Pointwise convergence of partial functions: The Gerlits-Nagy Problem
aa r X i v : . [ m a t h . GN ] O c t POINTWISE CONVERGENCE OF PARTIAL FUNCTIONS:THE GERLITS–NAGY PROBLEM
TAL ORENSHTEIN AND BOAZ TSABAN
Abstract.
For a set X ⊆ R , let B ( X ) ⊆ R X denote the space of Borel real-valuedfunctions on X , with the topology inherited from the Tychonoff product R X . Assumethat for each countable A ⊆ B ( X ), each f in the closure of A is in the closure of A underpointwise limits of sequences of partial functions. We show that in this case, B ( X ) iscountably Fr´echet–Urysohn, that is, each point in the closure of a countable set is alimit of a sequence of elements of that set. This solves a problem of Arnold Miller. Thecontinuous version of this problem is equivalent to a notorious open problem of Gerlitsand Nagy. Answering a question of Salvador Herna´ndez, we show that the same resultholds for the space of all Baire class 1 functions on X .We conjecture that, in the general context, the answer to the continuous version ofthis problem is negative, but we identify a nontrivial context where the problem has apositive solution.The proofs establish new local-to-global correspondences, and use methods of infinite-combinatorial topology, including a new fusion result of Francis Jordan. Introduction and basic results
Let X ⊆ R . C( X ) is the family of all continuous real-valued functions on X . Weconsider C( X ) with the topology inherited from the Tychonoff product R X . A basis ofthe topology is given by the sets[ f ; x , . . . , x k ; ǫ ] := { g ∈ C( X ) : ( ∀ i = 1 , . . . , k ) | g ( x i ) − f ( x i ) | < ǫ } , where f ∈ C( X ), k ∈ N , x , . . . , x k ∈ X , and ǫ is a positive real number. This is the topology of pointwise convergence , where a sequence (more generally, a net) f n convergesto f if and only if for each x ∈ X , the sequence of real numbers f n ( x ) converges to f ( x ).By definition, the (topological) closure A of a set A ⊆ C( X ) is the set of all f ∈ C( X )such that, for all k ∈ N , x , . . . , x k ∈ X , and positive ǫ , there is an element g ∈ A suchthat | g ( x i ) − f ( x i ) | < ǫ for i = 1 , . . . , k . (Equivalently, there is a net in A convergingpointwise to f .) C( X ) is metrizable only when X is countable, and thus it makes sense toask, when X is not countable, when do limits of sequences determine the closure of sets.For a topological space Y and A ⊆ Y , the closure of A under limits of sequences isthe smallest set C ⊆ Y containing A , such that for each convergent (in Y ) sequence ofelements of C , the limit of this sequence is also in C . The closure of A under limits ofsequences is contained in the topological closure A of A in Y .Gerlits [6], and independently Pytkeev [19], proved that if limits determine the closurein C( X ), then indeed it suffices to take limits once. Theorem 1.1 (Gerlits, Pytkeev) . Let X be a Tychonoff space. Assume that, for each A ⊆ C( X ) , each f ∈ A (closure in C( X ) ) belongs to the closure (in C( X ) ) of A under limits of sequences. Then, for each A ⊆ C( X ) , each f ∈ A is a limit of a sequence ofelements of A . The properties of C( X ) in the premise and in the conclusion of Theorem 1.1 are oftennamed sequential and Fr´echet–Urysohn , respectively.Consider now partial functions f : X → R , that is, functions whose domain is a (notnecessarily proper) subset of X . Definition 1.2.
Let f , f , · · · : X → R be partial functions. The partial limit function f = lim n f n is the partial real-valued function on X , with dom( f ) being the set of all x such that f n ( x ) is eventually defined and converges, defined by f ( x ) = lim n f n ( x ) for each x ∈ dom( f ).Thus, for f , f , · · · ∈ C( X ), the ordinary limit lim n f n exists in C( X ) if and only if thedomain of the partial limit function f = lim n f n is X , and f is continuous. The partiallimit of a sequence of partial functions always exist, though it may be the empty function. Definition 1.3.
For a set A of partial functions f : X → R , the closure of A under partiallimits of sequences , partlims( A ), is the smallest set C of partial functions f : X → R ,such that A ⊆ C and for each sequence in C , the partial limit of this sequence is also in C .Thus, the closure, in C( X ), of a set A ⊆ C( X ) under limits of sequences is a subset ofC( X ) ∩ partlims( A ). Lemma 1.4.
For each A ⊆ C( X ) , C( X ) ∩ partlims( A ) is contained in A , the closure of A in C( X ) .Proof. The definition of basic open sets in C( X ) (or R X ) may be extended to partialfunctions, by letting [ f ; x , . . . , x k ; ǫ ] be the set of all partial g : X → R such that x , . . . , x k ∈ dom( g ) and | g ( x i ) − f ( x i ) | < ǫ , for all i = 1 , . . . , k .Assume that f / ∈ A . Take x , . . . , x k ∈ X and ǫ >
0, such that A ∩ [ f ; x , . . . , x k ; ǫ ] = ∅ . Then A ⊆ [ f ; x , . . . , x k ; ǫ ] c , and [ f ; x , . . . , x k ; ǫ ] c is closed under limits of partialfunctions: Assume and g = lim n g n ∈ [ f ; x , . . . , x k ; ǫ ]. Then x , . . . , x k ∈ dom( g ), and | g ( x i ) − f ( x i ) | < ǫ , and therefore the same holds for g n , for all but finitely many n . Inparticular, it cannot be the case that g , g , · · · ∈ [ f ; x , . . . , x k ; ǫ ] c .It follows that f is not in the closure of A under partial limits of sequences. (cid:3) In 1982, Gerlits and Nagy published their seminal paper [7]. This paper has generatedover 200 subsequent papers and a rich theory. Among the problems posed in [7], onlyone remains open. On its surface, the
Gerlits–Nagy Problem is a combinatorial one, andwe defer its combinatorial formulation to Section 4, where we prove that the Gerlits–Nagy Problem is equivalent to the following fundamental problem, dealing with pointwiseconvergence of real-valued functions.
Problem 1.5 (Gerlits–Nagy [7]) . Assume that, for each A ⊆ C( X ) , each f ∈ A belongs tothe closure of A under partial limits of sequences. Does it follow that, for each A ⊆ C( X ) ,each f ∈ A is a limit of a sequence of elements of A ? HE GERLITS–NAGY PROBLEM 3
In the
Second Workshop on Coverings, Selections, and Games in Topology (Lecce,Italy, 2005), Arnold Miller delivered a plenary lecture, where he posed the variant of theGerlits–Nagy Problem, dealing with
Borel rather than continuous functions [16].Let B( X ) ⊆ R X be the family of all Borel real-valued functions on X . One may considerthe questions discussed above also for B( X ), with the following reservation: Here, onemust restrict attention to countable A ⊆ B( X ), as we now show.Each of the properties mentioned in the above discussion implies that C( X ) is countablytight , that is, each point in the closure of a set is in the closure of a countable subset ofthat set. The standard proof would be by transfinite induction on the countable ordinals,but we adopt here an argument given in [2]. Proposition 1.6.
