FRÉCHET-URYSOHN FOR FINITE SETS, II

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1 FRÉCHET-URYSOHN FOR FINITE SETS, II GARY GRUENHAGE AND PAUL J. SZEPTYCKI Abstract. We continue our study [10] of several variants of the property of the title. We answer a question in [10] by showing that a space defined in a natural way from a certain Hausdorff gap is a Fréchet α 2 space which is not Fréchet-Urysohn for 2-point sets (FU 2 ), and answer a question of Hrusak by showing that under MA ω1, no such gap space is FU 2. We also introduce versions of the properties which are defined in terms of selection principles, give examples when possible showing that the properties are distinct, and discuss relationships of these properties to convergence in product spaces, to the α i -spaces of A.V. Arhangel skii, and to topological games. 1. Introduction For a space X and a point x X, a family P of subsets of X is said to be a π-network at x iff for each open U containing x, there is P P such that P U. We will say that a sequence P n, n ω, converges to x, and write P n x, iff every neighborhood of x contains P n for all but finitely many n. We will also say that a countably infinite family P of subsets of X converges to x iff the sequence formed by any one-to-one enumeration of its elements converges to x; equivalently, for each open U containing x, the set {P P : P U} is finite. A space X is said to be Fréchet-Urysohn for finite sets (respectively, n-sized sets), which we will denote by FU fin (respectively, FU n ), if for each x X and each P [X] <ℵ0 (resp, P [X] n ), if P forms a π-network at x, then P contains a subfamily that converges to x. We also say that X is boundedly FU fin if X is FU n for every n ω. Though the concept appeared earlier (without being named), Reznichenko and Sipacheva[20] were the first to undertake a detailed investigation of the FU fin property. A primary motivation was the problem due to Malychin whether there could be in ZFC a separable Fréchet topological group which is not metrizable. They showed that if there is a countable FU fin space which is not first countable, then there is such a group. Whether or not there is in ZFC such a group or a countable FU fin space which is not first countable is still an open question. In [10], we continued their investigation of FU fin spaces, and also introduced the related FU n and boundedly FU fin spaces. Sipacheva s paper[23] contains some characterizations of these properties in terms of Fréchetness of products, and shows that standard Fréchetness in groups and some other spaces with structure often implies some of the stronger Fréchetness with respect to finite set properties. Date: 01/29/ Mathematics Subject Classification. Primary 54A20, Secondary 54D55, 54H11. Key words and phrases. Fréchet-Urysohn, π-network, α i -spaces. The first author acknowledges support from National Science Foundation grant DMS The second author acknowledges support from NSERC grant

2 2 GARY GRUENHAGE AND PAUL J. SZEPTYCKI This paper is a natural continuation of [10]. In Section 2, we consider a class of countable spaces obtained in a natural way from Hausdorff gaps. Motivated by a connection with a certain topological game, we had asked in [10] if there is a Fréchet α 2 -space which is not FU fin. (See below for the definition of the α i -spaces.) Here we show that there is a gap space that is a Fréchet α 2 space which is not even FU 2. While all such gap spaces are Fréchet α 2 in ZF C, we show that in fact none are FU 2 under MA ω1. This answers a question of M. Hrusak, who had asked if there is in ZF C a gap space which is FU fin. On the other hand, we show that the generic gap added by Hechler forcing always produces a FU fin space. In [10], we showed that the following selection principle versions of the FU fin notion were equivalent to FU fin : whenever P 0, P 1, P 2,... is a sequence of π-nets of finite sets at the point x, there exist F n P n for every (respectively, infinitely many) n ω such that the F n s converge to x. However, for the other FU-properties, the selection versions are not necessarily the same. Let us call the for all (respectively, for infinitely many ) version the σ- (respectively, weakly σ-) version. Note that even for singleton sets, the selection versions are different, for in this case the σ-version is equivalent to Fréchet α 2 and the weakly σ-version to Fréchet α 4, which are distinct properties stronger than Fréchet-Urysohn. (The Fréchet-Urysohn property, i.e., FU 1, is often called simply Fréchet.) So we make the following definiton: Definition. Given a π-net P (at x) of finite sets with property, we write: (a) FU to mean for every π-net P with property, there are P n P with P n x; (b) (weakly) σ-fu to mean if P n,n ω, is a sequence of π-nets having property, then for every (for infinitely many) n in ω, there are P n P n with P n x. So, e.g., (weakly) σ-fu 5 is: Given π-nets P n, n ω, where each P P n has cardinality 5, then n( n) there are P n P n with P n x. Also, (weakly) σ-boundedly-fu fin is: Given π-nets P n, n ω, where { P : P P n } is bounded for each n, then for every (infinitely many) n, there are P n P n with P n x. As noted above, (weakly)-σ-fu 1 is equivalent to Fréchet α 2 (α 4 ). In Section 3 we discuss the inter-relationships of these properties, giving examples separating the properties when possible. In several cases, we can give consistent examples but do not know if there may be ZF C examples. In Section 4 the relationships of these properties to a certain game, and to convergence in product spaces and the α i -properties are given. We end our introduction by reminding the reader of the definition of α i -spaces, introduced by Arhangel skii [1]. Let X be a space, and x X. Suppose that for any countable family {A n } n ω of sequences converging to x, there is a sequence A converging to x such that: (1) A n \ A < ω for every n ω, then x is an α 1 -point; (2) A n A = ω for every n ω, then x is an α 2 -point; (3) A n A = ω for infinitely many n ω, then x is an α 3 -point; (4) A n A = for infinitely many n ω, then x is an α 4 -point. X is an α i -space if every point is an α i -point. 2. Gap spaces An example due to J. Isbell, appearing in [O], produces two countable Fréchet α 2 -spaces whose product is not Fréchet. The assumption 2 ℵ0 < 2 ℵ1 is used in [O]

