing it, there is assigned anopen (in X ) set G ( x, U ) such that x ∈ G ( x, U ) ⊆ U , and if G ( x, U ) ∩ G ( y, V ) = ∅ ,then x ∈ V or y ∈ U . Observe that for any base for X we only need define G ( x, U ) for basic U , and we may assume that G ( x, U ) is in the base. This willbe used frequently in the sequel. If A is monotonically normal in some X thenclearly A is monotonically normal.A function F on A is a neighborhood assignment (or, neighbornet ) for A (in X ) if F ( x ) is a neighborhood of x , for every x in A . (A neighbornet for thewhole space is just called a ‘neighbornet’.) A neighbornet T on A is halvable in X if there is a neighbornet S for A such that: if S ( x ) ∩ S ( y ) = ∅ then x ∈ T ( y )or y ∈ T ( x ). Note that we may assume that every S ( x ) comes from any givenbase for X . The subspace A of X is halvable in X if every neighbornet of A in X is halvable. A space is halvable if it is halvable in itself (every neighbornetcan be halved).Observe that if A is monotonically normal in X then it is halvable in X ,and so every monotonically normal space is halvable. The converse is false. Forexample, every countable (Tychonoff) space is halvable [5], but there are count-able spaces that are not monotonically normal (for example, all polynomialswith rational coefficients with the topology of pointwise convergence, see [3]).However, it turns out that in certain cases nabla products are monotonicallynormal if they are halvable, indeed it suffices that just one specific neighbornetbe halvable. Lemma 3.
Let X be a space with partial order (cid:22) and neighborhood bases, B x ,for each x ∈ X , such that: (a) ↓ x = { y : y (cid:22) x } is a neighborhood of x for all x ,and (b) if y ∈ B ⊆ ↓ x , where B ∈ B x , then the interval [ y, x ] = { z : y (cid:22) z (cid:22) x } is contained in B . Let A be a subspace of X . Then, A is monotonically normalin X if and only if the neighbornet T A ( x ) = ↓ x , for x in A , is halvable in X . roof. We only need to show that if S halves T A ( x ) = ↓ x , then A is monoton-ically normal in X . For any element x of A in some B ∈ B x , where B ⊆ ↓ x ,define G ( x, B ) = S ( x ) ∩ B . We prove that this is a monotone normality opera-tor for A in X . Suppose z ∈ G ( x, B ) ∩ G ( x ′ , B ′ ). Then S ( x ) meets S ( x ′ ), andsuppose without loss of generality that x ′ ∈ ↓ x , that is, x ′ (cid:22) x . As z ∈ B ⊆ ↓ x ,we have z (cid:22) x . Hence, [ z, x ] is contained in B . But as z ∈ B ′ ⊆ ↓ x ′ , we have z (cid:22) x ′ . Hence x ′ is in [ z, x ], and so in B . (cid:4) A space X is κ -metrizable (for a cardinal κ ≥ ω ) if it has an open base B = { U x,α : α < κ, x ∈ X } so that { U x,α : α < κ } is a neighborhood base at x , and given two points x, y and two ordinals α ≤ β < κ then (i) if y ∈ U x,α then U y,β ⊆ U x,α ; and (ii) if y / ∈ U x,α then U y,β ∩ U x,α = ∅ . Every κ -metrizablespace is paracompact and monotonically normal. We observe that certain spaces are not hereditarily normal. These will be usedas test spaces to show certain nabla products are not hereditarily normal. Theresults are probably folklore, so we sketch just enough of their proofs for thefull argument to be reconstructed by the reader.If λ, κ are cardinals, denote by D ( κ ) the discrete space of size κ and let L λ ( κ )be the space with underlying set D ( κ ) ∪ { κ } , and topology where points of D ( κ )are isolated and neighborhoods around κ have the form { κ } ∪ ( D ( κ ) \ C ) for C ⊆ D ( κ ) of size less < λ . Write A ( κ ) for L ω ( κ ), the one-point compactificationof D ( κ ), and L ( κ ) = L ω ( κ ) the one-point Lindelofication of D ( κ ). Lemma 4. L ( ω ) is not hereditarily normal.Proof. More precisely, Y = L ( ω ) × L ( ω ) \ { ( ω , ω ) } is not normal, becausethe sets H = ( L ( ω ) \ { ω } ) × { ω } and K = { ω } × ( L ( ω ) \ { ω } ) are disjointand closed in Y , and can not be separated by disjoint open sets.Indeed suppose U and V be any open neighborhoods around H and K ,respectively. For every ( α, ω ) ∈ H choose A α ∈ [ D ( ω )] ω such that { α } × ( L ( ω ) \ A α ) ⊆ U , and similarly, for ( ω , β ) ∈ K choose sets B β such that( L ( ω ) \ B β ) × { β } ⊆ V . Then there is δ ≥ ω such that for every α ≤ δ , A α ⊆ δ . (To see this, let M ≺ H ( ω ) be an elementary submodel of size ω such that { A α : α < ω } ∈ M , set δ = M ∩ ω ∈ ω , and now, for α ∈ M , M thinks ‘ A α is contained in M ’, and so does H ( ω ).) Now a counting argumenteasily shows U and V meet. (cid:4) Lemma 5.
Let S be a stationary subset of a regular uncountable cardinal κ .Then, S × ( S ∪ { κ } ) (as a subspace of ( κ + 1) ) is not normal.Proof. Consider the diagonal H = { ( α, α ) : α ∈ S } and the top-edge K = { ( α, κ ) : α ∈ S } . Note that H and K are closed disjoint sets. Now, if U, V are neighborhoods of
H, K , respectively, a standard Pressing Down Lemmaargument shows that U and V must meet. (cid:4) Embeddings into Nabla Products
The following simple embedding result will be used frequently in the sequel. Fora space X let X δ be the G δ -modification of X (the space with underlying set X and topology generated by all G δ subsets of the space X ). Lemma 6 (Williams [13], Lemma 4.4) . Let X be a space. Then X δ embeds asa closed subspace in ∇ X ω via the map x c x , where c x is constantly equal to x . However the main technical result of this section is about non-embedding.