Let X be a topological space. Assume that, for each A ⊆ C( X ) , each f ∈ A belongs to the closure of A under partial limits of sequences. Then C( X ) iscountably tight.Proof. Let A ⊆ C( X ). By Lemma 1.4, partlims( A ) ∩ C( X ) ⊆ A . Thus, it suffices to showthat for each f ∈ A , there is a countable D ⊆ A such that f ∈ partlims( D ).Let B = S { partlims( D ) : D ⊆ A is countable } . Then B is closed under partial limitsof sequences: Let f , f , · · · ∈ B . Then there are countable D , D , . . . ⊆ A , such that f n ∈ partlims( D n ) for all n . Let D = S n D n . Then f , f , · · · ∈ partlims( D ), and thereforelim n f n ∈ partlims( D ) ⊆ B .Thus, partlims( A ) ⊆ B , as required. (cid:3) By a classical result of Arhangel’ski˘ı, C( X ) is countably tight for all X ⊆ R (indeed,for all topological spaces X such that all finite powers of X are Lindel¨of). However,B( X ) is not countably tight, unless X is countable (in which case, R X , and thus B( X ),is metrizable).We denote by the constant function identically equal to 1 on X . Proposition 1.7.
Let X be an uncountable space, where each singleton is Borel. Then B( X ) is not countably tight.Proof. Take A = { χ F : F ⊆ X finite } ⊆ B( X ), where χ F denotes the characteristicfunction of F . Then the constant function is in A . Let D = { χ F n : n ∈ N } ⊆ X . Take a ∈ X \ S n F n . Then χ F n ( a ) = 0 for all n , and thus / ∈ D . (cid:3) Problem 1.8 (Miller 2005 [16]) . Assume that, for each countable A ⊆ B( X ) , each f ∈ A belongs to the closure of A under partial limits of sequences. Does it follow that, for eachcountable A ⊆ B( X ) , each f ∈ A is a limit of a sequence of elements of A ? Our main result (Section 2) is a solution, in the affirmative, of Miller’s problem. At theend of the second named author’s talk in the conference
Functional Analysis in Valencia2010 , Salvador Herna´ndez asked what is the solution to Miller’s Problem when consideringBaire class 1 functions (i.e., functions which are pointwise limits of sequences of continuousfunctions). We solve Herna´ndez’s problem in Section 3. Finally, we establish severalresults concerning the original Gerlits–Nagy Problem, and pose some related problems.
TAL ORENSHTEIN AND BOAZ TSABAN Borel functions (Miller’s Problem)
We solve Miller’s Problem 1.8 in the affirmative. Indeed, we do so not only for sets X ⊆ R , but for all topological spaces X . Theorem 2.1.
Let X be a topological space. Assume that, for each countable A ⊆ B( X ) ,each f ∈ A belongs to the closure of A under partial limits of sequences. Then for eachcountable A ⊆ B( X ) , each f ∈ A is a limit of a sequence of elements of A . The proof is divided naturally into four steps. For brevity, we make the followingconvention, that will hold throughout the paper.
Convention 2.2.
Let X be a topological space. We say that U is a cover of X if X = S U ,but X / ∈ U . By
Borel cover of X we always mean a countable family U of Borel subsetsof X , such that the union of all members of U is X . Step 1: Local to global.
We deduce from the given local property of B( X ), a global prop-erty of X . Definition 2.3 (Gerlits–Nagy [7]) . A cover U of X is an ω -cover of X if each finite F ⊆ X is contained in a member of U .For sets B , B , . . . , let liminf n B n = [ m \ n ≥ m B n , that is, the set of all x which belong to B n for all but finitely many n . Let LI( U ) be theclosure of U under the operator liminf.A basic property of liminf n B n is that it does not depend on the first few sets B n . Lemma 2.4.
Let X be a topological space. Assume that, for each countable A ⊆ B( X ) ,each f ∈ A belongs to partlims( A ) . Then for each Borel ω -cover U of X , X ∈ LI( U ) .Proof. Let U be a Borel ω -cover of X . Take A = { χ U : U ∈ U } . Then A ⊆ B( X ) iscountable, and ∈ A . Thus, ∈ partlims( A ).As each f ∈ A is { , } -valued, and limits of convergent sequences of 0’s and 1’s mustbe either 0 or 1, each f in partlims( A ) is { , } -valued. Let C be the set of all partial { , } -valued functions f on X , such that f − (1) ∈ LI( U ). Then A ⊆ C , and C isclosed under partial limits of sequences. Indeed, let f , f , · · · ∈ C , and f = lim n f n . Aslim n f n ( x ) = f ( x ) and the functions f n are { , } -valued, f − (1) = liminf n f − n (1) ∈ LI( U ).Therefore, partlims( A ) is contained in C , and in particular ∈ C , that is, there is B ∈ LI( U ) such that X = − (1) ⊆ B . Thus, X = B ∈ LI( U ). (cid:3) Step 2: A selective property.
Definition 2.5.
For a family F of subsets of X , let F ↓ = { B ⊆ X : ( ∃ A ∈ F ) B ⊆ A } , the closure of F under taking subsets. HE GERLITS–NAGY PROBLEM 5
For a family F of sets, S F (without running index) denotes the union of all membersof F . We say that a family of sets V refines another family U if each V ∈ V is containedin some U ∈ U . The following result may be obtained by following arguments of Gerlitsand Nagy [7] and arguments of Nowik, Scheepers, and Weiss [17], proved for open covers(under certain hypotheses on the space X ). We provide a different, direct proof, whichmakes no assumption on X . Proposition 2.6.
