Combinatorial properties on nodec countable spaces with analytic topology
aa r X i v : . [ m a t h . GN ] J a n COMBINATORIAL PROPERTIES ON NODEC COUNTABLE SPACESWITH ANALYTIC TOPOLOGY
JAVIER MURGAS AND CARLOS UZC ´ATEGUI
Abstract.
We study some variations of the product topology on families of clopen subsetsof 2 N × N in order to construct countable nodec regular spaces (i.e. in which every nowheredense set is closed) with analytic topology which in addition are not selectively separableand do not satisfy the combinatorial principle q + . Keywords: nodec countable spaces; analytic sets, selective separability, q + MSC: 54G05, 54H05, 03E15 Introduction
A topological space X is selectively separable ( SS ), if for any sequence ( D n ) n of densesubsets of X there is a finite set F n ⊆ D n , for n ∈ N , such that S n F n is dense in X . Thisnotion was introduced by Scheepers [13] and has received a lot of attention ever since (seefor instance [1, 2, 3, 4, 5, 6, 8, 12]). Bella et al. [4] showed that every separable space withcountable fan tightness is SS . On the other hand, Barman and Dow [1] showed that everyseparable Fr´echet space is also SS (see also [6]).A topological space is maximal if it is a dense-in-itself regular space such that any strictlyfiner topology has an isolated point. It was shown by van Douwen [17] that a space ismaximal if, and only if, it is extremely disconnected (i.e. the closure of every open set isopen), nodec (i.e. every nowhere dense set is closed) and every open set is irresolvable (i.e. if U is open and D ⊆ U is dense in U , then U \ D not dense in U ). He constructed a countablemaximal regular space.A countable space X is q + at a point x ∈ X , if given any collection of finite sets F n ⊆ X such that x ∈ S n F n , there is S ⊆ S n F n such that x ∈ S and S ∩ F n has at most onepoint for each n . We say that X is a q + -space if it is q + at every point. Every countablesequential space is q + (see [15, Proposition 3.3]). The collection of clopen subsets of 2 N withthe product topology is not q + at any point. This notion is motivated by the analogousconcept of a q + filter (or ideal) from Ramsey theory.A problem stated in [4] was to analyze the behavior of selective separability on maximalspaces. The existence of a maximal regular SS space is independent of ZFC. In fact, inZFC there is a maximal non SS space [1] and it is consistent with ZFC that no countablemaximal space is SS [1, 12]. On the other hand, it is also consistent that there is a maximal,countable, SS regular space [1].In this paper we are interested in these properties on countable spaces with an analytictopology (i.e. the topology of the space X is an analytic set as a subset of 2 X [14]). Maximaltopologies are not analytic. In fact, in [16] it was shown that there are neither extremely The second author thanks Vicerrector´ıa de Investigaci´on y Extensi´on de la Universidad Industrial deSantander for the financial support for this work, which is part of the VIE project isconnected nor irresolvable analytic topologies, nevertheless there are nodec regular spaceswith analytic topology. In view of the above mentioned results about maximal spaces, itseems natural to wonder about the behavior of selective separability on nodec spaces withan analytic topology. Nodec regular spaces are not easy to construct. We continue the studyof the method introduced in [16] in order to construct similar nodec regular spaces withanalytic topology that are neither SS nor q + . A countable regular space has an analytictopology if, and only if, it is homeomorphic to a subspace of C p ( N N ) [14]. Thus our examplesare constructed using some special topologies on a collection of clopen subsets of 2 N × N . Itis an open question whether there is a nodec SS regular space with analytic topology.2. Preliminaries An ideal on a set X is a collection I of subsets of X satisfying: (i) A ⊆ B and B ∈ I ,then A ∈ I . (ii) If A, B ∈ I , then A ∪ B ∈ I . (iii) ∅ ∈ I . We will always assume that anideal contains all finite subsets of X . If I is an ideal on X , then I + = { A ⊆ X : A
6∈ I} . Fin denotes the ideal of finite subsets of the non negative integers N . An ideal I on X is tall , if for every A ⊆ X infinite, there is B ⊆ A infinite with B ∈ I . We denote by A <ω thecollection of finite sequences of elements of A . If s is a finite sequence on A and i ∈ A , | s | denotes its length and s b i the sequence obtained concatenating s with i . For s ∈ <ω and α ∈ N , let s ≺ α if α ( i ) = s ( i ) for all i < | s | and[ s ] = { α ∈ N : s ≺ α } . If α ∈ N and n ∈ N , we denote by α ↾ n the finite sequence ( α (0) , · · · , α ( n − n > α ↾ s ] with s ∈ <ω is a basis of clopensets for 2 N . As usual we identify each n ∈ N with { , · · · , n − } .The ideal of nowhere dense subsets of X is denoted by nwd ( X ). Now we recall somecombinatorial properties of ideals. We put A ⊆ ∗ B if A \ B is finite.( p + ) I is p + , if for every decreasing sequence ( A n ) n of sets in I + , there is A ∈ I + suchthat A ⊆ ∗ A n for all n ∈ N . Following [9], we say that I is p − , if for every decreasingsequence ( A n ) n of sets in I + such that A n \ A n +1 ∈ I , there is B ∈ I + such that B ⊆ ∗ A n for all n .( q + ) I is q + , if for every A ∈ I + and every partition ( F n ) n of A into finite sets, there is S ∈ I + such that S ⊆ A and S ∩ F n has at most one element for each n . Such sets S are called (partial) selectors for the partition.A point x of a topological space X is called a Fr´echet point , if for every A with x ∈ A there is a sequence ( x n ) n in A converging to x . We will say that x is a q + - point , if I x is q + .We say that a space is a q + -space, if every point is q + . We define analogously the notion ofa p + and p − points. Notice that if x is isolated, then I x is trivially q + as I + x is empty. Thusa space is q + if, and only if, I x is q + for every non isolated point x . The same occurs withthe other combinatorial properties defined in terms of I x .We say that a space Z is wSS if for every sequence ( D n ) n of dense subsets of Z , there is F n ⊆ D n a finite set, for each n , such that S n F n is not nowhere dense in Z . In the terminology ofselection principles [13], wSS corresponds to S fin ( D , B ) where D is the collection of densesubsets and B the collection of non nowhere dense sets. Seemingly this notion has not beenconsidered before. Notice that if Z is SS and W is not SS , then the direct sum of Z and W is wSS but not SS . subset A of a Polish space is called analytic , if it is a continuous image of a Polish space.Equivalently, if there is a continuous function f : N N → X with range A , where N N is thespace of irrationals. For instance, every Borel subset of a Polish space is analytic. A generalreference for all descriptive set theoretic notions used in this paper is [10]. We say that atopology τ over a countable set X is analytic , if τ is analytic as a subset of the cantor cube2 X (identifying subsets of X with characteristic functions) [14, 15, 16], in this case we willsay that X is an analytic space . A regular countable space is analytic if, and only if, it ishomeomorphic to a subspace of C p ( N N ) (see [14]). If there is a base B of X such that B isan F σ (Borel) subset of 2 X , then we say that X has an F σ (Borel) base . In general, if X hasa Borel base, then the topology of X is analytic.We end this section recalling some results about countable spaces that will be used in thesequel. Theorem 2.1. [6, Corollary 3.8]
Let X be a countable space with an F σ base, then X is p + . Next result is essentially Lemma 4.6 of [16].