3 FRÉCHET-URYSOHN FOR FINITE SETS, II 3 in describing Isbell s example, and it is only claimed that the spaces are countably bi-sequential (=Fréchet α 4 ). But P. Nyikos [18] noticed that the examples are α 2 - spaces, and that what is needed to construct the examples is a Hausdorff gap, so they exist in ZFC. Recall that an ω 1 -sequence (a 0 α, a 1 α : α < ω 1 ) of pairs of infinite subsets of ω is a Hausdorff gap if (a) a 0 α a 0 β a 1 β a 1 α for all α < β < ω 1 ; (b) There is no c such that a 0 α c a 1 α for all α < ω 1. (Recall a b means a \ b < ω.) Given a Hausdorff gap as above, let I 0 = {a ω : a a 0 α < ω for all α < ω 1 }, I 1 = {a ω : a (ω \ a 1 α) < ω for all α < ω 1 }. Then Nyikos s observation is that the spaces X e = ω { }, where neigborhoods of are complements of members of the ideal I e, are the same as Isbell s spaces and are Fréchet α 2. We call a space obtained from (either the left or right side of) a Hausdorff gap g in this way a gap space, and the corresponding filter F g on ω a gap filter. (We will also say that the gap filter has a certain covergence property iff the corresponding gap space does.) It is not hard to show that the product X 0 X 1 of the above gap spaces is not Fréchet. (See [O] and [18], or Example 2.4 in [7].) We asked in [10] if there is a Fréchet α 2 -space which is not FU fin. (See Section 4 for a game theoretic motivation for this question.) We will show that the Isbell- Nyikos example can be modified to produce, in ZFC, a gap space which is not FU 2. Example 2.1. There is a gap space X = ω { } which is Fréchet α 2 but not FU 2. Proof. Let X 0 and X 1 be the gap spaces as described above. Let Y 1 be the space X 1 using a disjoint copy ω of ω, and let X be the space obtained by identifying the points of X 0 and Y 1. Note that X is also a gap space via the Hausdorff gap (a 0 α (ω \ a 1 α), a 1 α (ω \ a 0 α) : α < ω 1 ) in ω ω. (Here c denotes the copy in ω of a subset c of ω.) So X is Fréchet α 2. Let P = {{n, n } : n ω}. Any neighborhood U of has to almost contain every a 0 α, so there is β < ω 1 such that U \a 1 β is infinite. Then (U \a1 β ) is convergent. Hence we can find n U \ a 1 β such that n U, and so {n, n } [U] 2 P. Thus P is a π-net at. But no infinite subcollection C of P converges to. Suppose otherwise, and consider the set c = {n : n, n C}. Then c and c are convergent. Now, c convergent implies c (ω \ a 1 α) is infinite for some α. But then c is not convergent, contradiction. In the other direction, noting that it is not difficult to construct gap spaces that are FU fin under CH or p = c, M. Hrusak asked if there could be in ZFC a gap space which is FU fin. We will show that the answer is no: under MA ω1, no gap space is FU 2. Theorem 2.2. MA(ω 1 ) implies that no gap space is FU 2. Proof. Given a gap g = (a α, b α : α ω 1 ), let P g be the set of pairs (p, F ) such that

4 4 GARY GRUENHAGE AND PAUL J. SZEPTYCKI (1) p F n(ω, ω) and p is one-to-one and the domain and range of p are disjoint. (2) F [ω 1 ] <ω The ordering on P g is defined by 3. (p, F ) < (q, G) if p extends q, F G, and for each n dom(p) \ dom(q) and each α G, (a) n a α if and only if p(n) b α. (b) n b α if and only if p(n) a α. It is a standard argument to show that P g is σ-centered. For each α ω 1 and each n ω, let and let D α,n = {(p, F ) : α F and m (dom(p) a α+1 ) \ (a α n)}, E α,n = {(p, F ) : α F and m (dom(p b α ) \ (b α+1 n)}. Lemma 2.3. D α,n and E α,n are dense in P g for every α and n. Proof.We present the proof for D α,n. By the symmetry in the definition of the order on P g, density of E α,n follows. Fix (p, F ) P g. Without loss of generality {α, α + 1} F. Fix N large enough so that n dom(p) ran(p) N and {a ξ, b ξ : ξ F } is totally ordered by N, where a N b a \ N b \ N. Choose m a α+1 \ N such that m a ξ for each ξ F α + 1. And choose k ω \ (b α+1 N) such that k b ξ for each ξ F (α + 1). Then by choice of N we have for every ξ F (1) m b ξ. (2) k a ξ. (3) m a ξ if and only if ξ > α (4) k b ξ if and only if ξ α Thus (p {(m, k)}, F ) D α,n and, by definition of the order on P g, we have that (p {(m, k)}, F ) < (p, F ). If G P g is {D α,n, E α,n : n, α}-generic and if f : dom(f) ran(f) is the generic function, then both g d = g dom(f) and g r = g ran(f) are pregaps, i.e., satisfy condition (a) of the definition of a gap. Moreover, the function f defines an isomorphism between the pregap g d and the complement of the pregap g r. Either of these pregaps is a gap if and only if the other is a gap. Indeed, the pregap g d is filled by a set x dom(f) if and only if g r if filled by ran(f) \ f(x). Moreover, if dom(f) ran(f) = ω, then both pregaps must in fact be gaps. Let D n = {(p, F ) : n dom(p) ran(p)}. If we could prove that the sets D n are dense, then our theorem would easily follow. However, these sets need not be dense, so our proof is a bit more complicated. The rest of the proof depends on the following property of gaps. Lemma 2.4. For every gap (a α, b α : α ω 1 ) there are α < β such that both a α \b β and a β \ b α are not empty. Proof.Let C = {n ω : α n ω 1 [ α α n (n b α )]}. Claim 1. {(a α \ C, b α \ C) : α < ω 1 } is also a Hausdorff gap.