Let X be a topological space. Assume that for each Borel ω -cover U of X , X ∈ LI( U ) . Then for each sequence U , U , . . . of Borel covers of X , there are finitesets F ⊆ U , F ⊆ U , . . . , such that for each x ∈ X , x ∈ S F n for all but finitely many n .Proof. By moving to refinements, we may assume that for each n , the elements of U n arepairwise disjoint, and U n +1 refines U n . This way, if there are infinitely many n such that U n contains a finite subcover F n of X , then this is true for all n and the required assertionfollows immediately. Thus, we may assume that for each n , U n does not contain a finitesubcover of X .Let B = n liminf n [ F n : ( ∀ n ) F n is a finite subset of U n o . We must prove that X ∈ B .LI( B ↓ ) = B ↓ : For each k , assume that B k ⊆ liminf n S F kn , with each F kn a finite subsetof U n . Take F n = F n ∪ F n ∪ · · · ∪ F nn for each n . Thenliminf k B k ⊆ liminf k liminf n [ F kn ⊆ liminf n [ F n ∈ B , and thus liminf n B n ∈ B ↓ .Thus, LI( B ) ⊆ B ↓ , and therefore if X ∈ LI( B ) then X ∈ B . B is an ω -cover of X andits elements are Borel, but B is in general not countable, and thus we cannot apply thepremise of the lemma. To overcome this problem, we use a trick similar to one in [7]:Define A = [ n ∈ N n[ F : F ⊆ U n , |F | = n o . A is a Borel ω -cover of X , and therefore by the premise of the lemma, X ∈ LI( A ) ⊆ LI( A ↓ ∪ B ↓ ). As X / ∈ A , it remains to show that LI( A ↓ ∪ B ↓ ) = A ↓ ∪ B ↓ .Let B , B , · · · ∈ A ↓ ∪ B ↓ . As liminf n B n ⊆ liminf n B m n for each increasing sequence m n , and A ↓ ∪ B ↓ is closed downwards, we may move to subsequences at our convenience.If B n ∈ B ↓ for infinitely many n , then by moving to a subsequence we may assume that B n ∈ B ↓ for all n , and therefore liminf n B n ∈ LI( B ↓ ) = B ↓ ⊆ A ↓ ∪ B ↓ . In the remainingcase, by moving to a subsequence, we may assume that B n ∈ A ↓ for all n .Consider first the case where, after moving to an appropriate subsequence of B , B , . . . ,there is an increasing sequence k n such that B n ⊆ S F k n , F k n ⊆ U k n with |F k n | = k n , for Given a Borel cover U = { U n : n ∈ N } , the Borel cover { U n \ ( U ∪ · · · ∪ U n − ) : n ∈ N } refines U , andits elements are pairwise disjoint. Given two Borel covers U , V whose elements are pairwise disjoint, theBorel cover { U ∩ V : U ∈ U , V ∈ V} refines U and V , and in particular its elements are pairwise disjoint. TAL ORENSHTEIN AND BOAZ TSABAN all n . As the covers U n are getting finer with n , for each i / ∈ { k n : n ∈ N } there is a finite F i ⊆ U i such that S F i contains S F k n for the first n with i < k n . Thenliminf n B n ⊆ liminf n [ F n ∈ B , as required.Finally, there remains the case where, after moving to an appropriate subsequence of B , B , . . . , there is k such that for each n , there is F n ⊆ U k with |F n | = k , such that B n ⊆ S F n . Let B = lim inf B n . We will show that B ∈ A ↓ . We may assume that B = ∅ . Take x ∈ B , and U ∈ U k such that x ∈ U . If B ⊆ U , then B ∈ A ↓ .Otherwise, take x ∈ B \ U , and U ∈ U k such that x ∈ U . Continue in the samemanner until it is impossible to proceed, but not more than k steps, to have x , . . . , x i ∈ B , where i ≤ k , and distinct (and therefore disjoint) U , . . . , U i ∈ U k . If i < k , then B ⊆ U ∪ · · · ∪ U i , a union of less than k elements of U k , and thus B ∈ A ↓ . Otherwise i = k , and for all but finitely many n , x , . . . , x k ∈ B n ⊆ S F n , and as the elements of U k are pairwise disjoint, F n = { U , . . . , U k } for all but finitely many n . Consequently, B ⊆ liminf n S F n = U ∪ · · · ∪ U k ∈ A , and therefore B ∈ A ↓ . (cid:3) Step 3: A stronger selective property.
The selective property in the following theorem isstronger ([23], or Lemma 4.1) than the one introduced in the previous step. In its originalformulation [16], Miller’s Problem 1.8 asks whether the following theorem is true.
Theorem 2.7.
Assume that for each Borel ω -cover U of X , X ∈ LI( U ) . Then in fact,for each Borel ω -cover U of X , there are U , U , . . . ∈ U such that X = liminf n U n .Proof. Let B = { liminf n U n : U , U , · · · ∈ U } ↓ . It suffices to show that LI( B ) = B . Let B , B , · · · ∈ B , and B = liminf n B n . For each n ,take U n , U n , · · · ∈ U such that B n ⊆ liminf m U nm . Then for each n , the sets V nm = T k ≥ m U nk are increasing to B n , and therefore the sets V nm ∪ ( X \ B n ) are increasing to X .Applying Proposition 2.6 to the covers U n = { V nm ∪ ( X \ B n ) : m ∈ N } , there are m n such that X = liminf n V nm n ∪ ( X \ B n ) (since the covers are increasing, it suffices to pickone element from each cover). As liminf n B n = B , we have that B ⊆ (liminf n V nm n ∪ ( X \ B n )) ∩ B ⊆ liminf n V nm n ⊆ liminf n U nm n , and therefore B ∈ B . (cid:3) Step 4: Global to local.
The following lemma and its proof are, in the open/continuouscase, due to Gerlits and Nagy [7]. Their argument also applies to the Borel case.
Lemma 2.8.
Assume that for each Borel ω -cover U of X , there are U , U , · · · ∈ U suchthat X = liminf n U n . Then for each countable A ⊆ B( X ) , each f ∈ A is a pointwise limitof a sequence of elements of A .Proof. We may assume, by adding the function − f to all considered functions, that f = , the constant 1 function. For each n , let U n = { g − [(1 − /n, /n )] : g ∈ A } .As ∈ A , U n is a (Borel) ω -cover of X . By Theorem 2.7, there are g n ∈ A such that X = liminf n g − n [(1 − /n, /n )]. Then = lim n g n . (cid:3) This completes the proof of Theorem 2.1.
HE GERLITS–NAGY PROBLEM 7 Baire class 1 functions (Herna´ndez’s Problem)
The following Theorem, which strengthens Theorem 2.1 (in the realm of perfectly nor-mal spaces), answers in the positive a question of Salvador Hern´andez.A topological space X is perfectly normal if it is normal (any two disjoint closed sets havedisjoint neighborhoods), and each open subset of X is F σ , that is, a union of countablymany closed subsets of X . For example, metric spaces are perfectly normal.A function f : X → R is of Baire class 1 if f is the pointwise limit of a sequence ofcontinuous real-valued functions on X . Let Baire ( X ) ⊆ R X denote the subspace of allBaire class 1 functions f : X → R . Theorem 3.1.