Proposition 7.
Let { X n : n ∈ ω } be a family of metrizable spaces and S astationary subset of a regular uncountable cardinal κ . Then S does not embedinto ∇ n X n . Since Balogh and Rudin [1] showed that a monotonically normal space isparacompact if and only if it does not contain closed copies of stationary subsetsof regular uncountable cardinals, we deduce:
Theorem 8.
Let { X n : n ∈ ω } be a family of metrizable spaces. If a subspace A of ∇ n X n is monotonically normal then A is hereditarily paracompact.Proof of Proposition 7. Suppose, for a contradiction, ϕ : S → ∇ n X n is an em-bedding. We split the proof in two cases depending on the size of κ . If κ ≤ d ,then any α in Lim ( S ) has cf ( α ) = d , so S , and A = ϕ ( S ), have limit pointsbut no points of character d , contradicting Lemma 9. If κ > d , by Lemma 10the map ϕ is eventually constant, thus it cannot be an embedding. (cid:4) Recall that the character of a topological space X at a point x is the cardi-nality χ ( x, X ) of the smallest local base for x . The tightness at a point x in X ,denoted t ( x, X ), is the smallest cardinal κ such that whenever x ∈ Y for some Y ⊆ X , there exists a subset Z ⊆ Y , with x ∈ Z and | Z | ≤ κ . Lemma 9.
Let { X n : n ∈ ω } be a family of first countable spaces. For any point x ∈ A \ A , where A ⊆ ∇ n X n , we have that t ( x, A ∪ { x } ) = d = χ ( x, A ∪ { x } ) .Proof. It is easy to check that t ( x, A ∪ { x } ) ≤ χ ( x, A ∪ { x } ) ≤ χ ( x, ∇ n X n ) = d (the last equality holds because a local basis of x can be represented by adominating family of ω ω ). We only have to prove that t ( x, A ∪ { x } ) ≥ d .If x is a limit point of A , then for infinitely many n ∈ ω , x ( n ) is non-isolatedin X n . Thus, without loss of generality we may assume that for all n ∈ ω , x ( n )is non-isolated in X n . Let { B m ( n ) : m ∈ ω } be a decreasing countable localbasis for x ( n ), for every n ∈ ω .Suppose for a contradiction that there is Z ⊆ A with | Z | < d and x ∈ cl A ∪{ x } ( Z ). For every z ∈ Z , there is an infinite set a z ⊆ ω such that for n ∈ a z , z ( n ) = x ( n ). Also, for every z ∈ Z there is a function f z ∈ ω ω such thatfor n ∈ a z , z ( n ) / ∈ B f z ( n ) ( n ). Thus, z / ∈ N ( x, f z ). Let G = { f z : z ∈ Z } and A = { a z : z ∈ Z } . By Lemma 1, there is f ∈ ω ω diagonalizing the families G and A . Then, for any z ∈ Z , z / ∈ N ( x, f ), contradicting that x ∈ cl Z ∪{ x } ( Z ). (cid:4) emma 10. Let { ( X n , d n ) : n ∈ ω } be a family of metric spaces and S a sta-tionary subset of a regular uncountable cardinal κ > d . Then, every continuousfunction ϕ : S → ∇ n X n is eventually constant.Proof. For each n ∈ ω , write B n ( a, ε ) for the ε -ball around a in X n with respectto the metric d n . Let { f µ : µ < d } ⊆ ω ω be a dominating family. For every x ∈∇ n X n and µ < d , define N ( x, f µ ) = ∇ n ∈ ω B n ( x ( n ) , / f µ ( n ) ); { N ( x, f µ ) : µ < d } is a local basis at x .Fix, for the moment, µ < d . For every α ∈ Lim ( S ), pick g µ ( α ) < α , g µ ( α ) ∈ S , such that ϕ [( g µ ( α ) , α ]] ⊆ N ( ϕ ( α ) , f µ ). Then, g µ is a regressivefunction and by the Pressing Down Lemma, there is α µ in S and a stationaryset S µ ⊆ S such that for any β ∈ S µ , g µ ( β ) = α µ .We claim that for all δ, γ > α µ , where δ and γ are in S , ϕ ( γ ) is in N ( ϕ ( δ ) , f µ ).To see this, take any δ and γ strictly larger than α µ in S . Since S µ is stationary,there is a β in Lim ( S µ ) with β > max { γ, δ } . Then, { ϕ ( γ ) , ϕ ( δ ) } ⊆ ϕ [( α µ , β ]] = ϕ [( g µ ( β ) , β ]] ⊆ N ( ϕ ( β ) , f µ ). By definition of N ( ϕ ( β ) , f µ ), for all n ∈ ω wehave that ϕ ( γ )( n ) and ϕ ( δ )( n ) are in B n ( ϕ ( β )( n ) , / f µ ( n ) ). Then by symmetryand the triangle inequality, for all n ∈ ω , we get d n ( ϕ ( γ )( n ) , ϕ ( δ )( n )) < / f µ ( n ) .This implies ϕ ( γ ) ∈ N ( ϕ ( δ ) , f µ ), as claimed.Now, as we let µ run over all values below d , since κ > d , there is a leastupper bound α ∞ of { α µ : µ < d } in S . Notice that by the claim above, forany µ < d and γ, δ ∈ S \ α ∞ , we have ϕ ( γ ) ∈ N ( ϕ ( δ ) , f µ ), and so ϕ ( γ ) = ϕ ( δ ).Hence ϕ is constant from α ∞ on, as desired. (cid:4) ∆ and ∇ ( ω + 1) ω ∆ In order to state our parametrized versions of Roitman’s ∆ principle we in-troduce some specific definitions and notation, naturally extending those ofRoitman, for partial functions.For any function x : N → ω , where N ⊆ ω , consider x to be a partialfunction from ω to ω and write dom x for N , the domain of x . We identify apartial function with its graph, which is a subset of ω × ω . Then two partialfunctions, x and y , are almost equal , x = ∗ y , if x \ y and y \ x are both finite. Let ω ⊆ ω be the set of all partial functions, including the empty function. Denote by ω ⊂ ω the set of all partial functions whose domain is infinite and co-infinite. Forany subset A of ω ⊆ ω let A ∗ = { y ∈ ω ⊆ ω : y = ∗ x for some x ∈ A } . For k ≤ ω ,let c k be the constant k -valued function in ( ω + 1) ω . Let x be a partial functionand h ∈ ω ω , then we say that x > ∗ h if for all but finitely many n ∈ dom ( x ), x ( n ) > h ( n ). Definition 11.