5 FRÉCHET-URYSOHN FOR FINITE SETS, II 5 Proof of Claim 1. Suppose f ω fills this gap. Then f C is easily seen to fill the original gap. By Claim 1, there is k ω \ C such that A k = {α : k a α } is uncountable. Since k C, the set B k = {β < ω 1 : k b β } is also uncountable. Note that for any α A k and β B k, we have a α \ b β. So the proof is complete once the following claim is established. Claim 2. α A k β B k [(α β) (a β \ b α )]. Proof of Claim 2. Suppose not. Then for each α A k and β B k \ {α}, we have a β \ b α =, i.e., a β b α. Consider the set A = β B k a β. We aim for a contradiction by showing that A fills the gap. Clearly A a α for all α. Now fix α ω 1. Then a α b α and β B k \{α} a β b α, so A a α a β b α, which completes the proof. We now need the following lemma: β B k \{α} Lemma 2.5. If G P g is V-generic and Γ : ω ω is the generic function then P g forces that both of the following sequences are gaps: (a α dom(γ), b α dom(γ) : α ω 1 ) and (a α ran(γ), b α ran(γ) : α ω 1 ). Proof.By Lemma 2.3 they are both pregaps. To see that they are not filled, by symmetry it suffices to prove that the pregrap restricted to the domain of Γ is not filled. Suppose not and let τ be a name for a subset of ω that fills the gap. For each α, fix r α and n α such that r α a α dom(γ) \ n α τ \ n α b α There is an uncountable set X ω 1 and a p F n(ω, ω) and n ω such that for each α X, p α = (p, F α ) and n α = n. Moreover, we may assume α F α for each α and that the F α form a -system with root F. So {p α : α X} is a centred family. Let N be large enough so that dom(p) ran(p) n N. Consider the following gap (A α, B α : α X) defined as follows: Let (1) A = {a ξ : ξ F }. (2) B = {b ξ : ξ F }. (3) A α = {a ξ : ξ F α \ F } (B \ (A N)). (4) B α = {b ξ : ξ F α \ F } (B \ (A N)). Clearly, (A α, B α : α X) is a gap. So by Lemma 2.4, we may fix α < β in X, m A α \ B β, and k A β \ B α. Note that m and k satisfy the following for each ξ F α F β : (a) m b ξ ξ F α (b) m a ξ ξ F α \ F (c) k b ξ ξ F β (d) k a ξ ξ F β \ F. Fix ξ F α F β. Note that if (b) and (c) we have that m a ξ if and only if k b ξ. Also, by (a) and (d) we have that m b ξ if and only if k a ξ.

6 6 GARY GRUENHAGE AND PAUL J. SZEPTYCKI Thus r = (p {(m, k)}, F α F β ) extends both r α and r β. Clearly, α F α and β F β so we have that m a α \ b β. But we have that r m dom(γ) a α \ N But since r extends r α r m τ \ N and since r extends r β we have that r m b β But this contradicts m a α \ b β. Next we need some basic results about indestructibility of (ω 1, ω1) gaps from [11]. Theorem 2.6. (Kunen) Assume that g is an (ω 1, ω 1) pregap, such that a α b α ; then (e) g is a gap in every forcing extension if for all α < β a α \ b β. (f) If g is a gap, let Q be the set of finite subsets s of ω 1 such that a α \ b β for each α < β in s. Then Q is ccc. (g) If g is a gap, then there is α < ω 1 so that for each s Q such that α < min(s), {β > max(s) : s {b} Q} is uncountable. The poset Q described in the above theorem makes any gap indestructible. Hence, MA(ω 1 ) implies that all (ω 1, ω1) gaps are indestructible. Now we are ready to apply MA(ω 1 ) to prove the theorem: We apply MA(ω 1 ) to the iteration of two ccc-posets: P g Q, where P g is the poset described above, and Q is a P g -name for the ccc poset that makes g dom(γ) indestructible. We assume that Q is the subposet that satisfies (g) of Kunen s theorem. Without loss of generality we may assume that for all (p, q) P g Q, (h) there is a finite subset s of ω 1 such that q = š, and (i) if p = (r p, F p ), and α < β are elements of s, then dom(r p ) a α \ b β (Clearly, the set of such conditions is dense). Now we fix a subset G P g Q that is generic for the family of dense sets D α,n, E α n and C α, where C α = {(p, q) : β > α(β q)}. and D α,n = {(p, q) : p D α,n } and E α,n is defined similarly. If f is the generic function defined from the first coordinate, we first verify that g dom(f) is a gap. Density of D α,n and E α,n imply that it is a pregap. To verify that it is a gap, it suffices to verify that Kunen s condition (e) holds. Let X = {q : q [ω 1 ] < ω p(p, q) G} Density of the C α s imply that X is uncountable. It is easy to see that dom(f) a α \ b β for each α < β in X. Indeed this follows from item (i) above. So g dom(f) and thus g ran(f) are both gaps. Now, consider the family H = {{n, f(n)} : n dom(f)}. Claim 1. H is a π-net for the gap filter F g. Proof.Suppose X is a subset of ω which is in the filter for the gap. Thus, a α X for all α. Since g dom(f) is a gap, there are α < β such that Y = X dom(f) (b α \ b β ) is infinite. Since f(b α \ b β a β, it follows that f(y ) a β. Thus, since a β X, there is n Y such that f(n) X. Thus {n, f(n)} X. So H is a π-net.

7 FRÉCHET-URYSOHN FOR FINITE SETS, II 7 Claim 2. For each α, {n : {n, f(n)} a α } is finite. Hence, no subset of H converges in F g. Proof.Fix (p, F ) P g such that α F and ((p, F ), r) G. Then for each n dom(f) \ dom(p), if n a α, then f(n) b α. Thus, for all n dom(f) \ domp we have that {n, f(n)} a α as required. Thus, F g is not F U 2, completing the proof of Theorem 2.2. Theorem 2.7. If g is any gap added generically by Hechler s poset Then F g is a F U fin filter. Proof. Let I = ω 1 {0, 1} have the order < making it order isomorphic to ω 1 + ω1. Recall that Hechler s poset P is the set of all finite partial functions p : I 2 <ω such that (a) For all α, (α, 0) dom(p) iff (α, 1) dom(p). For any such α we also require that p(α, 0)(k) = 1 implies that p(α, 1)(k) = 1. (b) There is n ω (called the height of p and denoted ht(p)) such that p(i) 2 n for all i dom(p). The ordering on P is defined by p q if (c) ht(q) ht(p) (d) dom(q) dom(p) (e) q(i) p(i) for all i domq (f) For all i < j in the domain of q, p(i)(k) = 1 implies that p(j)(k) = 1 for every ht(q) k < ht(p). If G is P -generic over some model M, in M[G] we define sets a α and b α as follows: k a α if p(α, 0)(k) = 1 for some p G. And k b α if p(α, 1)(k) = 1 for some p G. It is clear from the definition of the order on P and by a standard density argument that (a α, b α ) is a pre-gap. The fact that it is a gap is also standard. For now, let a α and b α also denote the P -names for these elements. We claim that the filter F g determined by this generic gap is always FU fin. To see this, suppose that σ is a P -name for a subset of [ω] <ω. And suppose that p P and p forces that σ is a π-net at. I.e., p α n y σ such that y b α \ n. We now claim that p also forces that there is some α such that for every n some element of σ is a subset of a α \ n. I.e., we claim that p α n y σ such that y a α \ n. Assuming this is the case, it is easy to recursively define in M[G] a sequence y n σ G such that y n a α \ n. Then {y n : n ω} is necessarily infinite and converges to. Proving that the filter is F U fin. So, by way of contradiction assume not. I.e., there is some extension q p such that for all α, q n y σ(y (a α \ n)). For each α fix q α < q and n α ω such that q α y σ(y (a α \ ň α )). Without loss of generality we may assume that α dom(q α ) for each α. P is ccc, so without loss of generality there is a β < ω 1 such that for every q that appears in the name σ, dom(q) β {0, 1}. In particular, if for some