Let X be a perfectly normal topological space. Assume that, for eachcountable A ⊆ Baire ( X ) , each f ∈ A (closure in Baire ( X ) ) belongs to the closure of A under partial limits of sequences. Then for each countable A ⊆ Baire ( X ) , each f ∈ A (closure in Baire ( X ) ) is a limit of a sequence of elements of A .Moreover, for each countable A ⊆ B( X ) , each f ∈ A (closure in B( X ) ) is a limit of asequence of elements of A .Proof. As the closure in a subspace Y of R X is equal to the intersection of the closure in R X and Y , and Baire ( X ) ⊆ B( X ), it suffices to prove the second assertion. We followthe proof steps of Theorem 2.1, and modify them when needed.A set A ⊆ X is ∆ if both A and X \ A are F σ . The family ∆ ( X ) of all ∆ subsetsof X forms an algebra of sets, that is, it is closed under finite unions and complements(and therefore also under finite intersections and set differences). This fact is appliedrepeatedly when following the steps in the proof of Theorem 2.1.A function f : X → R is ∆ -measurable if for each open U ⊆ R , f − [ U ] is ∆ . For each∆ set U ⊆ X , χ U is ∆ -measurable.The following lemma is proved for the metrizable case in [12, Lemma 24.12]. The proofthere uses only Urysohn’s Lemma, which applies for all normal spaces. Lemma 3.2 (folklore) . Let X be a normal space, and U be a ∆ subset of X . Then χ U is of Baire class 1.Proof. Let F n ⊆ X be closed, and G n ⊆ X be open, such that F n ⊆ F n +1 ⊆ U ⊆ G n +1 ⊆ G n for all n , and U = S n F n = T n G n . By Urysohn’s Lemma, there is for each n acontinuous function f n : X → R such that f n ( x ) = 1 for all x ∈ F n and f n ( x ) = 0 for all x / ∈ G n . Then lim n f n ( x ) = χ U ( x ) for all x ∈ X . (cid:3) Thus, arguing as in Step 1 of Theorem 2.1, we have that for each countable ∆ ω -cover U of X , X ∈ LI( U ).The arguments of Step 2 show the following. Proposition 3.3.
Assume that for each countable ∆ ω -cover U of X , X ∈ LI( U ) . Thenfor each sequence U , U , . . . of countable ∆ covers of X , there are finite sets F ⊆U , F ⊆ U , . . . , such that for each x ∈ X , x ∈ S F n for all but finitely many n . (cid:3) In particular, as X is perfectly normal, X has the property in the conclusion of Propo-sition 3.3 for closed sets. We use the following strong result of Bukovsk´y, Rec law, andRepick´y. TAL ORENSHTEIN AND BOAZ TSABAN
Lemma 3.4 (Bukovsk´y–Rec law–Repick´y [1]) . Let X be a perfectly normal space. Assumethat for each sequence U , U , . . . of countable closed covers of X , there are finite sets F ⊆ U , F ⊆ U , . . . , such that for each x ∈ X , x ∈ S F n for all but finitely many n .Then the same holds for each sequence U , U , . . . of Borel covers of X . The property established in Lemma 3.4 implies that every Borel subset of X is F σ (e.g.,[23]), and thus every Borel set is ∆ . By the property established before Proposition 3.3,we have that, for each countable Borel ω -cover U of X , X ∈ LI( U ).Thus, theorem 2.7 and Step 4 apply, and the proof is completed. (cid:3) Remark . The proof of Theorem 3.1 shows that it suffices to assume that for eachcountable set A of ∆ -measurable real-valued functions on X , the closure of A in thespace of all ∆ -measurable real-valued functions on X is contained in partlims( A ).4. Continuous functions (Gerlits–Nagy’s Problem)
Thus far, we have refrained from using the notation of the field of selective properties,despite their playing important role in the proofs. However, as we are about to makea more extensive use of the theory, we give here the necessary introduction. Readerswho wish to learn more on the topic and its history are referred to any of its surveys[22, 13, 24].Let X be a topological space. Let O( X ) be the family of all open covers of X . Definethe following subfamilies of O( X ): U ∈ Ω( X ) if U is an ω -cover of X . U ∈ Γ( X ) if U isinfinite, and each element of X is contained in all but finitely many members of U .Some of the following statements may hold for families A and B of covers of X . (cid:0) AB (cid:1) : Each element of A contains an element of B . S ( A , B ) : For all U , U , · · · ∈ A , there are U ∈ U , U ∈ U , . . . such that { U n : n ∈ N } ∈ B . S fin ( A , B ) : For all U , U , · · · ∈ A , there are finite F ⊆ U , F ⊆ U , . . . such that S n F n ∈ B . U fin ( A , B ) : For all U , U , · · · ∈ A , none containing a finite subcover, there are finite F ⊆ U , F ⊆ U , . . . such that { S F n : n ∈ N } ∈ B .We say, e.g., that X satisfies S (O , O) if the statement S (O( X ) , O( X )) holds. This way, S (O , O) is a property of topological spaces, and similarly for all other statements andfamilies of covers. Under some mild hypotheses on the considered topological spaces, eachnontrivial property among these properties, where A , B range over O , Ω , Γ, is equivalentto one in Figure 1, named after Scheepers in recognition of his seminal contribution tothe field. In this diagram, an arrow denotes implication.Other types of covers, most notably
Borel covers, were also considered in this context.We say, for example, that X satisfies S (Ω , Ω) for Borel covers if S (Ω( X ) , Ω( X )) holds,when redefining Ω( X ) to consist of all countable Borel ω -covers of X .For clarity of notation, we identify a property with the family of topological spaces (ofa certain type, which should be clear from the context) satisfying it. HE GERLITS–NAGY PROBLEM 9 U fin (O , Γ)Hurewicz [9] / / U fin (O , Ω) / / S fin (O , O)Menger [14] S fin (Γ , Ω) ❥❥❥❥❥❥❥❥❥❥ S (Γ , Γ) / / ❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥ S (Γ , Ω) / / ❥❥❥❥❥❥❥❥ S (Γ , O) ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥ S fin (Ω , Ω) O O S (Ω , Γ)Gerlits–Nagy [7] / / O O S (Ω , Ω) O O / / ❧❧❧❧❧❧❧❧ S (O , O)Rothberger [20] O O Figure 1.
The Scheepers DiagramThe property deduced in Theorem 2.6 is U fin (O , Γ) for Borel covers. For Borel cov-ers, U fin (O , Γ) = S (Γ , Γ) [23], and using this the proof of Theorem 2.7 can be slightlysimplified.Gerlits and Nagy [7] proved the following lemma for Hausdorff spaces. We will see thatit holds for arbitrary topological spaces.