Lemma 2.2.
Let X be a σ -compact space and W a countable collection of clopen subsets of X . Then W , as a subspace of X , has an F σ base. Theorem 2.3. [6, Theorem 3.5]
Let X be a countable space. If X is p − , then X is SS . Inparticular, if X has an F σ base, then X is SS . A space X is discretely generated (DG) if for every A ⊆ X and x ∈ A , there is E ⊆ A discrete such that x ∈ E . This notion was introduced by Dow et al. in [7]. It is not easyto construct spaces which are not DG, the typical examples are maximal spaces (which arenodec). Theorem 2.4.
Let X be a regular countable space. Suppose every non isolated point is p − ,then X is discretely generated.Proof. Let A ⊂ X with x ∈ A . Fix a maximal family ( O n ) n of relatively open disjointsubsets of A such that x O n . Let B = S n O n . From the maximality we get that x ∈ B .Since each O n does not accumulate to x and x is a p − -point, there is E such that x ∈ E and E ∩ O n is finite for every n . Clearly E is a discrete subset of A . (cid:3) Theorem 2.5. (Dow et al [7, Theorem 3.9] ) Every Hausdorff sequential space is discretelygenerated.
In summary, we have the following implications for countable regular spaces (see [6]). F σ -base ւ Fr´echet p + ւ ց ւ Sequential p − ↓ ց ւ ց q + DG SS ↓ ↓ non nodec wSS .1. A SS , q + nodec analytic non regular topology. As we said in the introduction,nodec regular spaces are not easy construct. However, non regular nodec spaces are fairlyeasy to define. We recall a well known construction given in [11]. Let τ be a topology anddefine τ α = { V \ N : V ∈ τ and N ∈ nwd ( τ ) } . Then τ α is a topology finer than τ (see [11]). Lemma 2.6. [11]
Let ( X, τ ) be a space. (i) V ∈ τ α iff V ⊆ int τ ( cl τ ( int τ ( V ))) . (ii) Let A ⊆ X and x A . Then x ∈ cl τ α ( A ) if, and only if, x ∈ cl τ ( int τ ( cl τ ( A ))) . (iii) ( X, τ α ) is a nodec space. Proposition 2.7.
Let ( X, τ ) be a countable space. (i) If ( X, τ ) is Fr´echet, then ( X, τ α ) is a q + -space. (ii) ( X, τ ) is SS if, and only if, ( X, τ α ) is SS .Proof. (i) Suppose x ∈ cl α ( A ) \ A and ( F n ) n is a partition of A with each F n finite. Let V = int τ ( cl τ ( A )). By Lemma 2.6 we have that x ∈ cl τ ( V ). Let ( y m ) m be an enumeration of V . Since A is τ -dense in V , for every m there is a sequence ( x mi ) i in A such that x mi → y m when i → ∞ (with respect to τ ). Since each F n is finite, we can assume (by passing to asubsequence if necessary) that each ( x mi ) i is a selector for the partition ( F n ) n . Let S m bethe range of ( x mi ) i . Notice that x cl τ ( S m ) and every infinite subset of S m is also a selectorfor ( F n ) n . By a straightforward diagonalization, for each m , there is T m ⊆ S m such thateach T m is a selector and moreover S m T m is also a selector. Hence we can assume that S = S m { x mi : i ∈ N } is a selector for the partition. But clearly S is τ -dense in V and thus V ⊆ int τ ( cl τ ( S )). Hence x ∈ cl α ( S ) (by Lemma 2.6(i)).(ii) By Lemma 2.6(ii), a set is τ -dense iff it is τ α -dense. (cid:3) Let τ be the usual metric topology on the rational Q . It is not difficult to verify that τ α isanalytic (in fact, it is Borel) and non regular (see [16]). Thus ( Q , τ α ) is a SS, q + and nodecnon regular space with analytic topology. It is not known if there is a regular space with thesame properties. 3. The spaces X ( I ) and Y ( I )We recall the definitions of the spaces X ( I ) and Y ( I ) for an ideal I , which were introducedin [16].For each non empty A ⊆ N , let ρ A be the topology on 2 N × N generated by the followingsets: ( α, p ) + = { θ ∈ N × N : θ ( α, p ) = 1 } , ( α, p ) − = { θ ∈ N × N : θ ( α, p ) = 0 } , with α ∈ A . A basic ρ A -open set is as follows: V = m \ i =1 ( α i , p i ) + ∩ n \ i =1 ( β i , q i ) − or some α , · · · , α m , β , · · · , β n ∈ A , p , , ..., p m , q , ..., q n ∈ N . We always assume that( α i , p i ) = ( β j , q j ) for all i and j , which is equivalent to saying that any set V as above is notempty.Let X be the collection of all finite unions of clopen sets of the form [ s ] × { n } with n ∈ N and s ∈ <ω . We also include ∅ as an element of X . As usual, we regard X as a subset of2 N × N . Let { ϕ n : n ∈ N } be an enumeration of X and for convenience we assume that ϕ is ∅ . Each ϕ n , regarded as a function from 2 N × N to { , } , is continuous. Notice that X is agroup with the symmetric difference as operation.Let ψ n : 2 N × N → { , } be defined by ψ n ( α, m ) = (cid:26) ϕ n ( α, m ) , if α ( n ) = 0 . , if α ( n ) = 1 . Then ψ n is a continuous function. Let Y = { ψ n : n ∈ N } . Given
I ⊆ N , we define X ( I ) = ( X , ρ I ) , Y ( I ) = ( Y , ρ I ) . Also notice that X ( I ) is a topological group.To each F ⊆ N , we associate two sets F ′ ⊆ X and b F ⊆ Y : F ′ := { ϕ n : n ∈ F } , b F := { ψ n : n ∈ F } . The topological similarities between F ′ and b F are crucial to establish some properties of Y ( I ).As usual, we identify a subset A ⊆ N with its characteristic function. So from now on,an ideal I over N will be also viewed as a subset of 2 N . The properties of Y ( I ) naturallydepend on the ideal I . Lemma 3.1. If I is analytic, then X ( I ) and Y ( I ) have analytic topologies.Proof. It is easy to see that the standard subspace subbases for X ( I ) and Y ( I ) are alsoanalytic when I is analytic. Thus the topology is analytic (see [14, Proposition 3.2]). (cid:3) Theorem 3.2. If I is an F σ ideal over N , then Y ( I ) has an F σ base and thus it is SS andDG.Proof. It follows from Lemma 2.2 and Theorems 2.3 and 2.4. (cid:3)
The reason to study the space Y ( I ) is the following theorem. Let I nd := { F ⊆ N : { ϕ n : n ∈ F } is nowhere dense in X } . Theorem 3.3. [16] Y ( I nd ) is a nodec regular space without isolated points and with ananalytic topology. Y ( I nd ) was so far the only space we knew with the properties stated above. We willpresent a generalization of this theorem showing other ideals I such that Y ( I ) has the sameproperties. .1. The space X ( I ) . We present some properties of the space X ( I ) that will be neededlater. We are interested in whether X ( I ) is DG, SS or q + . We start with a general resultwhich is proven as Theorem 3.2. Theorem 3.4. If I is an F σ ideal over N , then X ( I ) has an F σ base, and thus it is SS andDG. We will show that X ( I ) is not q + except in the extreme case when I is Fin . The keylemma to show this is the following result.