Two partial functions, x and y , switch if | x \ y | = | y \ x | = ω and x ( n ) = y ( n ) for all but finitely many n in dom x ∩ dom y . efinition 12. Let A be any subset of ω ⊆ ω . Then ∆( A ) is the statement:there exists F : A → ω ω such that if x, y ∈ A switch, then x \ y ≯ ∗ F ( y ) or y \ x ≯ ∗ F ( x ) . Lemma 13.
Let A and B be subsets of ω ⊆ ω . Then:(1) if A ⊆ B ⊆ ω ⊆ ω then ∆( B ) = ⇒ ∆( A ) , and(2) ∆( A ) ⇐⇒ ∆( B ) when A ∩ ω ⊂ ω ⊆ B ⊆ ( A ∪ ω ω ∪ {∅} ) ∗ .Proof. Claim (1) is clear – simply restrict an F witnessing ∆( B ) to A to get awitness of ∆( A ).For claim (2) it suffices to show: ∆( A ∩ ω ⊂ ω ) = ⇒ ∆(( A ∪ ω ω ∪ {∅} ) ∗ ).Fix F as in ∆( A ∩ ω ⊂ ω ). Note that if x is in ω ω ∪ {∅} then it does not switchwith any y in ω ⊆ ω . So we can extend F over ω ω ∪ {∅} completely arbitrarily,and it witnesses ∆( A ∪ ω ω ∪ {∅} ). Note that if x = ∗ x ′ and y = ∗ y ′ then x, y switch if and only if x ′ , y ′ switch, and similarly for the conclusion of ∆. So wecan extend F over A ∗ in the natural way (if x ∈ A , x = ∗ x ′ and x ′ / ∈ A then set F ( x ′ ) = F ( x )) to get a witness of ∆(( A ∪ ω ω ∪ {∅} ) ∗ ), as required. (cid:4) Abbreviate ∆( ω ⊂ ω ) to ∆, this is Roitman’s combinatorial principle in [9]and [12]. It is known to be consistently true (under b = d , d = c , MH , and inany forcing extension obtained by adding cofinally many Cohen reals) but it isunknown if it can be consistently false, or it is true in ZFC .In [9] Roitman showed that ∆ implies the subspace ∇ ∗ = ∇ ( ω +1) ω \ ( ∇ ω ω ∪{ c ω } ) of ∇ ( ω +1) ω is paracompact. Then, she claimed, without proof, that ‘ ∇ ∗ isparacompact if and only if ∇ ( ω +1) ω is paracompact’. Here lies the gap. Addingisolated points (like those of ∇ ω ω ) to even the best behaved of spaces (discrete,for example) frequently destroys normality and paracompactness (indeed, manyclassical counter-examples related to normality have this form). ∆( A ) -Principles That Hold in ZFC
While we only know of consistency proofs for ∆ = ∆( ω ⊂ ω ), there are, however,interesting A such that ∆( A ) is true in ZFC . We present here an example.Let INC be the set of all increasing partial functions (so x ∈ ω ⊂ ω is in INCif whenever m ≤ n are in dom x we have x ( m ) ≤ x ( n )). Let FI be all partialfunctions which are the finite disjoint union of increasing functions (so x is inFI if dom x can be partitioned into S , . . . , S k such that each x ↾ S i is in INC).We will show: the combinatorial principle ∆(FI) is true in ZFC .First we need to show that elements of FI have a nice representation. Let x be in ω ⊆ ω . Define ⊥ x = { m ∈ dom x : ∀ n > m, x ( m ) ≤ x ( n ) } . Set x ⊥ = x ↾ ( ⊥ x ). Observe that x ⊥ is increasing. Set x = x ⊥ ; and inductively, x n = ( x \ S i Let x be in FI , say x = x ∪· · ·∪ x ℓ where each x i is increasing.Then x = x ∪ · · · ∪ x ℓ . roof. Evidently each x k is a subset of x , so we need to show x ⊆ x ∪ · · · ∪ x ℓ .We do so by breaking x into finite pieces, i x , each of which is contained in S k ≤ ℓ x k .For any y ∈ ω ⊆ ω set Dec ( y ) ⊆ dom y such that min ( dom y ) ∈ Dec ( y ) and n ∈ Dec ( y ) with n > min ( dom y ) if and only if for all m < n and m ∈ dom y , y ( m ) > y ( n ). Note that Dec ( y ) is finite and non-empty, and y ↾ Dec ( y ) is strictly decreasing. Set x = x ↾ Dec ( x ); and inductively, n x = x ↾ Dec ( x \ S i 1. Also, notice that for every s ≤ ℓ i − j , if n iℓ i − ( j + s ) ∈ dom x k , then n i ′ ℓ i ′ − j / ∈ dom x k , because x k is increasing and x ( n iℓ i − j ) > x ( m )(our assumption). Thus, there is a one-to-one correspondence between the last j points of i x and the last j + 1 points of i ′ x , the desired contradiction. (cid:4) We will say that two elements z, w of ω ⊆ ω are compatible if for all but finitelymany n in dom z ∩ dom w we have z ( n ) = w ( n ). Now let x be an element of ω ⊆ ω with infinite domain. Note that x ⊥ also has infinite domain. For each n / ∈ dom x let n + x be the minimal element of dom x larger than n . Let INC + be the set of members of INC with infinite domain. For x in INC + , define F ( x )by F ( x )( n ) is x ( n ) when n ∈ dom x and is x ( n + x ) + 1 otherwise. Lemma 15. Let z, w be compatible elements of ω ⊆ ω with infinite domain. If z \ w > ∗ F ( w ⊥ ) and w \ z > ∗ F ( z ⊥ ) then w ⊥ = ∗ z ⊥ Proof. We will show that z ⊥ = ∗ u ⊥ where u = ( w \ z ) ∪ z . Then symmetrically, w ⊥ = ∗ v ⊥ where v = ( z \ w ) ∪ w , hence by compatibility of z, w we have u = ∗ v ,and so z ⊥ = ∗ w ⊥ , as claimed. For z ⊥ = ∗ u ⊥ it suffices that ⊥ z = ∗ ⊥ u , because9 equals z on ⊥ z ⊆ dom