8 8 GARY GRUENHAGE AND PAUL J. SZEPTYCKI k [ω] <ω and some q P we have that q k σ, then q (dom(q) β {0, 1}) also forces k σ. Henceforth, we will abbreviate restrictions of conditions like q (dom(q) β {0, 1}) as just q β. Now, choose α > β. Without loss of generality assume that n α = 0. Let n > ht(q α ). Since q α < p, we have that q α y σ such that y b α \ n. Thus we may fix r q α and a y [ω \ n] <ω such that r ˇy σ and ˇy b α In particular r forces that y is an element of σ. Thus, r = r β also forces that y σ. Note that the height of r and r are the same and is greater than the maximum of y which of course is greater than n, the height of q α. We also have that r q α β and that r and q α are compatible (r is a common extension). We now define a possibly different common extension q. We require that dom(q) = dom(r ) dom(q α ) and that m = ht(q) is equal to the height of r. We define each q(i) as follows: (g) For i dom(r ) let q(i) = r (i). Let i 0 the the <-largest element of dom(q α ) below (β, 0). Thus, i 0 is also in the domain of r and so q(i 0 ) is defined by item (g). (h) If i dom(q α ) \ dom(r ) then (β, 0) < i < (β, 1) and we let q(i) n = q α (i). (i) For i dom(q α ) \ dom(r ) and for k m \ n let q(i)(k) = 1 if either k y or q(i 0 )(k) = 1. Items (g), (h) and (i) completely define the condition q. Claim 1: q ˇy a α \ n α Proof. Fix k y. Since (α, 0) dom(q α ) \ dom(r ), q(α, 0)(k) is defined to be 1 by clause (i) in the definition of q. Thus for each k y we have that q k a α. Claim 2: q q α Proof. Certainly (c), (d) and (e) in the definition of are satisfied. To verify (f) we need to fix i < j in the domain of q α and fix k in the interval [ht(q α ), ht(q)).we need to check that q(i)(k) = 1 q(j)(k) = 1. Assume that q(i)(k) = 1. (1) If both i, j dom(r ). Then by clause (g) in the definition of q and since r < q α β we have that q(j)(k) = 1. (2) Both i, j dom(q α )\dom(r ). Then i 0 < i. So either k y, or q(i 0 )(k) = 1. In either case q(j)(k) = 1. (3) If i i 0 and (β, 0) < j < (β, 1). Then q(i)(k) = 1 implies that q(i 0 )(k) = 1 since r < q α β. Thus q(j)(k) = 1 by (i). (4) If (β, 0) < i < (β, 1) < j. In the case that q(i 0 )(k) = 1, note that r (i 0 ) = q(i 0 ) and that j dom(r ). Thus r (j) = q(j). Also, r (i 0 )(k) = 1. Thus, the fact that r < q α β allows us to to conclude that r (j)(k) = 1. Thus q(j)(k) = 1 as required. In the case that k y, we need to use the facts that (α, 1) < j and that r ˇy b α. Thus, since k y we have that r(α, 1)(k) = 1. Also, since both (α, 1) and j are in the domain of q α and r < q α we may conclude that r(j)(k) = 1. But q(j) = r (j) = r(j) since j β {0, 1}. Thus q(j)(k) = 1 as required.

9 FRÉCHET-URYSOHN FOR FINITE SETS, II 9 Claim 3: q r Proof. Trivial since q dom(r ) = r. We now finish the proof of the theorem: by Claim 3, we have that q y σ. So Claims 1 and 2 give us a contradiction since q α forces there is no such y in σ. 3. Selection principle versions The table below shows the relationships among the properties we are considering (see introduction for the definitions). FU fin σ-fu fin weakly σ-fu fin 1 σ-bddly-f U weakly-σ-bddly-f U 2 n(σ-fu n ) 3 bddly-fu = n(fu n ) n(weakly-σ FU n ) 4 5 σ-fu n FU n weakly-σ-fu n 8 FU n All but one of the implications in the chart are trivial or proven in [10]. The exception is the equivalence of σ-boundedly FU fin and weakly σ-boundedly FU fin. Theorem 3.1. A space X is σ-boundedly FU fin iff it is weakly σ-boundedly FU fin. Proof. Since the other direction is trivial, we need only show that if X is weakly σ-boundedly FU fin, then it is is σ-boundedly FU fin. Suppose P n, n ω, is a sequence of π-nets at x X, such that each P n consists of sets bounded in cardinality by some k n ω. Let Pn = { i n P i : P i P i }. It is easy to check that each Pn is a π-net at x consisting of sets bounded in cardinality by Σ i n k i. By the weak boundedly FU fin property, there are P n Pn for each n in some infinite set A ω such that P n x. Let P n = i n P ni. If {n 0, n 1,... } is the increasing enumeration of A, then Q m = P nmm P m for each m, and Q m x. Next we present a series of examples showing that implications in the chart need not reverse. Example 3.6 below shows in ZFC that implication 3 does not reverse. All of the other examples are consistent ones, constructed using CH, though p = c would do. Example 3.2 gets implication 8, Example 3.3 gets 6 and 7, Example 3.4 gets 4, Example 3.7 gets 2, and Example 3.9 gets 1. Not only do we not know ZFC examples showing 1,2,4,6,7, or 8 do not reverse, but for all we know, substantial portions of the chart could collapse in some models. Question 1. Is there in ZFC an FU 2 -space which is not boundedly FU fin? σ-fu 2 space which is not FU fin? The Fréchet fan is a FU 1, not weakly σ-fu 1 space in ZFC. However, for higher n, we only have the following examples under CH (p = c is enough). Example 3.2. (CH) A FU n, not weakly σ-fu n+1 space. A