Lemma 3.5.
There is a pairwise disjoint family { A n : n ∈ N } of finite subsets of X suchthat S k ∈ E A k is dense in X (with the product topology) for any infinite E ⊆ N . Moreover, foreach infinite set E ⊆ N , each selector S for the family { A n : n ∈ E } and each ϕ / ∈ S ∪ {∅} ,there is p ∈ N and α ∈ N such that α − (1) ⊆ ∗ E , ϕ ∈ ( α, p ) + and ( α, p ) + ∩ S is finite.Proof. We say that a ϕ ∈ X has the property ( ∗ m ), for m ∈ N , if there are k ∈ N and finitesequences s i , for i = 1 , ..., k , of length m + 1 such that ϕ = S ki =1 [ s i ] × { m i } , m i ≤ m and s i ↾ m = s j ↾ m , whenever m i = m j (i.e. [ s j ] ∪ [ s i ] is not a basic clopen set). Let A m = { ϕ ∈ X : ϕ has the property ( ∗ m ) } . Let E ⊆ N be an infinite set. We will show that A := S k ∈ E A k is dense in 2 N × N . Let V bea basic open set of 2 N × N , let us say V = m \ i =1 ( α i , p i ) + ∩ n \ i =1 ( β i , q i ) − for some α , · · · , α m , β , · · · , β n ∈ N , p , · · · , p m , q , · · · , q n ∈ N . We need to show that V ∩ A is not empty. Pick l large enough such that l + 1 ∈ E , l + 1 > max { p i , q j : i ≤ m, j ≤ n } , α i ↾ l = α j ↾ l for all i and j such that α i = α j , β i ↾ l = β j ↾ l for all i and j such that β i = β j and α i ↾ l = β j ↾ l for all i and j such that α i = β j . Let ϕ = S mi =1 [ α i ↾ ( l + 2)] × { p i } .Then ϕ belongs to A l +1 ∩ V .To see the second claim, let E ⊆ N be an infinite set and let S = { z n : n ∈ E } be aselector, that is, z n ∈ A n for all n ∈ E . Fix ϕ / ∈ S ∪ {∅} , say ϕ = S li =1 [ t i ] × { p i } for some t i ∈ <ω and p i ∈ N . The required α is recursively defined as follows: α ( n ) = t ( n ) , if n < | t | , , if n ≥ | t | , n ∈ E and [( α ↾ n ) b × { p } ⊆ z n , , otherwise.From the definition of the sets A m , it is easily shown that ( α, p ) / ∈ S { z k : k ≥ | t | and k ∈ E } . Clearly ( α, p ) + ∩ S ⊆ { z k : k < | t | and k ∈ E } is finite and ϕ ∈ ( α, p ) + . Finally, it isalso clear from the definition of α that α − (1) ⊆ ∗ E . (cid:3) Theorem 3.6.
Let I be an ideal on N . Then X ( I ) is q + at some (every) point if, and onlyif, I = Fin .Proof. If I = Fin , then X ( I ) has a countable basis and thus it is q + at every point. Since X ( I ) is homogeneous (as it is a topological group), then if X ( I ) is q + at some point, then itis q + at every point. Suppose now that there is E ∈ I \ Fin . We will show that I is not q + at some point. Let { A n : n ∈ N } be the sequence, given by Lemma 3.5, of pairwise disjoint nite subsets of X such that A := S k ∈ E A k is dense in X . Since the topology of X is finerthan the topology of X ( I ), then A is dense in X ( I ). Let ϕ A ∪ {∅} . We will show that X ( I ) fails the property q + at ϕ . Let S be a selector of { A n : n ∈ E } . Let α ∈ N and p ∈ N be as in the conclusion of Lemma 3.5, that is, α − (1) ⊆ ∗ E , ϕ ∈ ( α, p ) + and ( α, p ) + ∩ S isfinite. Notice that α ∈ I and hence ϕ is not in the ρ I -closure of S . Hence X ( I ) is not q + at ϕ . (cid:3) Now we look at the SS property. The following result provides a method to constructdense subsets of X ( I ). Lemma 3.7.