Lemma 4.1. (cid:0) ΩΓ (cid:1) = S (Ω , Γ) (for general topological spaces).Proof. Assume that X satisfies (cid:0) ΩΓ (cid:1) , and let U , U , . . . be open ω -covers of X . We mayassume that for each n , U n +1 refines U n .For each n , enumerate U n = { U nm : m ∈ N } . Let V m = U m for all m . Define W = [ n ∈ N { V n ∩ U nm : m ∈ N } . W is an open ω -cover of X . Thus, there are W , W , · · · ∈ W such that X = liminf k W k .Fix n . As V n = X , it is not possible that W k ∈ { V n ∩ U nm : m ∈ N } for infinitely many k .Since the sets U nm are increasing with m , we may assume that there is at most one W k ineach set W n = { V n ∩ U nm : m ∈ N } . For each n , let r n ≥ n be the first such that there issome W k in W r n . Since the covers U n get finer with n , we can pick for each n an element U nm n ∈ U n containing the W k which is in W r n . Then X = liminf k W k ⊆ liminf n U nm n , andtherefore liminf n U nm n = X . (cid:3) Using Lemma 4.1, Gerlits and Nagy proved the following fundamental local-to-globalcorrespondence result.
Theorem 4.2 (Gerlits–Nagy [7]) . For Tychonoff spaces X , the following properties areequivalent: (1) For each A ⊆ C( X ) , each f ∈ A is a limit of a sequence of elements of A (i.e., C( X ) is Fr´echet–Urysohn). (2) X satisfies (cid:0) ΩΓ (cid:1) . We establish a similar result for the other major property studied in the present paper.To this end, we need the following definition and a lemma.
Definition 4.3. L( X ) is the family of open covers of X such that X ∈ LI( U ). Theorem 2.7 tells that (cid:0) ΩL (cid:1) = (cid:0) ΩΓ (cid:1) for Borel covers. In particular, using that (cid:0) ΩΓ (cid:1) = S (Ω , Γ) for Borel covers, we have that (cid:0) ΩL (cid:1) = S (Ω , L) for Borel covers. The last assertionalso holds in the open case, but a different proof is required.
Lemma 4.4. (cid:0) ΩL (cid:1) = S (Ω , L) = S fin (Ω , L) .Proof. As S (Ω , L) implies S fin (Ω , L), which in turn implies (cid:0) ΩL (cid:1) , it remains to prove that (cid:0) ΩL (cid:1) implies S (Ω , L). To this end, it suffices to prove that (cid:0) ΩL (cid:1) implies S (Ω , Ω). S (Ω , Ω)is equivalent to having all finite powers of X satisfy S (O , O) [21]. Gerlits and Nagy [7]proved that (cid:0) ΩL (cid:1) implies S (O , O). Thus, it remains to prove that (cid:0) ΩL (cid:1) is preserved by finitepowers.Assume that X satisfies (cid:0) ΩL (cid:1) , and let k ∈ N . Let U be an open ω -cover of X k . Thenthere is an open ω -cover V of X such that V ′ = { V k : V ∈ V} refines U [11]. Then X ∈ LI( V ). For arbitrary sets B , B , . . . , liminf n ( B n ) k = (liminf n B n ) k . Thus, X k ∈{ B k : B ∈ LI( V ) } = LI( V ′ ), and therefore X k ∈ LI( U ). (cid:3) Theorem 4.5.
For Tychonoff spaces X , the following properties are equivalent: (1) For each A ⊆ C( X ) , A ⊆ partlims( A ) . (2) X satisfies (cid:0) ΩL (cid:1) (that is, for each open ω -cover U of X , X ∈ LI( U ) ).Proof. (1 ⇒
2) For partial functions f and g , g ◦ f is the partial function with domain { x ∈ dom( f ) : f ( x ) ∈ dom( g ) } , defined as usual by g ◦ f ( x ) = g ( f ( x )).For a surjection ϕ : X → Y and partial functions f n : Y → R , the domain of lim n ( f n ◦ ϕ )is ϕ − [dom(lim n f n )], and lim n ( f n ◦ ϕ ) = (lim n f n ) ◦ ϕ . Thus, we have the following. Lemma 4.6.
Assume that for each A ⊆ C( X ) , each A ⊆ partlims( A ) . Then everycontinuous image of X has the same property. (cid:3) A topological space is zero-dimensional if its clopen (simultaneously closed and open)sets form a base for its topology. An argument similar to one in [7] gives the following.
Lemma 4.7.
Let X be a Tychonoff space. Assume that for each A ⊆ C( X ) , each f ∈ A belongs to the closure of A under partial limits of sequences. Then X is zero-dimensional.Proof. It suffices to prove that [0 ,
1] is not a continuous image of X . Indeed, for eachopen U ⊆ X and each a ∈ U , let Ψ : X → [0 ,
1] be continuous, such that Ψ( a ) = 0and Ψ( x ) = 1 for all x ∈ X \ U . Take r ∈ [0 ,
1] which is not in the image of Ψ. ThenΨ − [[0 , r )] is a clopen neighborhood of x contained in U .Assume that [0 ,
1] is a continuous image of X . Let A ⊆ C([0 , f : [0 , → [0 ,
1] such that the Lebesgue measure of f − [(1 / , /
2. Then is in the closure of A . Let C be the set of all partial f : [0 , → [0 ,
1] suchthat f − [(1 / , / C is closedunder partial limits of sequences and contains A , but / ∈ C ; a contradiction. (cid:3) Let U be an open ω -cover of X . As X is zero-dimensional, U can be refined to a clopen ω -cover of X by replacing each U ∈ U with all finite unions of clopen subsets of U . Now,for each clopen U the function χ U is continuous, and is in the closure of { χ U : U ∈ U } .By (1), is in the closure of { χ U : U ∈ U } under partial limits of sequences. Continue asin the proof of Lemma 2.4. HE GERLITS–NAGY PROBLEM 11 (2 ⇒
1) In (1), by adding − f to all of the involved partial functions, it sufficesto consider the case f = . Let A ⊆ C( X ), and assume that ∈ A . For each n , let U n = { f − [(1 − /n, /n )] : f ∈ A } . U n is an open ω -cover of X . By Lemma 4.4,there are f , f , · · · ∈ A such that X ∈ LI( { f − n [(1 − /n, /n )] : n ∈ N } ).We claim that A = ( { f − n [(1 − /n, /n )] : n ∈ N } ∪ { f − (1) : f ∈ partlims( A ) } ) ↓ is closed under the operator liminf. Indeed, assume that we are given a sequence ofelements of A . By thinning it out, and replacing each element by an appropriate elementcontaining it, we may assume that this sequence is all in { f − n [(1 − /n, /n )] : n ∈ N } or all in { f − (1) : f ∈ partlims( A ) } . In the first case, by thinning out further we mayassume that the sequence is either constant (in which case we are done), or consistsof distinct elements f − m n [(1 − /m n , /m n )] with m n increasing. In this case, let f = lim n f m n . For each x ∈ liminf n f − m n [(1 − /m n , /m n )], f ( x ) = lim n f m n ( x ) = 1,and thus liminf n f − m n [(1 − /m n , /m n )] is in A . The second case is similar (and slightlyeasier).Thus, X ∈ A , which means that there is f ∈ partlims( A ) such that X = f − (1), thatis, = f ∈ partlims( A ). (cid:3) Clearly, (cid:0) ΩΓ (cid:1) implies (cid:0) ΩL (cid:1) . The original Gerlits–Nagy Problem, posed in [7], asks whetherthese properties are in fact equivalent (for Tychonoff X , or even for X ⊆ R ). Theorems4.2 and 4.5 justify the reformulation given in Problem 1.5.Originally, Gerlits and Nagy [7] studied five properties, numbered α, β, γ, δ, ǫ , whereeach property implies the subsequent one. (cid:0) ΩΓ (cid:1) and (cid:0) ΩL (cid:1) were numbered γ and δ , respec-tively, and are often named accordingly in the literature. Their problem was originallystated as whether property δ implies (and is therefore equivalent to) property γ .A topological space X is said to satisfy a property P hereditarily if each Y ⊆ X satisfies P . Pushing our methods further, we can solve the Gerlits–Nagy Problem in the affirmativefor spaces X satisfying S (Γ , Γ) hereditarily. We will use the following result of FrancisJordan [10] (see also [18]), proved using a new fusion argument of his.