z , and so u ⊥ = u ↾ ( ⊥ u ) = ∗ z ↾ ( ⊥ z ) = z ⊥ . Fix N such that for all n ≥ N we have ( w \ z )( n ) > F ( z ⊥ )( n ).First a general observation: if k ∈ dom ( w \ z ) and k ≥ N , then k ≤ k + z ⊥ and u ( k ) = ( w \ z )( k ) > F ( z ⊥ )( k ) ≥ z ⊥ ( k + z ⊥ ) = u ( k + z ⊥ ) . Now take any n in ⊥ u with n ≥ N . By the observation n must be in dom z (not dom ( w \ z )) and clearly (as z ⊆ u ) n is in ⊥ z .For the other inclusion, take any n ≥ N in ⊥ z . Take any m ∈ dom u , m ≥ n . If m ∈ dom z then u ( n ) ≤ u ( m ) because n ∈⊥ z . Otherwise m ∈ dom ( w \ z ), then by the observation, u ( m ) > z ⊥ ( m + z ⊥ ) ≥ z ( n ) (as n ∈⊥ z ),and z ( n ) = u ( n ). Either way, u ( n ) ≤ u ( m ). Thus n ∈⊥ u , as required. (cid:4) We now show ∆(FI) is true. From Lemma 13 we deduce that in fact ∆((FI ∪ ω ω ∪ {∅} ) ∗ ) is true. Note that a partial function is in FI ∗ if it is eventually in FI,or equivalently, if it is the finite union of eventually increasing partial functions. Proposition 16. In ZFC we have ∆(FI) .Proof. Take any x ′ in FI. Then by Proposition 14, x ′ is the disjoint union x ′ ∪ x ′ ∪ · · · ∪ x ′ ℓ ′ . Note that x ′ = ∗ x where x is either the empty set, or thedisjoint union x ∪ x ∪ · · · ∪ x ℓ , and each of x , . . . , x ℓ is in INC + . Let FI + be all x in FI such that all of x , . . . , x ℓ are in INC + . Then we have just saidthat FI ⊆ ( {∅} ∪ FI + ) ∗ , hence, by Lemma 13, to show ∆(FI) it suffices to prove∆(FI + ). For x in FI + define F ( x ) to be the maximum of F ( x ) , . . . , F ( x ℓ ).Take any x, y in FI + . They have representation x = x ∪ · · · ∪ x ℓ and y = y ∪ · · · ∪ y m . Assume, without loss of generality, that ℓ ≤ m . Suppose x and y are compatible, x \ y > ∗ F ( y ) and y \ x > ∗ F ( x ). To establish ∆(FI + )we show x \ y is finite (so x, y do not switch), because x n = ∗ y n for all n ≤ ℓ ,which we verify by induction on n .Inductively, suppose x i = ∗ y i for all i < n . Let z = x \ S i Let x, y be in ω ⊆ ω .(i) x and y are compatible if and only if for all but finitely many n in dom x ∩ dom y we have x ( n ) = y ( n ) .(ii) If x and y are compatible then they have a greatest lower bound, z = x ∧ y ,where z = { ( n, k ) : x ( n ) = k = y ( n ) } ∪ ( x \ y ) ∪ ( y \ x ) .(iii) y (cid:14) x if and only if x and y are not compatible or x, y are compatiblebut x \ y infinite. For any x in ω ⊆ ω basic neighborhoods around x are of the form N ( x, h ) = { y ∈ ∇ ( ω + 1) ω : y (cid:22) x and y \ x > ∗ h } , where h ∈ ω ω . Lemma 18. Take any x, y in ω ⊆ ω and f x , f y in ω ω .(i) (a) x ∈ N ( x, c ) ⊆↓ x and (b) if y ∈ N ( x, f x ) and y (cid:22) z (cid:22) x then z ∈ N ( x, f x ) .(ii) N ( x, f x ) ∩ N ( y, f y ) = ∅ if and only if x ∧ y ∈ N ( x, f x ) ∩ N ( y, f y ) , if andonly if x, y are compatible, and y \ x > ∗ f x and x \ y > ∗ f y .Proof. Claim (i) (a) is evident. Towards (i) (b), suppose y (cid:22) z . Then for allbut finitely many n in dom ( z \ x ) ⊆ dom z we have ( z \ x )( n ) = z ( n ) = y ( n ) =( y \ x )( n ). Hence if y \ x > ∗ f x then also z \ x > ∗ f x . And (i) (b) follows.For the first equivalence of (ii), note that if z is in N ( x, f x ) ∩ N ( y, f y ) then z is (cid:22) -below both x and y . Now apply (i) (b). The second equivalence followsfrom the definitions. (cid:4) If A is any subset of ω ⊆ ω , write ∇ ( A ) for the subspace { x : x ∈ A } of ∇ ( ω +1) ω , set ∇ ∗ ( A ) = ∇ ( A ∩ ω ⊂ ω ) and set ∇ + ( A ) = ∇ (( A ∪ ω ω ∪{∅} ) ∗ ). Then ∇ ∗ ( ω ⊆ ω ) = ∇ ( ω ⊂ ω ) is ∇ ∗ from above, and abbreviate ∇ ( ω ⊆ ω ) = ∇ + ( ω ⊂ ω ) = ∇ ( ω + 1) ω to ∇ . Theorem 19. Let A be a subset of ω ⊆ ω . Then the following are equivalent:(1) ∆( A ) holds, (2) ∇ ∗ ( A ) is halvable in ∇ , and (3) ∇ + ( A ) is monotonicallynormal in ∇ .Proof. From Lemma 18(i) we see that ∇ with (cid:22) and the standard basic neigh-borhoods satisfies conditions (a) and (b) of Lemma 3, and any subspace of ∇ ismonotonically normal in ∇ if and only a specific neighbornet is halvable. Com-bining this with Lemma 13 we see that to prove the equivalence of (1) through(3) it is sufficient to show: ∆( A ) holds if and only if the neighbornet T ( x ) = ↓ x for x in A is halvable in ∇ .Suppose F is a function from A into ω ω . Define the neighbornet S of ∇ ( A )in ∇ by S ( x ) = N ( x, f x ) where f x = F ( x ). On the other hand, suppose S is a neighbornet of ∇ ( A ) in ∇ . We may assume each S ( x ) is basic, say11 ( x ) = N ( x, f x ). Define F : A → ω ω by F ( x ) = f x . We show F witnesses∆( A ) if and only S halves in ∇ the neighbornet T ( x ) = ↓ x for x in A .First let us