For each A ⊆ N infinite, let D ( A ) be the following subset of X : ( k [ i =0 [ s i ] × { m i } ∈ X : A ∩ s − i (0) = ∅ for all i ∈ { , ..., k } , k ∈ N , s i ∈ <ω ) ∪ {∅} . Then A ∈ I if, and only if, D ( A ) is not dense in X ( I ) if,and only if, D ( A ) is nowhere denseand closed in X ( I ) .Proof. We first show that D ( A ) is closed for every A ∈ I . We shall show that the complementof D ( A ) is open in X ( I ). Let ϕ ∈ X \ D ( A ). Since ϕ = ∅ , we have that ϕ = S ki =1 [ s i ] × { m i } and we can assume that A ∩ s − (0) = ∅ . Let B = A ∪ s − (1). Notice that B ∈ I . Let β be the characteristic function of B . Clearly β ∈ [ s ] and thus ϕ ∈ ( β, m ) + . On the otherhand, suppose that ϕ ′ = S li =1 [ t i ] × { p i } ∈ ( β, m ) + . Assume that β ∈ [ t ] and p = m ,then t − (0) ⊂ β − (0) and hence t − (0) ∩ A = ∅ . This shows that ϕ ′ D ( A ) and thus( β, m ) + ∩ D ( A ) = ∅ .Now we show that if A ∈ I , then D ( A ) is nowhere dense. Since D ( A ) is closed, it sufficesto show that it has empty interior. Let V be a basic ρ I -open set. Let us say(1) V = m \ i =1 ( α i , p i ) + ∩ n \ i =1 ( β i , q i ) − for some α , · · · , α m , β , · · · , β n ∈ I , p , , ..., p m , q , ..., q n ∈ N . Recall that ( α i , p i ) = ( β j , q j )for all i = j . Since β i ∈ I , then β − i (0) = ∅ for all i . Let l = max { min( β − i (0)) : 1 ≤ i ≤ n } and t be the constant sequence 1 of length l . Since X is clearly ρ I -dense, let ϕ ∈ V ∩ X .Then ϕ ∪ ([ t ] × { } ) ∈ V \ D ( A ).Finally, we show that if A
6∈ I , then D ( A ) is dense. Let V be a basic ρ I -open set asgiven by (1). Pick l large enough such that α i ↾ l = α j ↾ l for i = j , β i ↾ l = β j ↾ l for i = j and α i ↾ l = β j ↾ l for all i and j such that α i = β j . Then pick k ≥ l such that k ≥ min( α − i (0) ∩ A ) for all i ≤ m (notice that α − i (0) ∩ A = ∅ as A
6∈ I and α i ∈ I ). Let s i = α i ↾ k for i ≤ m and ϕ = S mi =1 [ s i ] × { p i } . Then ϕ ∈ V ∩ D ( A ). (cid:3) We remind the reader that F ′ denotes the set { ϕ n : n ∈ F } for each F ⊆ N . Theorem 3.8.
Let I be an ideal over N . If I is not p + , then X ( I ) is not wSS .Proof. Suppose that I is not p + and fix a sequence ( A n ) n ∈ N of subsets of N such that A n / ∈ I , n ∈ N , and S n ∈ N F n ∈ I for all F n ⊆ A n finite. et D n = D ( A n ) as in Lemma 3.7. We show that the property wSS fails at the sequence( D n ) n . Let K n ⊆ D n be a finite set for each n , we need to show that S n K n is nowhere densein X ( I ). Let us enumerate each K n as follows: K n = k n,l [ i =0 [ s n,li ] × { p n,li } : l < | K n | . Let q n > max {| s n,li | : l < | K n | , i ≤ k n,l } . By hypothesis, B = S n ∈ N ( A n ∩ { , · · · , q n } ) ∈ I .Let β be the characteristic function of B . We claim that for all m ∈ N ( β, m ) + ∩ ( [ n ∈ N K n ) = ∅ . Otherwise, there are n ∈ N , l < | K n | and i ≤ k n,l such that β ∈ [ s n,li ], that is, s n,li (cid:22) β . Butthis contradicts the fact that ( A n ∩ { , · · · , q n } ) ∩ (cid:16) s n,li (cid:17) − (0) = ∅ for all i and l (recall that D n = D ( A n )). Thus ( S n ∈ N K n ) ∩ ( S m ( β, m ) + ) = ∅ . Since S m ( β, m ) + is ρ I -open dense, S n K n is ρ I -nowhere dense. (cid:3) Proposition 3.9.
Let I be an ideal over N . Any element of X ( I ) is a limit of a non trivialsequence.Proof. Since X ( I ) is a topological group, it suffices to show that there is a sequence converg-ing to ∅ (i.e. to ϕ ).Let ( α n ) n ∈ N be a sequence in 2 N such that α k ↾ ( k + 1) = α l ↾ ( k + 1) for each k < l .Let ( x n ) n be defined by x n = [ α n ↾ ( n + 1)] × { } . Let V be a neighborhood of ∅ , namely, V = T mi =1 ( β i , n i ) − for some β i ∈ I and n i ∈ N . We have that α n ↾ ( n + 1) β i for almostevery n , therefore x n ∈ V and x n → ∅ . (cid:3) Question 3.10.
When is X ( I ) discretely generated? The space c ( I ) . It is natural to wonder what can be said if instead of X we use themore familiar space CL (2 N ) of all clopen subsets of 2 N .Exactly as before we can define a space c ( I ) as follows. Definition 3.11.
Let I be an ideal over N and c ( I ) be ( CL (2 N ) , τ I ) , where τ I is generatedby the following subbasis: α + = { x ∈ CL (2 N ) : α ∈ x } and α − = { x ∈ CL (2 N ) : α / ∈ x } , where α ∈ I . In fact, it is easy to see that c ( I ) is homeomorphic to { S ki =0 [ s i ] ×{ } ∈ X : k ∈ N , s i ∈ <ω } and by a simple modification of the proofs above we have the following. Theorem 3.12.
Let I be an ideal over N . Then c ( I ) is q + at some (every) point if, andonly if, I = Fin . Theorem 3.13.
Suppose that I is an ideal over N . If I is not p + , then c ( I ) is not wSS . .3. The space Y ( I ) . In this section we work with the space Y ( I ) in order to constructnodec spaces. To that end we introduce an operation ⋆ on ideals. We remind the readerthat to each F ⊆ N we associate the sets F ′ = { ϕ n : n ∈ F } and b F = { ψ n : n ∈ F } . Definition 3.14.
Let I be a nonempty subset of N . We define: I ⋆ = { F ⊆ N : F ′ is nowhere dense in X ( I ) } . Notice that I ⋆ is a free ideal and I nd = (2 N ) ⋆ . We are going to present several results thatare useful to compare X ( I ) and Y ( I ).The following fact will be used several times in the sequel. Lemma 3.15.
Let I be an ideal over N . Let V be a basic ρ I -open set. Then { n ∈ N : ϕ n ∈ V }△{ n ∈ N : ψ n ∈ V } ∈ I . Proof.