Lemma 4.8 (Jordan) . Let B = S n B n ⊆ X be an increasing union, where each B n satisfies S (Γ , Γ) . For all open sets U nm ⊆ X , n, m ∈ N , with B n ⊆ liminf m U nm for each n , there are m , m , · · · ∈ N such that B ⊆ liminf n U nm n . Theorem 4.9.
For topological spaces X satisfying S (Γ , Γ) hereditarily, the following areequivalent: (1) X satisfies (cid:0) ΩL (cid:1) . (2) X satisfies (cid:0) ΩΓ (cid:1) .Proof of (1) ⇒ (2) . Lemma 4.10.
Assume that X satisfies S (Γ , Γ) hereditarily. Then X satisfies (cid:0) LΓ (cid:1) .Proof. Let U be an open cover of X with X ∈ LI( U ). Define B = { liminf n U n : U , U , · · · ∈ U } . We will prove that X ∈ B . To this end, it suffices to show that LI( B ↓ ) = B ↓ . Let B , B , · · · ∈ B ↓ , and B = liminf n B n . Replacing each B n with T m ≥ n B m , we mayassume that B ⊆ B ⊆ . . . , and S n B n = B . For each n , take U n , U n , · · · ∈ U suchthat B n ⊆ liminf m U nm . By the premise of the proposition, each B n satisfies S (Γ , Γ). ByJordan’s Lemma 4.8, there are m , m , · · · ∈ N such that B ⊆ n liminf n U nm n ∈ B ↓ , andtherefore B ∈ B ↓ . (cid:3) It remains to note that the conjunction of (cid:0) LΓ (cid:1) and (cid:0) ΩL (cid:1) implies (cid:0) ΩΓ (cid:1) . (cid:3) Remark . For each topological space X , Γ( X ) ⊆ L( X ) ⊆ Ω( X ). To see the secondinclusion, assume that there is a finite F ⊆ X not covered by any U ∈ U . Then F is notcovered by any element of LI( U ), and in particular, X / ∈ LI( U ). Thus, the implication atthe end of the proof of Theorem 4.9 is in fact an equivalence, that is, (cid:0) ΩL (cid:1) ∩ (cid:0) LΓ (cid:1) = (cid:0) ΩΓ (cid:1) . Corollary 4.12.
For Tychonoff spaces X satisfying S (Γ , Γ) , the following are equivalent: (1) X satisfies (cid:0) ΩL (cid:1) hereditarily. (2) X satisfies (cid:0) ΩΓ (cid:1) hereditarily.Proof of (1) ⇒ (2) . By Theorem 4.9, it suffices to prove that X satisfies S (Γ , Γ) heredi-tarily .Nowik, Scheepers and Weiss proved that (cid:0) ΩL (cid:1) implies U fin (O , Γ) [17]. Thus, if X satisfies (cid:0) ΩL (cid:1) hereditarily, then X satisfies U fin (O , Γ) hereditarily. Fremlin and Miller [4] proved thatin the latter case, X is a σ -space, that is, each Borel subset of X is F σ . This, togetherwith X ’s satisfying S (Γ , Γ), implies that X satisfies S (Γ , Γ) hereditarily [8, 18]. (cid:3)
Remark . The argument in the proof of Corollary 4.12 shows that, for Tychonoff σ -spaces X , (cid:0) ΩΓ (cid:1) = (cid:0) ΩL (cid:1) ∩ S (Γ , Γ). In this case, this joint property coincides with itshereditary version.Assuming that the answer to the Gerlits–Nagy Problem is negative , the results of thissection explain, to some extent, why no counter example was discovered thus far. Anatural strategy would be to begin with a set X ⊆ R satisfying (cid:0) ΩΓ (cid:1) , and then look fora subset of X , in a way which “destroys” (cid:0) ΩΓ (cid:1) , but not too much, so that (cid:0) ΩL (cid:1) still holds.There are several constructions of subsets of R satisfying (cid:0) ΩΓ (cid:1) . The first one is due toGalvin and Miller [5]. Here, X has a countable subset Q such that X \ Q does not satisfy (cid:0) ΩΓ (cid:1) . Unfortunately, X \ Q does not even satisfy U fin (O , Γ), and in particular not (cid:0) ΩL (cid:1) . Another, substantially different, construction is due to Todorˇcevic [5], but this X satisfies (cid:0) ΩΓ (cid:1) hereditarily. Finally, using a variation of Todorˇcevic’s method, Miller [15] constructed X ⊆ R satisfying (cid:0) ΩΓ (cid:1) for Borel covers, and a subset Y of X not satisfying (cid:0) ΩΓ (cid:1) . (cid:0) ΩΓ (cid:1) forBorel covers, implies S (Γ , Γ) for Borel covers, which is hereditary. Thus, in this case X satisfies S (Γ , Γ) hereditarily , and by Theorem 4.9 no subset of X would separate (cid:0) ΩL (cid:1) from (cid:0) ΩΓ (cid:1) .We conclude this section with a local reformulation of Theorem 4.9. A topological space Z has the Arhangel’ski˘ı property α if, for each z ∈ Z , whenever lim m z nm = z for all n , For a direct proof, see the proof of Proposition 2.6. On the other hand, we proved in [18] that any “natural” change of Galvin and Miller’s construction without moving to a subset at the end would keep X in (cid:0) ΩΓ (cid:1) . HE GERLITS–NAGY PROBLEM 13 there are m , m , . . . such that lim n z nm n = z . When Z = C( X ), we can take z = in thedefinition. Haleˇs proved that, for perfectly normal spaces X , the following properties areequivalent:(1) For each Y ⊆ X , C( Y ) is an α space.(2) X satisfies S (Γ , Γ) hereditarily.Collecting together the results of this section, we have the following.