note, ‘ x and y in A switch’, reinterpreted in terms of x and y via Lemma 17(iii), is equivalent to, ‘ x, y are compatible, but y (cid:14) x and x (cid:14) y ’.Next, taking the contrapositive, ‘ S ( x ) = N ( x, F ( x )) halves T ( x ) = ↓ x in ∇ ’ isequivalent to, ‘ y (cid:14) x and x (cid:14) y implies N ( x, F ( x )) ∩ N ( y, F ( y )) = ∅ ’.Now applying Lemma 17(iii) and Lemma 18(ii), we see that ‘ S ( x ) = N ( x, F ( x ))halves T ( x ) = ↓ x in ∇ ’ is equivalent to, ‘( x, y not compatible) or ( x, y compatibleand x \ y infinite and y \ x infinite) implies ( x, y not compatible) or ( x \ y ≯ ∗ F ( y )or y \ x ≯ ∗ F ( x ))’, which is equivalent to, ‘if ( x, y compatible and x \ y infi-nite and y \ x infinite) then ( x \ y ≯ ∗ F ( y ) or y \ x ≯ ∗ F ( x ))’, which (by thereinterpretation of switching above) is equivalent to ‘ F witnesses ∆( A )’. (cid:4) Theorem 20. Let A be a subset of ω ⊆ ω .(1) If ∆( A ) then ∇ ( A ) is monotonically normal and hereditarily paracom-pact.(2) If ∇ ( A ) is monotonically normal and, whenever x, y in ∇ ( A ) are com-patible then x ∧ y is in ∇ ( A ) , then ∆( A ) holds.Proof. For (1) note that if ∆( A ) holds, then by the preceding theorem ∇ + ( A )is monotonically normal, so its subspace ∇ ( A ) is monotonically normal.For (2) assume ∇ ( A ) is closed under ∧ . By Lemma 18(iii), for any x and y in ∇ ( A ) we have that one basic open set in ∇ ( A ), say N ∇ ( A ) ( x, f ) = N ( x, f ) ∩∇ ( A ), meets another, say N ∇ ( A ) ( y, g ), if and only if they both contain x ∧ y ;and so they meet (in ∇ ( A )) if and only if the corresponding open sets in ∇ , N ( x, f ) and N ( y, h ), meet (in ∇ ). Hence if ∇ ( A ) is monotonically normal thenit is monotonically normal in ∇ , and thus, by the preceding theorem, ∆( A )holds. (cid:4) From Proposition 16 we deduce: Example 21. Let FI be the family of finite disjoint unions of increasing par-tial functions. Then, in ZFC , we have ∇ ( F I ) is monotonically normal, andhereditarily paracompact. A space may be monotonically normal for ‘trivial’ reasons, such as beingdiscrete or metrizable. Indeed, it is not difficult to check that INC is a closedand discrete subspace of ∇ , and so monotonically normal ‘trivially’. Howeverthis is not the case for FI. For x in FI let ht ( x ) be the minimal number ofpartial functions in a representation of x as a disjoint union of increasing partialfunctions, and set FI n = { x : ht ( x ) = n } . Then FI = INC, the increasingpartial functions. One can verify that the closure of FI contains FI . FromLemma 9 it follows that every point of FI has uncountable character in FI.Hence FI is far from being metrizable (or discrete).12 .4 Another Not Hereditarily Normal Space Denote by X ( ω ω , ≤ ∗ ) the subspace ω ω ∪{ c ω } of ∇ ( ω +1) ω and write N ( c ω , f ) X = N ( c ω , f ) ∩ X ( ω ω , ≤ ∗ ), with f ∈ ω ω , the neighborhoods around c ω in X ( ω ω , ≤ ∗ ). Theorem 22. The space L ( ω ) × X ( ω ω , ≤ ∗ ) is hereditarily normal if and onlyif b = ω .Proof. Recall that a subset L of ω ω ( ω ω with the product topology) is a K -Luzin set if it is uncountable and meets every compact of ω ω in a countable set, orequivalently, for every g ∈ ω ω , the set { f ∈ L : f ≤ ∗ g } is countable. Observethat any uncountable subspace of K -Luzin is K -Luzin, hence the existence of a K -Luzin set is equivalent to b = ω . We prove the equivalence ‘ L ( ω ) × X ( ω ω , ≤ ∗ ) is hereditarily normal if and only if there is a K -Luzin set’.For the sufficiency, let p = ( ω , c ω ) be the top-right corner of the givenproduct. In X ′ = L ( ω ) × X ( ω ω , ≤ ∗ ) \ { p } the top edge, T = L ( ω ) × { c ω } \ { p } ,and right edge, R = { ω } × X ( ω ω , ≤ ∗ ) \ { p } , are disjoint closed sets. Hence,there are disjoint open sets U and V such that T ⊆ U and R ⊆ V . For each α < ω , pick f α such that { α } × N ( c ω , f α ) X ⊆ U . For each g in ω ω pickcountable C g ⊆ D ( ω ) such that ( L ( ω ) \ C g ) × { g } ⊆ V .Let A = { f α : α < ω } . The choice of the f α ’s can be in such way so they areall distinct, so the enumeration of A is injective. We check that A is K -Luzin.Take any g in ω ω , then for any α not in C g , as U and V are disjoint, ( α, g ) isnot in { α } × N ( c ω , f α ) X , so f α (cid:2) ∗ g . Hence, { α ∈ ω : f α ≤ ∗ g } is contained in C g , and so is countable.For the converse, note that L ( ω ) × X ( ω ω , ≤ ∗ ) is regular and points in( L ( ω ) \ { ω } ) × ω ω are isolated, and thus this product is hereditarily nor-mal provided: whenever A ⊆ T , B ⊆ R (where T and R are as above), thenthere are sets U, V open