Let V be a non empty basic open set, that is,(2) V = m \ i =1 ( α i , p i ) + ∩ l \ j =1 ( β j , q j ) − . From the very definition of ψ n and viewing it as a clopen set, we have that ψ n = ϕ n ∪ ( { α ∈ N : α ( n ) = 1 } × N ) . From this we have the following: { n ∈ N : ϕ n ∈ V } \ { n ∈ N : ψ n ∈ V } ⊆ l [ j =1 β − j (1)and { n ∈ N : ψ n ∈ V } \ { n ∈ N : ϕ n ∈ V } ⊆ m [ i =1 α − i (1) . Thus when each α i and each β j belongs to I , the unions on the right also belong to I . (cid:3) In the following we compare X ( I ) and Y ( I ) in terms of their dense and nowhere densesubsets. Some results need that the ideals I and I ⋆ are comparable, i.e. I ⊆ I ⋆ or I ⋆ ⊆ I ,it is unclear whether this is always the case.We are mostly interested in crowded spaces. The following fact gives a sufficient conditionfor Y ( I ) to be crowded. Lemma 3.16.
Let I be an ideal on N . Then(1) X is dense in (2 N × N , ρ I ) .(2) int X ( I ) ( F ′ ) = ∅ , for all F ∈ I if, and only if, Y is dense in (2 N × N , ρ I ) .(3) If I ⊆ I ⋆ , then Y is dense in (2 N × N , ρ I ) .(4) If I ⋆ ⊆ I , then Y is dense in (2 N × N , ρ I ⋆ ) .Proof. (1) is clear. The only if part of (2) was shown in [16, Lemma 4.2], but we include aproof for the sake of completeness. Let V be a nonempty basic ρ I -open set. We need to find n such that ψ n ∈ V . From Lemma 3.15 we have that E = { n ∈ N : ϕ n ∈ V and ψ n V } ∈ I . ince int X ( I ) ( E ′ ) = ∅ , there is n such that ϕ n ∈ V and n E . Therefore ψ n ∈ V .For the if part, suppose that Y is dense in (2 N × N , ρ I ) and, towards a contradiction, thatthere is a nonempty basic ρ I -open set V such that F = { n ∈ N : ϕ n ∈ V } belongs to I .From this and Lemma 3.15 the following set belongs to I : E = F ∪ { n ∈ N : ϕ n V and ψ n ∈ V } . Let β be the characteristic function of E . Since V is a basic open set of the form (2), thereis m such that V ∩ ( β, m ) − = ∅ . Since Y is ρ I -dense, there is n such that ψ n ∈ V ∩ ( β, m ) − .Hence ψ n ( β, m ) + and, by the definition of ψ n , we have that β ( n ) = 0. Therefore n E and ψ n ∈ V , then ϕ n ∈ V . Thus n ∈ F , a contradiction.(3) follows immediately from (2). To see (4), it suffices to show that int X ( I ⋆ ) ( F ′ ) = ∅ ,for all F ∈ I ⋆ . Let F ∈ I ⋆ . By definition, F ′ is nowhere dense in X ( I ). In particular int X ( I ⋆ ) ( F ′ ) = ∅ , as I ⋆ ⊆ I . (cid:3) Now we show that the operation ⋆ is monotone. Lemma 3.17.
Let I and J be ideals over N with J ⊆ I . Then(1) For every basic ρ I -open set V of N × N there are sets W , U such that V = W ∩ U , W is a ρ J -open set and U is a basic ρ I -open set which is also ρ J -dense.(2) If A ⊆ N × N is ρ J -nowhere dense, then A is ρ I -nowhere dense.(3) J ⋆ ⊆ I ⋆ . Moreover, if J ( I , then J ⋆ ( I ⋆ .Proof. (1) Let V be a basic open set, that is, V = m \ i =1 ( α i , p i ) + ∩ l \ j =1 ( β j , q j ) − . Notice that if every α and β belongs to I \ J , then V is ρ J -dense. Thus given such basicopen set V where every α and β belongs to I , we can separate them and form W and U asdesired: For W , we use the α ’s and β ’s belonging to J (put W = 2 N × N in case there is nonein J ) and for U , we use the α ’s and β ’s belonging to I \ J .(2) Let A ⊆ N × N be a ρ J -nowhere dense set. Let V be a basic ρ I -open set of 2 N × N .Then V = W ∩ U where W and U are as given by part (1). As A is ρ J -nowhere dense, thereis a non empty ρ J -open set W ′ ⊆ W such that W ′ ∩ A = ∅ . Since W ′ is also ρ I -open and U is ρ J -dense, then U ∩ W ′ is a non empty ρ I -open set disjoint from A and contained in V .(3) Since X is dense in 2 N × N , then A ∈ nwd ( X ( I )) if, and only if, A is nowhere dense in(2 N × N , ρ I ). From this and (2) we immediately get that J ⋆ ⊆ I ⋆ . Finally, notice that fromLemma 3.7, we have that for A ∈ I \ J , the set D ( A ) is nowhere dense in X ( I ) and densein X ( J ). (cid:3) Next result gives a sufficient condition for Y ( I ⋆ ) to be nodec. It is a generalization of aresult from [16]. Lemma 3.18.
Let I be an ideal over N and F ⊆ N .(1) If F ∈ I , then b F is closed discrete in Y ( I ) .(2) Let I be such that I ⋆ ⊆ I . If b F is nowhere dense in Y ( I ⋆ ) , then F ∈ I ⋆ .(3) If I ⋆ ⊆ I , then Y ( I ⋆ ) is nodec. roof. (1) is Lemma 4.1 from [16] we include the proof for the reader’s convenience.Since I is hereditary, it suffices to show that b F is closed for every F ∈ I . Let F ∈ I andlet F denote also its characteristic function. Notice that for each m ∈ N , if C = { n ∈ N : ψ n ∈ ( F, m ) + } , then b C is closed in Y ( I ). We claim that F = \ m ∈ N { n ∈ N : ψ n ∈ ( F, m ) + } . From this it follows that b F is closed in Y ( I ). To show the equality above, let n ∈ F , then bythe definition of ψ n , we have that ψ n ∈ ( F, m ) + for all m ∈ N . Conversely, suppose n F and let ϕ n be [ s ] × { m } ∪ · · · ∪ [ s k ] × { m k } . Pick m
6∈ { m , · · · , m k } , then ϕ n ( F, m ) + and thus ψ n ( F, m ) + by the definition of ψ n .(2) is a generalization of Lemma 4.3 of [16]. Let b F be nowhere dense in Y ( I ⋆ ) and suppose,towards a contradiction, that F
6∈ I ⋆ . Let V be a basic ρ I -open set such that F ′ ∩ V is ρ I -dense in V . By Lemma 3.17, there are sets W and U such that V = W ∩ U , W is a ρ I ⋆ -open set, U is a basic ρ I -open set and U is also ρ I ⋆ -dense. Since b F is nowhere dense in Y ( I ⋆ ), there is a basic ρ I ⋆ -open set W ′ ⊆ W such that b F ∩ W ′ = ∅ , that is F ∩ { n ∈ N : ψ n ∈ W ′ } = ∅ . From Lemma 3.15 we know that { n ∈ N : ϕ n ∈ W ′ } \ { n ∈ N : ψ n ∈ W ′ } ∈ I ⋆ . From this and the previous fact we get F ∩ { n ∈ N : ϕ n ∈ W ′ } ∈ I ⋆ . This says that F ′ ∩ W ′ is nowhere dense in X ( I ), which is a contradiction, as by construction, F ′ ∩ V is ρ I -dense in V and W ′ ∩ U ⊆ V is a non empty ρ I -open set (it is non empty as U is ρ I ⋆ -dense).(3) follows immediately from (1) and (2). (cid:3) The natural bijection ψ n ϕ n is not continuous (neither is its inverse), however it hassome form of semi-continuity as we show below. Proposition 3.19.