Theorem 4.14.
Let X be a perfectly normal space, such that for each Y ⊆ X , C( Y ) isan α space. Then the following properties are equivalent: (1) For each A ⊆ C( X ) , A ⊆ partlims( A ) . (2) For each A ⊆ C( X ) , each f ∈ A is a limit of a sequence of elements of A (i.e., C( X ) is Fr´echet–Urysohn). (cid:3) Some results about the missing piece
The property (cid:0) LΓ (cid:1) was central, implicitly or explicitly, in our proofs, for the basic reasonthat (cid:18) ΩΓ (cid:19) = (cid:18) ΩL (cid:19) ∩ (cid:18) LΓ (cid:19) . To prove that (cid:0) ΩΓ (cid:1) = (cid:0) ΩL (cid:1) (the Gerlits–Nagy Problem), it is necessary and sufficient toprove that (cid:0) ΩL (cid:1) implies (cid:0) LΓ (cid:1) . We therefore describe some fundamental properties of (cid:0) LΓ (cid:1) ,and the ensuing open problems concerning it. Proposition 5.1. (cid:0) LΓ (cid:1) = S (L , Γ) = S fin (L , Γ) . In particular, (cid:0) LΓ (cid:1) implies S (Γ , Γ) .Proof. It suffices to prove the last assertion. Assume that for each n , U n = { U nm : m ∈ N } ∈ Γ( X ). We may assume that the covers U n get finer with n . Let V m = U m for all m . Define W = [ n ∈ N { V n ∩ U nm : m ∈ N } . Then liminf n liminf m ( V n ∩ U nm ) = liminf n V n = X, and therefore X ∈ LI( W ). By (cid:0) LΓ (cid:1) , there are W , W , · · · ∈ U such that X = liminf k W k .Fix n . As V n = X , it is not possible that W k ∈ { V n ∩ U nm : m ∈ N } for infinitely many k .Thus, by thinning out the sequence W k if needed, we may assume that there is at mostone W k in each set { V n ∩ U nm : m ∈ N } . Since the covers U n get finer with n , we can pickfor each n an element U nm n ∈ U n , such that X = liminf n U nm n . (cid:3) Proposition 5.2.
The property of satisfying S (Γ , Γ) hereditarily is strictly stronger than (cid:0) LΓ (cid:1) .Proof. Lemma 4.10 tells that hereditarily- S (Γ , Γ) implies (cid:0) LΓ (cid:1) . Assuming for example theContinuum Hypothesis, there is X ⊆ R and a subset Y of X such that X satisfies (cid:0) ΩΓ (cid:1) (and thus also (cid:0) LΓ (cid:1) ), and Y does not even satisfy S fin (O , O), and in particular not S (Γ , Γ)[5]. Apply Proposition 5.1. (cid:3) If { U n : n ∈ N } , { V n : n ∈ N } ∈ Γ( X ), then { U n ∩ V n : n ∈ N } ∈ Γ( X ) and is finer than both. If S (Γ , Γ) implies (cid:0) LΓ (cid:1) , then the word “hereditarily” can be removed from Theorem 4.9.However, we suspect that this is not the case. Conjecture 5.3. (cid:0) LΓ (cid:1) is strictly stronger than S (Γ , Γ) . To prove this conjecture, it suffices to construct (say using the Continuum Hypothesis)sets
X, Y ⊆ R satisfying (cid:0) LΓ (cid:1) , such that X ∪ Y does not satisfy (cid:0) LΓ (cid:1) , because S (Γ , Γ) is σ -additive. Problem 5.4. Is (cid:0) LΓ (cid:1) preserved by finite unions? If it is, then (cid:0) LΓ (cid:1) is in fact σ -additive, because of the following. Proposition 5.5. (cid:0) LΓ (cid:1) is linearly σ -additive, that is, is preserved by countable increasingunions.Proof. Assume that X ⊆ X ⊆ . . . all satisfy (cid:0) LΓ (cid:1) , S n X n = X , and X ∈ LI( U ). Thenfor each n , X n ∈ LI( { U ∩ X n : U ∈ U } ), and thus there are U nm ∈ U , m ∈ N , suchthat X n ⊆ liminf m U nm . By Jordan’s Lemma 4.8, there are m , m , · · · ∈ N such that X = liminf n U nm n . (cid:3) The proofs of the above results are also valid in the case of Borel covers, and since S (Γ , Γ) for Borel covers is hereditary, we have the following.
Corollary 5.6.
For Borel covers, (cid:0) LΓ (cid:1) = S (Γ , Γ) . (cid:3) Thus, none of the above-mentioned problems remains open in the Borel case.We conclude with a local characterization of (cid:0) LΓ (cid:1) . Theorem 5.7.
For perfectly normal spaces X , the following are equivalent. (1) For each A ⊆ C( X ) each f ∈ C( X ) ∩ partlims( A ) is a limit of a sequence ofelements of A . (2) X satisfies (cid:0) LΓ (cid:1) .Proof. (1 ⇒ Lemma 5.8.
Let X be a perfectly normal space. Assume that for each A ⊆ C( X ) , each f ∈ C( X ) ∩ partlims( A ) is a limit of a sequence of elements of A . Then each element of L( X ) has a clopen refinement in L( X ) .Proof. Indeed, this follows from a formally weaker property: Let P be the property that,for each A ⊆ C( X ), each f in the closure of A in C( X ) under limits of sequences, is alimit of a sequence of elements of A .Fremlin [3] proved that P is equivalent to the property named wQN in [1], where it isshown that for perfectly normal spaces, wQN implies that each open set is a countableunion of clopen sets [1, Corollary 4.6].Now, let U ∈ L( X ). For each U ∈ U , present U as an increasing union U = S n C n ( U )of clopen sets. Then U = liminf n C n ( U ). Let V = { C n ( U ) : U ∈ U , n ∈ N } . Then V is aclopen refinement of U , and X ∈ LI( U ) ⊆ LI( V ), that is, V ∈ L( X ). (cid:3) HE GERLITS–NAGY PROBLEM 15
Let
U ∈ L( X ). By Lemma 5.8, we may assume that the elements of U are clopen.Let A = { χ U : U ∈ U } . A ⊆ C( X ). Let V = { f − (1) : f ∈ partlims( A ) } . U ⊆ V ,and V is closed under the operator liminf. Indeed, Let f , f , · · · ∈ partlims( A ), and B = liminf n f − n (1). As f = lim n f n ∈ partlims( A ), B = f − (1) ∈ V .Thus, X ∈ V , and therefore ∈ partlims( A ). By (1), there are U n ∈ U such thatlim n χ U n = , that is, liminf n U n = X .(2 ⇒
1) Assume that ∈ partlims( A ). For each n , let U n = { f − [(1 − /n, /n )] : f ∈ A } . U n ∈ L( X ). Indeed, let C be the family of all partial f : X → R , such that f − [(1 − /n, /n )] ∈ LI( U n ). Then A ⊆ C , and C is closed under partial limits ofsequences. Thus, ∈ C , that is, X = − [(1 − /n, /n )] ∈ LI( U n ).By Proposition 5.1, there are f , f , · · · ∈ A such that liminf n f − n [(1 − /n, /n )] = X . In particular, lim n f n = . (cid:3) The notation used below is available, e.g., in the survey [24].