in L ( ω ) × X ( ω ω , ≤ ∗ ) separating A and B .We show this latter condition holds if there is a K -Luzin set. Write A = { ( α, c ω ) : α ∈ S } , where S ⊆ ω . If A is countable, then the result is clear.Hence, suppose S is uncountable. Let L = { f α : α ∈ S } ⊆ ω ω be a K -Luzinset such that the enumeration is bijective. For every g ∈ ω ω , C g = { α ∈ S : f α ≤ ∗ g } is countable. Hence the open sets U = S α ∈ S { α } × N ( c ω , f α ) X and V = S ( g,ω ) ∈ B ( L ( ω ) \ C g ) × { g } separate A and B . (cid:4) A ( κ ) ’s ∆ -like Characterizations of Monotone Normality Denote by D ( κ ) ⊆ ω the set of partial functions from ω to D ( κ ), and D ( κ ) ⊂ ω for the subset of partial functions with infinite and co-infinite domain. Twoelements x, y ∈ D ( κ ) ⊆ ω switch , if | x \ y | = | y \ x | = ω , and |{ n ∈ ω : x ( n ) , y ( n ) ∈ D ( κ ) and x ( n ) = y ( n ) }| < ω . Definition 23. ∆( A ( κ )) is the statement: there is F : D ( κ ) ⊂ ω → ([ κ ] <ω ) ω suchthat if x, y ∈ D ( κ ) ⊂ ω switch, then ( x \ y )( n ) ∈ F ( y )( n ) or ( y \ x )( n ) ∈ F ( x )( n ) for infinitely many n ∈ ω . ∇ ∗ A ( κ ) = { x ∈ ∇ A ( κ ) ω : x ∈ D ( κ ) ⊂ ω } . For x and y in D ( κ ) ⊆ ω write y (cid:22) x if and only if for all but finitely many n in dom x we have y ( n ) = x ( n ). Abasic neighborhood of an x in ∇ A ( κ ) ω is N ( x, f ) = { y ∈ ∇ A ( κ ) ω : y (cid:22) x andfor all but finitely many n ∈ dom ( y \ x ) we have ( y \ x )( n ) / ∈ f ( n ) } , where f isin ([ κ ] <ω ) ω .Observe that ω + 1 is A ( ℵ ) and that all definitions here reduce in thecase κ = ℵ to those in Section 4. The natural analogues of Lemma 17 andLemma 18 hold. Their proofs, and that of the following theorem follow, mutatismutandis , those for ∆( ω ⊂ ω ) and ∇ ( ω + 1) ω in Section 4, and so are omitted. Theorem 24. ∆( A ( κ )) holds if and only if ∇ A ( κ ) ω is monotonically normal ifand only if ∇ ∗ A ( κ ) is monotonically normal if and only if ∇ A ( κ ) ω is halvable. When can we deduce from ∆( A ( κ )) that ∇ A ( κ ) ω is (hereditarily) paracom-pact? Note that, we can not simply apply Theorem 8. However, for all κ wesee that ∇ A ( κ ) ω is homeomorphic to its square, and the second author [4] hasshown that if the square of a space is monotonically normal then all finite powersare monotonically normal and hereditarily paracompact. Corollary 25. If ∆( A ( κ )) holds then ∇ A ( κ ) ω is hereditarily paracompact, and (cid:3) A ( κ ) ω is paracompact. Williams’ result in [13] that under d = ω , countable nabla products of compactspaces of weight no more than ℵ are ω -metrizable, and hence monotonicallynormal, implies, in particular, that consistently ∇ A ( ω ) ω is monotonically nor-mal. We now see that this last statement is independent, and ω is the largestcardinal such that ∇ A ( κ ) ω can be monotonically normal. Theorem 26 (Roitman [10]) . ∇ A ( ω ) ω is not hereditarily normal.Proof. Since A ( ω ) δ = L ( ω ), this latter space embeds into ∇ A ( ω ) ω . Now, as ∇ A ( ω ) ω is homeomorphic to its square, Lemma 4 applies. (cid:4) Roitman, in [10], asked: is ∇ A ( ω ) ω consistently non hereditarily normal? Theorem 27. If b > ω , then ∇ A ( ω ) ω is not hereditarily normal.Proof. Since L ( ω ) = A ( ω ) δ , both spaces L ( ω ) and ∇ ( ω + 1) ω embed into ∇ A ( ω ) ω , and the latter is homeomorphic to its square. Hence, Theorem 22applies. (cid:4) Remark We observe here that a claim of Roitman is incorrect. Theorem 6.1and Proposition 6.4 in [9] claim: (1) if b = d < ℵ ω and each X n is compact andhas weight ≤ d then ∇ n X n is b -metrizable (and hence monotonically normal);and (2) if κ < b = d < ℵ ω and the nabla product of countably many compactspaces of weight κ is b -metrizable, then the nabla product of countably manycompact spaces of weight κ + is b -metrizable (and hence monotonically normal).14laim (2) implies claim (1) by finite induction. But both are false. Indeed, thecompact spaces A ( ω ) and ( ω +1) have weight ω , but ∇ A ( ω ) ω and ∇ ( ω +1) ω are not hereditarily normal as shown in Theorem 26 and Theorem 34. Hence,they cannot be κ -metrizable. In the attempted proof of claim (2) it is assumedthat the nabla product under consideration is P b , but this is false, in general,when the factors are not first countable. ∆ -like Characterizations of Monotone Normality In this section we uncover a ∆-like combinatorial principle, namely ∆( α ), whichcharacterizes the monotone normality of a nabla product of ordinals, ∇ ( α + 1) ω .