10 10 GARY GRUENHAGE AND PAUL J. SZEPTYCKI Proof. Let n > 1, and let Y = n ω ω. Let d j,k = n {j} {k}, D k = {d j,k : j ω}, and D = k ω D k. We are going to make each D k a π-net (in fact a convergent sequence) of n-element sets, but no selection of one member of each of infinitely many D k s is going to be convergent. Of course, to make the space F U n, this means that no subset of D meeting each D k in a finite set can be a π-net. Let P α and E α, α < ω 1, list [[Y ] n \ D]] ω and {E [D] ω : k( E D k < ω)}, respectively. For S Y, let D(S) = {d D : d S }. We inductively define I α, α < ω 1, the complements of which will be a subbase for in X. Let T α be the topology on X such that {X \ β F I β : F [α] <ω } is a base at. We want the following conditions to be satisfied: (1) I α ( D n ) is finite for all n; (2) If β α and P β is a π-net at in T β, then there are P β k P β such that, if P β = k ω P β k, then P β I γ < ω for each γ α; (3) Pk α = P k,0 α P k,1 α, and the following holds: Let P e α = k ω P k,e α. There are j α ω and a finite F α such that (i) P0 α [ i j α D i ] [ β F P β ], (ii) P1 α P β < ω for each β < α, (iii) for each j ω, {k : Pk,1 α D j } 1, and (iv) D(P1 α ) \ P1 α I α ; (4) If E α is a π-net in T α, then there is some finite F α + 1 such that d E α d ( β F I β). First let s see that if everything is constructed according to the above conditions, the resulting space has the desired properties. Clearly (1) gives that each D k is convergent, and hence D k is a π-net. Condition (4) shows that no choice of one member of D k for infinitely many k will be convergent, for otherwise the set of choices would appear as some E α and would be a π-net, but by (4) E α gets destroyed as a π-net at step α. Finally, we need to check that X is FU n. Suppose P is a π-net of n-element sets. Then either P D or P \ D is too. The latter case is taken care of by condition (2) for some β where P \ D = P β. In the former case, it follows from (4) that P D k is infinite for some k, and this would be a convergent sequence. Suppose we have defined everything satisfying the above conditions up to α. If P α is a π-net in the topology T α, find Pk α such that I γ ( k ω P k α ) < ω for each γ < α. To get (3), we thin out as follows. If there are x k Pk α S for infinitely many k, where S = D j for some j or S = P β for some β < α, then pass to that infinite subsequence. Then if there are y k Pk α \ {x k} S for infinitely many k and some S as before, do it again. Continue until there is no longer such an infinite selection. Since Pk α = n, this will occur in n steps. The set of points selected from the Pk α s is P k,0 α, and P k,1 α = P k α \ P k,0 α. Then it is easy to see that (3)(i) and (3)(ii) are satisfied, and that we may thin out again if necessary to get (3)(iii). Let s see that making sure (3)(iv) holds does not destroy the convergence of the previous P β s. Let I α,0 = D(P1 α ) \ P1 α, and suppose I α,0 P β is infinite for some β < α. Assume β is the least such; then by (3)(i) and (3)(iii), it must be that I α,0 P β 1 is infinite. But then D(P β 1 ) \ P β 1 P 1 α is an infinite subset of I β P1 α, contradicting P α convergent in T α. Now look at E α. If there is a finite F α such that d E α d [I α,0 ( β F I β)], then simply let I α = Iα, 0 and all conditions will be satisfied. Otherwise, let {J k } k ω list {I β } β<α {I α,0 } and let {S k } k ω list {P β : β α} { D k : k ω}. Then inductively construct J k J k such that J k \ J k < ω and

11 FRÉCHET-URYSOHN FOR FINITE SETS, II 11 J k i k S i =. Then we let I α,1 = I α,0 ( k ω J k, and I α,2 = {d E α : d I α,1 = }. Finally, let I α = I α,0 I α,1 I α,2. This satisfies condition (4) with F = {α}. Conditions (1) and (3) are clear, so it remains to check condition (2). We need to see that P β I α is finite for all β α. We already saw that I α,0 does not ruin this condition, and by construction neither does I α,1. So we want to show that I α,2 P β is finite for any β α. Suppose by way of contradiction that β is least such that this set is infinite. Then I α,2 P β 1 is infinite, and hence I α,2 I β is infinite. But then I α,2 J k is infinite for some k, contradicting I α,2 I α,1 =. Example 3.3. (CH) A weakly σ-fu n -space which is not FU n+1 or σ-fu n (or even α 2 =σ-fu 1 ). Proof. This construction is with very minor modifications the same as the construction of Example 16 in [10] of a FU n not FU n+1 -space. So here we will only indicate the necessary changes. In our construction in [10], all potential π-nets of n-sized sets are listed as T α, α < ω 1, and at stage α, a certain subset S α of T α is chosen so that S α will be convergent if T α happens to be a π-net. If we instead let the T α s index all potential sequences of π-nets, and choose S α to be a diagonalizing sequence through infinitely many terms of T α, the same proof goes through easily. Why is it that the same construction as in Example 6 is not adaptable to obtain our next example, a σ-fu n -space which is not FU n+1? For one thing, it is important in the above proof to be able to thin out at will a preliminary choice S for S α, which weakly σ-fu n allows. But in fact it cannot be σ-fu n by the neighborhood structure, which is generated by complements of an almost disjoint family. The next proposition shows that if such a space were σ-fu 1, or even Fréchet α 3, it would have to be boundedly FU fin. Proposition 3.4. Let A be an almost disjoint family of subsets of ω with ω = A, and let X = ω { } be the space in which ω is the set of isolated points, and neighborhoods of have the form X \ F, where F is a finite subset of A. Then if X is Fréchet α 3, it is also boundedly FU fin. Proof. Suppose that A is such that the corresponding space X is Fréchet α 3 but not boundedly FU fin. Let k ω be least such that X is not FU k. Note that k > 1. Let {{n i,0, n i,1,..., n i,k 1 }} i ω be a faithful indexing of a π-net at which has no convergent subsequence. Let p i = {n i,0, n i,1,..., n i,k 1 }, and p i = {n i,1,..., n i,k 1 }. Then the p i s form a π-net also. Claim. There are infinitely many members A of A such that, for some infinite B ω, we have n i,0 A for all i B, n i,0 n j,0 for i j B, and {p i } i B. To see this, suppose there were only finitely many such A. Let F be a finite subset of A containing them all. For each m ω \ F, let P m = {p i : n i,0 = m}. For each m such that P m is a π-net at, let S m P m be convergent (S m exists by minimality of k). If the set M = {m ω \ F : P m is a π-net} were infinite, then by repeated applications of the α 3 property on the terms of the p i s in m M S m, we could find an infinite M M and T m S m for each m M such that T = m M T m. Let T (0) = {n i,0 : p i T }. Since no subsequence of the