Proposition 5.9.
The minimal cardinality of a set X ⊆ R such that X does not satisfy (cid:0) LΓ (cid:1) is b (the minimal cardinality of a subset of N N which is not bounded, with respect toeventual dominance).Proof. If | X | < b , then X satisfies S (Γ , Γ) [11]. Thus, X satisfies S (Γ , Γ) hereditarily,and by Lemma 4.10, X satisfies (cid:0) LΓ (cid:1) . On the other hand, there is X ⊆ R with | X | = b ,such that X does not satisfy S (Γ , Γ) [11]. By Proposition 5.1, this X does not satisfy (cid:0) LΓ (cid:1) . (cid:3) The proof of the main theorem in [18], with trivial modifications, gives the first itemof the following theorem. The other items are easy consequences.
Theorem 5.10. (1)
For each unbounded tower T of cardinality b in [ N ] ∞ , T ∪ [ N ] < ∞ satisfies (cid:0) LΓ (cid:1) . (2) If t = b , then there are subsets of R of cardinality b , satisfying (cid:0) LΓ (cid:1) . (3) There are subsets of R of cardinality t , satisfying (cid:0) LΓ (cid:1) . (cid:3) The assumption t = b is known to be strictly weaker than the Continuum Hypothesisor even Martin’s Axiom, but it is open whether it is weaker than p = b , which impliesthat the sets mentioned in Theorem 5.10 actually have the stronger property (cid:0) ΩΓ (cid:1) [18]. Acknowledgments.
We thank Salvador Herna´ndez for inviting the second named authorto give a lecture on the topic in the conference
Functional Analysis in Valencia 2010 ,dedicated to the 80’th birthday of Manuel Valdivia. We also thank Lyubomyr Zdomskyyfor reading the paper and making useful comments.This work is an extension of a part of the first named author’s M.Sc. thesis at theWeizmann Institute of Science, supervised by Gady Kozma and the second named author.We thank Gady Kozma for useful discussions, and the Weizmann Institute of Science forthe stimulating atmosphere.
References [1] L. Bukovsk´y, I. Rec law, and M. Repick´y,
Spaces not distinguishing pointwise and quasinormal con-vergence of real functions , Topology and its Applications (1991), 25–41. [2] M. Chasco, E. Martin-Peinador, V. Tarieladze, A class of angelic sequential non-Fr´echet–Urysohntopological groups
Topology and its Applications (2007) 741–748.[3] D. Fremlin,
SSP and WQN , unpublished note, Januray 2003. [4] D. Fremlin and A. Miller,
On some properties of Hurewicz, Menger and Rothberger , FundamentaMathematica (1988), 17–33.[5] F. Galvin and A. Miller, γ -sets and other singular sets of real numbers , Topology and its Applications (1984), 145–155.[6] J. Gerlits, Some properties of C(X), II , Topology and its Applications (1983), 255–262.[7] J. Gerlits and Zs. Nagy, Some properties of C ( X ) , I , Topology and its Applications (1982),151–161.[8] J. Haleˇs, On Scheepers’ conjecture , Acta Universitatis Carolinae Mathematica et Physica (2005),27–31.[9] W. Hurewicz, ¨Uber eine Verallgemeinerung des Borelschen Theorems , Mathematische Zeitschrift (1925), 401–421.[10] F. Jordan, There are no hereditary productive γ -spaces , Topology and its Applications (2008),1786–1791.[11] W. Just, A. Miller, M. Scheepers, and P. Szeptycki, The combinatorics of open covers II , Topologyand its Applications (1996), 241–266.[12] A. Kechris, Classical Descriptive Set Theory , Graduate Texts in Mathematics , Springer–Verlag, 1994.[13] L. Koˇcinac,
Selected results on selection principles , in:
Proceedings of the 3rd Seminar onGeometry and Topology (Sh. Rezapour, ed.), July 15–17, Tabriz, Iran, 2004, 71–104.[14] K. Menger,
Einige ¨Uberdeckungss¨atze der Punktmengenlehre , Sitzungsberichte der Wiener Akademie (1924), 421–444.[15] A. Miller,
A Nonhereditary Borel-cover γ -set , Real Analysis Exchange (2003/4), 601–606.[16] A. Miller, On γ -sets , plenary lecture, Second Workshop on Coverings, Selections, and Games inTopology, Lecce, Italy, 19–22 Dec 2005.Lecture notes: http://u.cs.biu.ac.il/~tsaban/SPMC05/Miller.pdf [17] A. Nowik, M. Scheepers, and T. Weiss, The algebraic sum of sets of real numbers with strong measurezero sets , Journal of Symbolic Logic (1998), 301–324.[18] T. Orenshtein and B. Tsaban, Linear σ -additivity and some applications , Transactions of the Amer-ican Mathematical Society (2011), 3621–3637.[19] E. Pytkeev, On sequentiality of spaces of continuous functions , Russian Mathematical Surveys (1982), 190–191.[20] F. Rothberger, Sur des families indenombrables de suites de nombres naturels, et les probl´emesconcernant la propriet´e C , Proceedings of the Cambridge Philosophical Society (1941), 109–126.[21] M. Sakai, Property C ′′ and function spaces , Proceedings of the American Mathematical Society (1988), 917–919.[22] M. Scheepers, Selection principles and covering properties in topology , Note di Matematica (2003),3–41.[23] M. Scheepers and B. Tsaban, The combinatorics of Borel covers , Topology and its Applications (2002), 357–382.[24] B. Tsaban,
Some new directions in infinite-combinatorial topology , in:
Set Theory (J. Bagaria andS. Todorˇcevic), Trends in Mathematics, Birkh¨auser 2006, 225–255.
HE GERLITS–NAGY PROBLEM 17 (Orenshtein)
Department of Mathematics, Weizmann Institute of Science, Rehovot 76100,Israel
E-mail address : [email protected] URL : (Tsaban) Department of Mathematics, Bar-Ilan University, Ramat Gan 52900, Israel
E-mail address : [email protected] URL ::