(For an ordinal β , write Lim ( β ) for the set of limit ordinals of β .)Basic neigborhoods of an x in ∇ ( α + 1) ω have the form, N ( x, f ) = { y :for all but finitely many n we have f ( n ) ≤ y ( n ) ≤ x ( n ) if x ( n ) ∈ Lim ( α ) and y ( n ) = x ( n ) if x ( n ) isolated } , where f is in α ω and for all but finitely many n ,if x ( n ) is a limit then f ( n ) < x ( n ). Define a partial order (cid:22) on ∇ ( α + 1) ω bysaying y (cid:22) x if for all but finitely many n we have y ( n ) ≤ x ( n ) and if x ( n ) isisolated then y ( n ) = x ( n ). Note that ∇ ( α + 1) ω , the above basic neighborhoodsand (cid:22) satisfy conditions (a) and (b) of Lemma 3. Hence for ∇ ( α + 1) ω to bemonotonically normal it suffices to halve the neighbornet T ( x ) = ↓ x = N ( x, c ).Next we state the appropriate notion of ‘switching’ elements in this context andthen ∆( α ). Definition 28. Let α be any ordinal and x, y ∈ ∇ ( α + 1) ω . We say that x, y switch if for infinitely many n , x ( n ) < y ( n ) ∈ Lim ( α ) , for infinitely many n , y ( n ) < x ( n ) ∈ Lim ( α ) , and { n ∈ ω : x ( n ) , y ( n ) are isolated and x ( n ) = y ( n ) } is finite. Definition 29. ∆( α ) is the statement: there is F : ∇ ( α + 1) ω → α ω such thatif x, y ∈ ∇ ( α + 1) ω switch, then y ( n ) < F ( x )( n ) < x ( n ) for infinitely many n or x ( n ) < F ( y )( n ) < y ( n ) for infinitely many n . Now we characterize when ∇ ( α + 1) ω is monotonically normal. Proposition 30. The following are equivalent: (1) ∆( α ) holds, (2) ∇ ( α + 1) ω is halvable, and (3) ∇ ( α + 1) ω is monotonically normal.Proof. By the discussion above, it suffices to show the equivalence of (1) and(2 ′ ) ‘the neighbornet T ( x ) = ↓ x is halvable’.For (1) implies (2 ′ ), suppose F is a witness of ∆( α ). Define S ( x ) = N ( x, F ( x )).We check that S halves T ( x ) = ↓ x = N ( x, c ).Take any x and y . Suppose x / ∈ N ( y, c ) and y / ∈ N ( x, c ). Various casesarise, but in all of them we show S ( x ) and S ( y ) are disjoint. If the set { n ∈ ω : x ( n ) , y ( n ) are isolated and x ( n ) = y ( n ) } is infinite, then S ( x ) and S ( y ) aretrivially disjoint. Hence, suppose it is finite. Then the sets N y = { n ∈ ω :15 ( n ) < y ( n ) } and N x = { n ∈ ω : y ( n ) < x ( n ) } are both infinite. Now, if thereare infinitely many n ∈ N y such that y ( n ) is isolated, then [0 , x ( n )] ∩{ y ( n ) } = ∅ ,and thus, S ( x ) and S ( y ) are disjoint; and likewise if there are infinitely many n ∈ N x such that x ( n ) is isolated. Assume, then, that for all but finitely many n ∈ N y and m ∈ N x , x ( m ) , y ( n ) ∈ Lim ( α ). That is, x and y switch. By ∆( α ),we have that S ( x ) ∩ S ( y ) = ∅ .For (2 ′ ) implies (1), consider the neighbornet T ( x ) = N ( x, c ). Then, thereis a neighborhood assignment S that halves T . For x ∈ ∇ ( α +1) ω , let F ( x ) ∈ α ω such that N ( x, F ( x )) ⊆ S ( x ). To see that F satisfies ∆( α ), pick x, y that switch.This implies x / ∈ N ( y, c ) and y / ∈ N ( x, c ), hence by halvability, N ( x, F ( x )) ∩ N ( y, F ( y )) = ∅ . Now it is clear that for infinitely many n ∈ ω , y ( n ) < F ( x )( n ) Corollary 31. (1) If ∆( α ) holds then ∇ α ω is hereditarily paracompact.(2) If ∆( α + 1) holds then (cid:3) ( α + 1) ω is paracompact. It is important to understand the relationship between ∆( α ) and ∆( β ), andespecially the strength of Roitman’s ∆ = ∆( ω + 1). Clearly if β ≥ α then∆( β ) = ⇒ ∆( α ) (because monotone normality is hereditary and ∇ ( α + 1) ω embeds in ∇ ( β + 1) ω ). The next two lemmas give a way to step up. Lemma 32. Let α be an ordinal. Then ∇ (( α.ω )+1) ω = L {∇ n I n : ( I n ) n ∈ I ω } where I = { [0 , α ] } ∪ { ( α.n, α. ( n + 1)] : n ∈ ω } .Proof. The sets in I form an open partition of α.ω . From ‘open’ we see thateach ∇ n I n is open in ∇ (( α.ω ) + 1) ω . While from ‘partition’, and the factthat we take every sequence of members of I , we see that the ∇ n I n partition ∇ (( α.ω ) + 1) ω . (cid:4) Lemma 33. If ∆( α + 1) holds, then ∆( α.