12 12 GARY GRUENHAGE AND PAUL J. SZEPTYCKI p i s converges, it must be the case that is not in the closure of T (0), so T (0) is almost included in some finite subcollection of A. Thus there is some A A such that A contains infinitely many m M. It follows that there is an infinite B ω such that, for each i B, p i T and n i,0 A, and n i,0 n j,0 for i j B. So the Claim holds in case the set M above is infinite. If M is finite, then we can find an infinite B ω such that {p i } i B and {n i,0 } i B [M ( F)] =. Note that no integer is repeated infinitely often in {n i,0 } i B. So as above, we can again find some A A and infinite B B satisfying the conditions of the Claim. To complete the proof of the proposition, let A 0, A 1,... be distinct members of A satisfying the condition of the Claim. Let B n ω be infinite such that {n i,0 } i Bn A n, n i,0 n j,0 for i j B n, and S n = {p i } i B n. Applying the α 3 property as above, there is an infinite N ω and for each n N an infinite C n B n such that {p i : i n N C n }. Then is not in the closure of the set J = {n i,0 : i n N C n }. So J must be almost covered by a finite subcollection of A, and at the same time J meets infinitely many A n s in an infinite set. This is a contradiction to A being almost disjoint. Thus we need a different type of example to show that implication 4 in the chart need not reverse. It turns out that gap spaces work. Example 3.5. (CH) A σ-fu n space which is not FU n+1. Proof. We construct a gap (a α, b α : α < ω 1 ) on ω (n + 1) in such a way that the family {{m} (n + 1)} : m ω} is a π-net but no infinite subset converges in the corresponding gap space X = [ω (n + 1)] { }. Let {x α : α < ω 1 } be an enumeration of [[ω n + 1] n ] ω. By recursion we construct {a α : α < ω 1 } and {b α : α < ω 1 } such that (1) a α, b α ω n + 1 for all α < ω 1, (2) a α a β b β b α for all α < β, (3) a α+1 \ a α = b α \ b α+1 = ℵ 0 for all α < ω 1. (4) {m} (n + 1) a α for all m ω and all α < ω 1. (5) For each α < ω 1 there is an infinite b α ω such that b α = a α (b α (n+1)) and a α (b α (n + 1)) =. (6) If {s x α : s a α \ (m n + 1)} = for some m, then either (a) for all m, {s x α : s a α+1 \ (m n + 1)} is infinite, or (b) there is an m such that {s x α : s b α+1 \ (m n + 1)} =. Hypothesis 6 assures two things: Since all infinite subsets of ω n + 1 are included in the enumeration of the x α s, we have that {b α : α < ω 1 } generates the gap filter F g. Also, if, after the construction, any x α is a π-net at, then there is an infinite subset that converges. Indeed, if x α is a π-net, then by construction {s x α : s a α+1 \ (m n + 1)} is infinite for every m. So, we can easily extract a convergent sequence. Hence the gap space X is FU n. Hypotheses 4 and 5 assure that {{m} n + 1 : m ω} is a π-net with no convergent subset. So X is not FU n+1.

13 FRÉCHET-URYSOHN FOR FINITE SETS, II 13 It suffices to show how to carry out the construction. Suppose that γ < ω 1 and that {a β : β < γ} and {b β : β < γ} have been fixed so that the inductive hypotheses are satisfied. Case 1. γ is a successor. Let β be such that γ = β + 1. First suppose that {s x γ : s a β \ (m n + 1)} is infinite for every m. Then we need not worry about hypothesis 6. To define a γ and b γ, partition b β = B 0 B 1 into infinite pairwise disjoint sets. Let a γ = a β B 0 {0} and let b γ = a γ (B 1 n + 1). It is easy to see that the hypotheses 1-6 hold for {a β, b β : β γ}. In the case that {s x γ : s a β \ (m n + 1)} = for some m, the construction is similar: First note that we may assume that {s x γ : s b β \ (m n + 1)} is infinite for all m (if not, the previous construction may be used and hypothesis 6(b) is satisfied). By our assumption, we may recursively define an increasing sequence of k m ω and s m x γ such that for each m s m b β ((k m+1 \ k m ) (n + 1)) and s m (b β (n + 1)). For each m let s m = s m (b β (n + 1)). If we now let a γ = a β {s m : m even} and b γ = {b β (k m+1 \ k m ) : m odd }. Then for each even m we have that s m a γ \ (k m n + 1). And the inductive hypotheses are easily seen to be satisfied. In particular, 6 is satisfied by item 6(a). Case 2. γ is a limit. In this case we have nothing to do to preserve hypothesis 3 and 6. However, preserving the other hypotheses requires a little work. Choose {γ j : j ω} increasing and cofinal in γ. Let b γ be a pseudo-intersection of the b γ j. Thus b γ (n + 1) a β is finite for every β < γ. Recursively define a j as follows. Let a 0 = a γ0. Having defined a j = a γj, choose k j large enough so that a j \ (k j 3) a γj+1. Let a j+1 = a j (a γj+1 \ (k j (n + 1))). Let a γ = a j. The main property to note is that {m} (n + 1) a γ for every m ω. To see this, suppose by way of contradiction that m (n + 1) a γ. Let j be minimal such that m (n + 1) a j. Then j 0. So a j = a j 1 a γj \ (k j 1 n + 1). Thus, m > k j 1. It follows that m n + 1 a γj since a j 1 \ (k j 1 n + 1) a γj. But this contradicts our inductive hypothesis 4 for a γj. By going to a subset of a γ we preserve hypothesis 4 for a γ so we may assume that a γ b β for every β < γ, and a γ (b γ n + 1) =. We let b γ = a γ (b γ n + 1). Clearly the rest of the hypotheses are satisfied for all α γ. This completes the construction. Example 3.6. A boundedly FU fin space which is not α 2 (i.e., not σ-fu 1 ), and hence not σ-boundedly FU fin. Proof. This example is a modification of the example of Theorem 4 of [10]. Let Q denote the rationals in the unit interval I = [0, 1]. Our space X will be