Let I be an ideal over N . Let Γ : Y → X given by Γ( ψ n ) = ϕ n . Let α ∈ I and p ∈ N . Then Γ − (( α, p ) + ∩ X ) is open in Y ( I ) . In general, if V is a ρ I -basicopen set, then there is D ⊆ Y closed discrete in Y ( I ) and an ρ I -open set W such that Γ − ( V ∩ X ) = ( W ∩ Y ) ∪ D .Proof. Let α ∈ I and p ∈ N . Let O = { ψ n : ϕ n ∈ ( α, p ) + } . We need to show that O is openin Y ( I ). Let F = α − (1). Since (( α, p ) + ∩ Y ) \ b F ⊆ O ⊆ ( α, p ) + ∩ Y , there is A ⊆ F suchthat O = (( α, p ) + ∩ Y ) \ b A . As A ∈ I , then by Lemma 3.18, b A is closed discrete in Y ( I ).Thus O is open in Y ( I ). On the other hand, { ψ n : ϕ n ∈ ( α, p ) − } = (( α, p ) − ∩ Y ) ∪ ( { ψ n : ϕ n ∈ ( α, p ) − } ∩ b F ). (cid:3) The derivative operator on Y ( I ) can be characterized as follows. Proposition 3.20.
Let I be an ideal over N and A ⊆ N . Then ψ l is a ρ I -accumulationpoint of b A if, and only if, for every non empty ρ I -open set V with ψ l ∈ V we have { n ∈ A : ϕ n ∈ V } 6∈ I . roof. Let V be a ρ I -open set with ψ l ∈ V . Suppose F = { n ∈ A : ϕ n ∈ V } ∈ I . Then byLemma 3.18, b F is closed discrete in Y ( I ) which is a contradiction as ψ l is an accumulationpoint of b F . Conversely, let V be a basic ρ I -open set containing ψ l . By Lemma 3.15 thefollowing set belongs to I : E = { n ∈ N : ϕ n ∈ V and ψ n V } . We also have F = { n ∈ A : ϕ n ∈ V } ⊆ { n ∈ A : ϕ n ∈ V and ψ n ∈ V } ∪ E. Since E ∈ I and by hypothesis F
6∈ I , then there are infinitely many n ∈ A such that ψ n ∈ V and we are done. (cid:3) Now we show that the spaces X ( I ) and Y ( I ) are not homeomorphic in general. Proposition 3.21.
Let I be a tall ideal over N . There are no non trivial convergent se-quences in Y ( I ) . In particular, Y ( I ) is not homeomorphic to X ( I ) .Proof. Let A ⊆ N be an infinite set. We will show that b A = { ψ n : n ∈ A } is not convergentin Y ( I ). Since I is tall, pick B ⊆ A infinite with B ∈ I . Then b B is closed discrete in Y ( I )(by Lemma 3.18). Thus b A is not convergent. From this, the last claim follows since X ( I )has plenty of convergent sequences (see Proposition 3.9). (cid:3) Next result shows that our spaces are analytic.
Lemma 3.22.
Let I be an analytic ideal over N . Then I ⋆ is analytic.Proof. The argument is analogous to that of the Lemma 4.8 of [16]. We include a sketch ofit for the sake of completeness. First, we recall a result from [16] (see Lemma 4.7).
Claim:
Let J be an infinite set. Then M ⊆ J is nowhere dense if, and only if, there is C ⊆ J countable such that M ↾ C = { x ↾ C : x ∈ M } is nowhere dense in 2 C Let Z be the set of all z ∈ (2 N × N ) N such that z ( k ) = z ( j ) for all k = j and { z ( k ) : k ∈ N } ⊆ I × N . Since I is an analytic set, then Z is an analytic subset of (2 N × N ) N .Consider the following relation R ⊆ P ( N ) × (2 N × N ) N :( F, z ) ∈ R ⇔ z ∈ Z and { ϕ n ↾ { z ( k ) : k ∈ N } : n ∈ F } is nowhere dense in 2 { z ( k ): k ∈ N } . Then R is an analytic set. From the claim above, we have F ∈ I ⋆ ⇔ ( ∃ z ∈ (2 N × N ) N ) R ( F, z ) . Thus, I ⋆ is analytic. (cid:3) Finally, we can show one of our main results. Let us define a sequence ( I k ) k ∈ N of idealson N as follows: I k = (cid:26) N , if k = 0 , ( I k − ) ⋆ , if k > . Notice that I k +1 ( I k for each k ∈ N by Lemma 3.17. Theorem 3.23.
For all k > , Y ( I k ) is analytic, nodec and crowded. roof. That Y ( I k ) is analytic and nodec follows from Lemmas 3.22, 3.1, 3.18 and 3.17. Since I k ⊆ I nd , then by Lemma 3.16, Y ( I k ) is crowded. (cid:3) Thus we do not know whether Y ( I ⋆ ) is nodec for ideals such that I ⋆
6⊆ I . The reason isthat it is not clear if part (2) in Lemma 3.18 holds in general without the assumption that I ⋆ ⊆ I . In this respect, we only were able to show the following. Lemma 3.24.