ω ) holds.Proof. Observe that each ∇ n I n from the preceding lemma is homeomorphic to ∇ (( α + 1)+1) ω , which is monotonically normal under ∆( α +1) (Proposition 30).Since a disjoint sum of monotonically normal spaces is monotonically normal,we can apply Proposition 30 again to complete the proof. (cid:4) As seen above, under d = ω , we have that ∇ α ω is monotonically normal for all α < ω . The next two results provide a sharp contrast. Theorem 34. The space ∇ ( ω + 1) ω is not hereditarily normal.Proof. Let S = E ω ω = { α ∈ ω : cf ( α ) = ω } to ∇ ( ω + 1) ω . Then S is astationary subset of ω . Note that S = E ω ω ∪ { c ω } , and its G δ -modification, S δ are equal. Hence S and S both embed into ∇ ( ω + 1) ω . Since, ∇ ( ω + 1) ω is homeomorphic to its square, to complete the proof, apply Lemma 5. (cid:4) heorem 35. If b > ω , then ∇ ( ω + 1) ω is not hereditarily normal.Proof. Let L be the subspace of ω + 1 consisting of the isolated points alongwith ω . Then L δ = L is homeomorphic to L ( ω ), and so both L ( ω ) and ∇ ( ω + 1) ω embed into ∇ ( ω + 1) ω , which is homeomorphic to its square. Hence,Theorem 22 applies. (cid:4) ∆ -like Characterizations of Monotone Normality For this section, { ( X n , d n ) : n ∈ ω } will be a family of metric spaces. For x, y ∈ ∇ n X n and f ∈ ω ω , define N ( x, f ) = ∇ n B n ( x ( n ) , / f ( n ) ) and M ( x, f ; y ) = { n ∈ ω : y ( n ) / ∈ B n ( x ( n ) , / f ( n ) ) } , where B n ( a, ε ) is { a } if a is isolated, and isotherwise the ε -ball in the metric d n .We say that ( x, f ) , ( y, g ) ∈ ∇ n X n × ω ω switch if M ( x, f ; y ) and M ( y, g ; x )are almost disjoint infinite sets. Observe that switching elements ( x, f ) and( y, g ) imply y / ∈ N ( x, f ) and x / ∈ N ( y, g ). Definition 36. Let { ( X n , d n ) : n ∈ ω } be a family of metric spaces. Then ∆(( X n , d n ) n ) is the statement: there is F : ∇ n X n × ω ω → ω ω , write f x := F ( x, f ) , such that if ( x, f ) , ( y, g ) ∈ dom F switch, then / f x ( n ) + / g y ( n ) For b = d : Let { f α : α < d } be a scale such that f α ≤ ∗ -dominates { f β : β < α } .Let ↓ f α = { f ∈ ω : f ≤ ∗ f α } and define F : ∇ n X n × ω ω → ω ω as F ( x, f ) = 2 f α if and only if α is the least such that f ∈ ↓ f α .Pick switching elements ( x, f ) , ( y, g ) ∈ ∇ n X n × ω ω . We may assume that f ∈ ↓ f α and g ∈ ↓ f β , for minimum β, α and β ≤ α . Then, g ≤ ∗ f β ≤ ∗ f β ≤ ∗ f α . Now, if n ∈ M ( y, g ; x ) then x ( n ) / ∈ B n ( y ( n ) , g ( n )), that is, / g ( n ) Question 39. Is ¬ ∆ consistent? Or is ∆ true in ZFC ? Suppose ∆ = ∆( ω + 1) were true in ZFC , so ∇ ( ω + 1) ω is monotonicallynormal. Then it seems implausible to the authors that ∇ ( ω.ω + 1) ω would notalso be monotonically normal, in other words ∆( ω.ω + 1) could be consistentlyfalse. If that is correct then, in ZFC , it should be possible to deduce ∆( ω.ω + 1)from ∆( ω + 1). Problem 40. Show, in ZFC , that ∆( ω + 1) = ⇒ ∆( ω.ω + 1) . We know ∇ A ( ω ) ω is monotonically normal under d = ω ; and that ∇ A ( ω ) ω monotonically normal implies b = ω . This leaves a gap. Problem 41. What is the consistency strength of ‘ ∇ A ( ω ) ω is monotonicallynormal’. Does b = ω suffice? We have seen that Roitman’s ∆ is equivalent to ∇ ( ω + 1) ω being monotoni-cally normal. Hence ∆ is a sufficient condition for ∇ ( ω + 1) ω to be hereditarilyparacompact. A natural question then is it necessary? But a more fundamen-tal problem is to find necessary combinatorial conditions for ∇ ( ω + 1) ω to beparacompact. That would open a path to showing the independence of the boxproduct problem. Problem 42. Find combinatorial properties (in the style of ∆ ) implied by‘ ∇ ( ω + 1) ω is paracompact’. eferences [1] Z. Balogh and M.E. Rudin. Monotone normality. Topology and its Appli-cations , 47(2):115–127, 1992.[2] E. K. van Douwen. Covering and separation properties of box products. In Surveys in general topology , pages 55–129. Elsevier, 1980.[3] P.M. Gartside. Nonstratifiability of topological vector spaces. Topology andits Applications , 86(2):133–140, 1998.[4] P.M. Gartside. Monotone normality in products. Topology and its Appli-cations , 91(3):181–195, 1999.[5] K.P. Hart. More remarks on Souslin properties and tree topologies. Topol-ogy and its Applications , 15:151–158, 1983.[6] K. Kunen. Paracompactness of box products of compact spaces. Transac-tions of the American Mathematical Society , 240:307–316, 1978.[7] L.B. Lawrence. The box product of countably many copies of the rationalsis consistently paracompact. Transactions of the American MathematicalSociety , pages 787–796, 1988.[8] J. Roitman. More paracompact box products. Proceedings of the AmericanMathematical Society , 74(1):171–176, 1979.[9] J. Roitman. Paracompactness and normality in box products: old and new.In Set theory and its applications , pages 157–181, 2011.[10] J. Roitman. Box products of one point compactifications and related re-sults. In Topol. Proc. , volume 44, pages 197–206, 2014.[11] J. Roitman and P.M. Gartside. Personal communication, 2019-05-17.[12] J. Roitman and S. Williams. Paracompactness, normality, and relatedproperties of topologies on infinite products. Topology and its Applications ,195:79–92, 2015.[13] S. Williams. Box products. In