14 14 GARY GRUENHAGE AND PAUL J. SZEPTYCKI Q { }, where points of Q are isolated, and the neighborhood filter of will be generated by complements of finite subsets of Q, together with complements of certain sequences S x of rationals converging to x, for some points x I. We will choose at most one S x for each x; by Lemma 3 of [10], this will guarantee the space is boundedly FU fin. Let {D α } α<c index all countable dense subsets of Q. At stage α, choose an irrational point x α [0, 1] \ {x β : β < α} and a sequence S α D α such that S α converges to x α. Then no Euclidean dense subset of Q converges to in the resulting space X. This is enough to destroy α 2. For we can list Q as {q n } n ω, and choose a sequence T n in Q of diameter less than 1/2 n converging to q n in the Euclidean topology. Then T n in X, but any selection of a point from each T n is Euclidean dense, hence not convergent in X. Example 3.7. (CH) A boundedly FU space which is α 2 (and hence σ-fu n for each n) but is not σ-boundedly FU. Proof. This example has the same form as the previous one, but it will take a bit more work to choose the S x s so as to make it α 2 yet not σ-boundedly FU fin. Let H = {H nm : n, m ω} be a pairwise disjoint collection of subsets of Q such that for each n, H nm = n + 1, {H nm : m ω} q n, and diam( m ω H nm) < 1/2 n. We will make sure that, for each n, each S x meets only finitely many members of H n = {H nm : m ω}, and hence that each H n is a π-net. Let X α = {X αi } i ω and g α, for α < ω 1, list all sequences of infinite subsets of Q and all g : ω ω, respectively. Suppose at stage α we have chosen, for each β < α, subsets S β and T β of Q satisfying: (1) S β x β in [0, 1]; (2) x β x γ if β γ < α; (3) If γ < α and X γi S β is finite for all β < γ and i ω, then for all i ω, T γ X γi ; (4) β, γ α (S β T γ is finite); (5) For each n ω and β < α, S β meets only finitely many members of H n ; (6) For infinitely many n ω, S β H ngβ (n). First, suppose we can carry out the indicated construction. Then clearly condition (5) ensures that each H n is a π-net, and (6) ensures that no selection of a member of each H n converges to ; hence X is not σ-boundedly FU fin. On the other hand, X is boundedly FU fin for the same reason as the previous example, and conditions (3) and (4) ensure that X is α 2. So it remains to check that the inductive construction can be done. If X αi S β is infinite for some β < α, let T α =. Otherwise, let S n, n ω, index {S β } β<α. Since H is a pairwise-disjoint collection, it is easy to see that we may choose points t αn X αn \ i n S i such that no member of H contains more than one t αn. Let T α = {t αn } n ω. This gets conditions (3) and (4) with γ = α. Now we choose S α so that (1)-(6) hold with β = α. Let T n, n ω, index {T β : β α}. At step n, since each T n meets each H H in at most one point, and H ni = n + 1 for all i, we can choose s αn H ngα(n) \ j<n T j. Note that the s αn s are Euclidean dense in [0, 1]. Thus there is some x α {x β : β < α} and infinite A α ω such that {s αn : n A α } x α. Then setting S α = {s αn : n A α } clearly works.

15 FRÉCHET-URYSOHN FOR FINITE SETS, II 15 Let us recall the following construction due to Nyikos[18] (see also Example 1 of [10]). Let T = 2 <ω be the Cantor tree, and let A 2 ω. Let X A = T { } be the space with T as the set of isolated points, and subbasic neighborhoods of are complements of branches b x = {x n : n ω} of the tree for x A. Nyikos showed that X A is Fréchet α 2 if A is a λ -set (i.e., if B 2 ω is countable, then B is G δ in A B). The next result strengthens this just a bit, the proof being a mild extension of Nyikos s argument, and shows that there is in ZFC a countable σ-boundedly F U fin space which is not first-countable. Example 3.8. If A is a λ -set, then the space X A is σ-boundedly FU fin. Proof. Suppose P 0, P 1,... is a sequence of π-nets at, where sup{ P : P P i } k i. Since X A is boundedly FU fin (again, by Lemma 3 in [10]), we may assume P i = {P ij } j ω, and P ij = k i for all i, j. We are going to use the compact metrizable topology on T 2 ω generated by the basis T {σ : σ 2 <ω }, where σ = {x 2 ω : σ x} {t T : σ t}. Note that the subspace 2 ω inherits its usual product topology. For each i, j, let p ij = p ijm m<ki, where P ij = {p ijm } m<ki. W.l.o.g., we may assume that for each i, { p ij } j ω converges to a point x im } m<ki in the ki th power of the above compact metrizable topology, and that P ij ( {b xim : m < κ i and x im A} =. Let B = {x im : i, m ω}. Let U n, n ω, be a decreasing sequence of open sets in 2 ω with n ω U n (A B) = B. Let Un = {σ : σ U n }, and let A B = {y n : n ω}. It is easy to check that for each i we can find a large enough j i ω so that P iji Ui and P iji ( {b yk } k i ) =. Let Q = {P iji } i ω. We need to show that Q, i.e., that each b x, x A, meets P iji for at most finitely many i. For x A B, this is clear by the construction. Suppose x A \ B. Then for sufficiently large i, x U i. Note that in this case b x Ui =, and hence P ij i b x =. The previous example does not separate in ZFC the σ-boundedly FU fin property from FU fin. Nyikos showed that X A is FU fin iff A is a γ-set, and A. Miller[15] showed that it is consistent that every λ -set is γ. There are many models, however, which have λ -sets which are not γ-sets (e.g., any model of CH or MA), so it does give consistent examples showing that implication 1 need not reverse. Example 3.9. If A is a λ -set which is not a γ-set, then X A is σ-boundedly FU fin space but not FU fin. 4. Relationships to games, products, and α i spaces Let X be a space, x X, and k ω. We define the game G O,P (X, x) (respectively, G k O,P (X, x), Gfin O,P (X, x)) as follows. The players are O and P. In the nth round, O chooses an open neighborhood U n of x, and P responds with a singleton (respectively, k-sized set, finite set) P n i n U i. O wins if P n x. The game in which the P n s are singletons was introduced and studied in [5]. Of course, O has a winning strategy in any of these games if X is first-countable at x. It is also not very difficult to see that O has a winning strategy in any one of these games iff O has a winning strategy in all of the games. (This was proven