Let I be an ideal on N such that I ⊆ I ⋆ . Let A ⊆ N . Then(1) Let V be a non empty ρ I -open set. If A ′ is ρ I -dense in V , then b A is ρ I -dense in V .(2) If b A is nowhere dense in Y ( I ) , then A ′ is nowhere dense in X ( I ) (i.e., A ∈ I ⋆ ). Inparticular, if b A is nowhere dense in Y ( I ) , then b A is closed discrete in Y ( I ⋆ ) .Proof. (1) Let V be a non empty ρ I -open set and suppose A ′ is ρ I -dense in V . Let W be abasic ρ I -open set with W ⊆ V . We need to find n ∈ A such that ψ n ∈ W . By Lemma 3.15the following set belongs to I : E = { n ∈ N : ϕ n ∈ W and ψ n W } . As I ⊆ I ⋆ , then E ′ is nowhere dense in X ( I ). Since A ′ is dense in V , then A ′ ∩ W E ′ .Let n ∈ A \ E such that ϕ n ∈ W . As n E , then ψ n ∈ W .(2) Follows from (1) and part (1) in Lemma 3.18. (cid:3) Now we compare the dense sets in Y ( I ) and X ( I ). Lemma 3.25.
Let I be an ideal on N such that Y is dense in (2 N × N , ρ I ) and D ⊆ N . If b D is dense in Y ( I ) , then D ′ is dense in X ( I ) .Proof. Suppose b D is dense in Y ( I ). Let V be a basic ρ I -open set. We need to find n ∈ D such that ϕ n ∈ V . By Lemma 3.15 the following set belongs to I : E = { n ∈ N : ϕ n V and ψ n ∈ V } . Let F = { n ∈ D : ψ n ∈ V } . Since b D is ρ I -dense, then F
6∈ I (by part (1) of Lemma 3.18and the assumption that Y is dense in (2 N × N , ρ I )). Thus there is n ∈ F \ E . Then ψ n ∈ V and ϕ n ∈ V . (cid:3) Observe that
Fin ⊆ Fin ⋆ ⊆ Fin ⋆⋆ ⊆ · · · ⊆ I k for all k . Notice that Fin ⋆ is isomorphic to nwd ( Q ) as X ( Fin ) is homeomorphic to Q . The following is a natural and intriguing question. Question 3.26. Is Y ( Fin ⋆ ) nodec? It is unclear when an ideal I satisfies either I ⊆ I ⋆ or I ⋆ ⊆ I . The following questionasks a concrete instance of this problem. Question 3.27.
Two ideals that naturally extend
Fin are {∅} ×
Fin and
Fin × {∅} (where × denotes the Fubini product). Let I be any of those two ideals. Is I ⊆ I ⋆ ? .4. SS property in Y ( I ) . We do not know whether Y ( I nd ) is SS . However, we showbelow that Y ( I k ) is not wSS for all k >
1, this was the reason to introduce the ideals I ⋆ .We need an auxiliary result. Lemma 3.28.
Let I be an ideal over N such that I ⋆ ⊆ I . Let V be a non empty ρ I ⋆ -openset and D ⊆ N . If D ′ is ρ I -dense in V , then b D in ρ I ⋆ -dense in V .Proof. Let V be a non empty ρ I -open set and suppose that D ′ is ρ I -dense in V . Let W bea ρ I ⋆ -basic open set such that W ⊆ V . We need to show that there is n ∈ D such that ψ n ∈ W . By Lemma 3.15 the following set belongs to I ⋆ : E = { n ∈ N : ϕ n ∈ W and ψ n W } . Since W is also ρ I -open (as I ⋆ ⊆ I ) and E ′ is ρ I -nowhere dense, then there is a non empty ρ I -open set V ⊆ W such that V ∩ E ′ = ∅ . Since D ′ is ρ I -dense in V , there is n ∈ D suchthat ϕ n ∈ V . Notice that n E . Since ϕ n ∈ W , then ψ n ∈ W . (cid:3) Theorem 3.29.
Let I be an ideal over N such that I ⋆ ⊆ I . Then Y ( I ⋆⋆ ) is not wSS .Proof. Notice that X is ρ I crowded (see Lemma 3.16). Also, observe that I ⋆⋆ ⊆ I ⋆ (seeLemma 3.17). Let ( U n ) n ∈ N be a pairwise disjoint sequence of non empty ρ I ⋆⋆ -open sets. Let A n = { m ∈ N : ϕ m ∈ U n } . It is clear that A n / ∈ I ⋆ for each n ∈ N . It is easy to verify thatthe sequence ( A m ) m witnesses that I ⋆ is not p + . Let D n = D ( A n ), as defined in Lemma 3.7.Let E n = { ψ m ∈ Y : ϕ m ∈ D n } . We claim that the sequence ( E n ) n ∈ N witnesses that the space Y ( I ⋆⋆ ) is not wSS . In fact,since A n / ∈ I ⋆ , then D n is dense in X ( I ⋆ ) (by Lemma 3.7), so E n is dense in Y ( I ⋆⋆ ) (byLemma 3.28). Let K n ⊆ E n be a finite set and L n = { ϕ m : ψ m ∈ K n } for each m ∈ N . Since A n
6∈ I ⋆ and I ⋆ is not p + , then, by the proof of Theorem 3.8, L = S n ∈ N L n is nowhere densein X ( I ⋆ ). Thus L ∈ I ⋆⋆ . Therefore b L = S n ∈ N K n is closed discrete in Y ( I ⋆⋆ ) (by Lemma3.18). (cid:3) We have seen in Theorem 3.23 that Y ( I k ) is nodec for every k ≥
1. From Theorem 3.29we have the following.
Corollary 3.30. Y ( I k ) is not wSS for every k > . Recall that I is I nd . We do not know whether Y ( I nd ) is SS. We only know the following.Suppose c D n = { ψ m : m ∈ D n } is open dense in Y ( I nd ), for every n ∈ N . Then there is F n ⊆ D n finite for each n such that S n ∈ N c F n is dense. Question 3.31.
Is there an ideal I on N such that I ⊆ I ⋆ and Y ( I ⋆ ) is wSS ? In particular,is Y ( Fin ⋆ ) wSS ? q + in Y ( I ) . We shall prove that for certain kind of ideals, Y ( I ) is not q + . We use aconstruction quite similar to that in the proof of Theorem 3.6.We recall that in the proof of Lemma 3.5 we have introduced the following property: Let m ∈ N . We say that ϕ ∈ X has the property ( ∗ m ) if there are k ∈ N , s i ∈ m +1 ( i = 1 , ..., k )finite sequences and m i ≤ m ( i = 1 , ..., k ) natural numbers such that ϕ = S ki =1 [ s i ] × { m i } and if m i = m j with i = j , then s i ↾ m = s j ↾ m . emma 3.32. Let I be an ideal over N such that I ⊆ I nd . Let A m = { ϕ ∈ X : ϕ has the property ( ∗ m ) } and B m = { ψ n ∈ Y : ϕ n ∈ A m } . Let L = { n ∈ N : ϕ n / ∈ S m ∈ N A m } and suppose there is an infinite set L ′ = { m k : k ∈ N } ⊆ L such that L ′ ∈ I . Let B = [ k B m k . Let q ∈ N be such that ϕ q = 2 N × { } . Then(1) B is dense in Y ( I ) and, in particular, ψ q ∈ cl ρ I ( B ) .(2) Let S ⊆ B be such that S ∩ B m k has at most one element for each k , then ψ q cl ρ I ( S ) .Proof. (1) Let A = S k A m k . By Lemma 3.5, A is dense in X . Thus by Lemma 3.28, B isdense in Y ( I nd ) (recall that I nd = (2 N ) ⋆ ). As I ⊆ I nd , then B is also dense in Y ( I ).(2) Let S = { ψ n k : k ∈ N } be such that ψ n k ∈ B m k for all k ∈ N . We will show that ψ q cl ρ I ( S ).Let α ∈ N be defined as follows: If 0 ∈ L ′ and [ h i ] ×{ } ⊆ ϕ n , then α (0) = 1. Otherwise, α (0) = 0. For n > α ( n ) = , if n ∈ L ′ , n = m k for some k and [ h α (0) , ..., α ( n − , i ] × { } ⊆ ϕ n k . , otherwise.Observe that α ∈ I , as α − (1) ⊆ L ′ ∈ I .It is clear that ψ q ∈ ( α, + . To finish the proof, it suffices to show that ( α, / ∈ S k ∈ N ψ n k .Suppose, towards a contradiction, that there is l ∈ N such that ( α, ∈ ψ n l , that is, ( α, ∈ ϕ n l ∪ ([ n l ] × N ). There are two cases to be considered.(i) Suppose α ( n l ) = 1. Then n l ∈ L ′ and thus ϕ n l A m l which contradicts that ψ n l ∈ B m l .(ii) Suppose α ( n l ) = 0 and thus ( α, ∈ ϕ n l . Let ϕ n l = S ri =1 [ s i ] × { p i } with s i ∈ m l +1 .Then α ∈ [ s ], where s is s i for some i with p i = 0. Hence α ( n ) = s ( n ) for all n ≤ m l . Weconsider two cases. Suppose α ( m l ) = 1. Then s ( m l ) = 1. Let t be such that s = t b
1. Then bythe definition of α , we have that [ t b ×{ } ⊆ ϕ n l . But also [ s ] ×{ } = [ t b ×{ } ⊆ ϕ n l whichcontradicts that ϕ n l ∈ A m l (i.e. that it has property ( ∗ m l )). Now suppose that α ( m l ) = 0.Then [ s ] × { } = [ t b × { } 6⊆ ϕ n l , but this contradicts that [ s ] × { } is [ s i ] × { p i } for some i . (cid:3) From the previous lemma we immediately get the following.
Theorem 3.33.
Let I be a tall ideal over N such that I ⊆ I nd . Then Y ( I ) is not q + . Question 3.34.
Is there an ideal (necessarily non tall) different from
Fin such that Y ( I ) isq + ? Two natural candidates are {∅} × Fin and
Fin × {∅} . Finally, we have the following.
Theorem 3.35. Y ( I k ) is a non SS, non q + nodec regular space with analytic topology forevery k > . cknowledgment: We are thankful to the referee for his (her) comments that improvedthe presentation of the paper.
References [1] D. Barman and A. Dow. Selective separability and SS + . Topology Proc. , 37:181–204, 2011.[2] D. Barman and A. Dow. Proper forcing axiom and selective separability.
Top. and its Appl. , 159(3):806– 813, 2012.[3] A. Bella, M. Bonanzinga, and M. Matveev. Variations of selective separability.
Top. and its Appl. ,156(7):1241 – 1252, 2009.[4] A. Bella, M. Bonanzinga, M. Matveev, and V. Tkachuk. Selective separability: general facts and behaviorin countable spaces.
Topology Proceedings , 32:15–30, 2008.[5] A. Bella. When is a Pixley-Roy hyperspace SS + ? Top. and its Appl. , 160(1):99 – 104, 2013.[6] J. Camargo and C. Uzc´ategui. Selective separability on spaces with an analytic topology.
Topology andits applications , 248(1):176–191, 2018.[7] A. Dow, M. G. Tkachenko, V. V. Tkachuk, and R. G. Wilson. Topologies generated by discrete subspaces.
Glas. Math. Ser. III , 37(57):187–210, 2002.[8] G. Gruenhage and M. Sakai. Selective separability and its variations.
Top. and its Appl. , 158(12):1352– 1359, 2011.[9] M. Hru˘s´ak, D. Meza-Alc´antara, E. Th¨ummel, and C. Uzc´ategui. Ramsey type properties of ideals.
Annals of Pure and Applied Logic , 168(11):2022–2049, 2017.[10] A. S. Kechris.
Classical Descriptive Set Theory . Springer-Verlag, 1994.[11] O. Nj˚astad. On some classes of nearly open sets.
Pacific. J. Math. , 15:961–970, 1965.[12] D. Repovˇs and L. Zdomskyy. On M -separability of countable spaces and function spaces. TopologyAppl. , 157(16):2538–2541, 2010.[13] M. Scheepers. Combinatorics of open covers VI: Selectors for sequences of dense sets.
QuaestionesMathematicae , 22(1):109–130, 1999.[14] S. Todorˇcevi´c and C. Uzc´ategui. Analytic topologies over countable sets.
Top. and its Appl. , 111(3):299–326, 2001.[15] S. Todorˇcevi´c and C. Uzc´ategui. Analytic k -spaces. Top. and its Appl. , 146-147:511–526, 2005.[16] S. Todorˇcevi´c and C. Uzc´ategui. A nodec regular analytic space.
Top. and its Appl. , 166:85–91, 2014.[17] E. Van Douwen.
Applications of maximal topologies.
Top. and its Appl., 51:125-139, 1993.
Escuela de Matem´aticas, Facultad de Ciencias, Universidad Industrial de Santander,Ciudad Universitaria, Carrera 27 Calle 9, Bucaramanga, Santander, A.A. 678, COLOMBIA.
E-mail address : javier [email protected]. Escuela de Matem´aticas, Facultad de Ciencias, Universidad Industrial de Santander,Ciudad Universitaria, Carrera 27 Calle 9, Bucaramanga, Santander, A.A. 678, COLOMBIA.Centro Interdisciplinario de L´ogica y ´Algebra, Facultad de Ciencias, Universidad de LosAndes, M´erida, VENEZUELA.
E-mail address : [email protected]@saber.uis.edu.co.