16 16 GARY GRUENHAGE AND PAUL J. SZEPTYCKI in [5] for G O,P showed that X is FU fin at x iff P has no winning strategy in G fin O,P and G fin O,P.) The situation for P is different, however. In [10], we (X, x). What the proof really shows is that the game property is equivalent to σ-fu fin, and the same proof virtually word for word shows the following result: Theorem 4.1. Let be either k, where k ω \ {0}, or fin. winning strategy in G O,P (X, x) iff X is σ-fu at x. Then P has no It follows that if X is the gap space of Example 2.1, which is σ-fu 1 (=Fréchet α 2 ) but not FU 2, then P has no winning strategy in the singleton game, but P does have a winning strategy in the doubleton game. We remark that the case n = 1 in the above theorem, which boils down to P has no winning strategy in G O,P (X, x) iff X is Fréchet α 2 at x was proven by P.L. Sharma[22]. Next we discuss product spaces. Sipacheva[21] showed that X is FU n at x iff X n if Fréchet at the diagonal point x, x,..., x. Since the Fréchet property is a pointwise property, it follows, e.g., that there is a space X such that X n is Fréchet but X n+1 is not Fréchet iff there is one with only one non-isolated point iff there is an X which is FU n but not FU n+1. There is a similar relationship to Fréchetness in products for many of the other properties we are considering, summarized by the chart below. In the chart, X is a space with exactly one non-isolated point. This is a critical assumption. For example, in [10] we gave an example (under CH) of two countable FU fin spaces whose product is not FU fin, so if X is their topological sum, then X is FU fin but X 2 is not. X σ FU n X n Fréchet α 2 X FU n + α 2 X n Fréchet α 3 X FU n + α 3 X weakly σ-fu n X n Fréchet α 4 X FU n + α 4 X FU n X n Fréchet n(x σ-fu n ) X ω Fréchet α 2 n(x n Fréchet α 2 ) X ω Fréchet n(x FU n ) n(x n Fréchet(+α 4 )) X boundedly FU fin The down arrows in the chart are trivial. The equivalences in the bottom connected piece are immediate from those in the top piece. We should also mention Nogura s result[16] that, for i = 1, 2, 3, if X n is Fréchet α i for all n ω then X ω Fréchet α i. For i = 2 this is precisely one of the implications in the chart. The next three results establish the equivalences in the top piece. Theorem 4.2. The following are equivalent for a space X with one non-isolated point:

17 FRÉCHET-URYSOHN FOR FINITE SETS, II 17 (a) X is σ-fu n ; (b) X n is Fréchet α 2 ; (c) X is FU n and α 2. Proof. (b) (c) is immediate from Sipacheva s result that X FU n iff X n is Fréchet. If X satisfies (a), then X n is Fréchet for the same reason. That X n is α 2 is easily shown by translating convergent sequences of points in X n to the corresponding sequences of π-nets of n-sized sets consisting of the coordinates of the points, applying σ-fu n in X to these π-nets, and translating back again to points in X n. Suppose X satisfies (c), and let P 0, P 1,... be a sequence of π-nets of n-sized sets. By FU n, we may assume each P k = {P ki } i ω is convergent. Let P ki = {p kij : j < n}. Apply α 2 to obtain an infinite T k (0) of S k (0) = {p ki0 : i ω} such that T (0) = k ω T k(0) converges. Then apply α 2 again to obtain an infinite subset T k (1) of S k (1) = {p ki1 : p ki0 T (0)} such that T (1) = k ω T k(1) converges. Continue in the same manner for each j < n. Then the set T = {P ki : p kij T (j) for all j < n} is a convergent selection of infinitely many sets from each P k. So X satisfies (a). By essentially the same method as the proof of (c) (a) above, it is easy to establish the following: Theorem 4.3. For a space X with one non-isolated point, X n is Fréchet α 3 iff X is FU n and α 3. Theorem 4.4. For a space X with one non-isolated point, X is weakly σ-fu n iff X n is Fréchet and α 4. Proof. Translate back and forth between convergent sequences in X n and π-nets of n-sized sets. Note that the analogue of condition (c) in Theorem 4.2 cannot necessarily be added to the list of equivalences in the previous theorem. Of course, it is equivalent for n = 1, but Example 3.2 in the previous section shows this need not be the case (i.e., the down arrow third from the top need not reverse) for n 2. Regarding the non-reversibility of the other down arrows, any of the (known, ZFC) examples differentiating α 3 from α 2, and α 4 from α 3, show that the top two do not reverse (for n = 1), and any Fréchet non-α 4 space the fourth one (which does reverse for n 2, since FU 2 implies α 4 ). The example in [6] under MA shows that the bottom down arrow need not reverse. The remaining down arrow, and also the top one for n 2, is taken care of by the boundedly FU fin non-α 2 space X of Example 3.6 in the previous section. By Lemma 3 in [10], X is α 3. Then it follows from Theorem 4.3 above that X n is Fréchet α 3 for all n, so by Nogura s result mentioned earlier, X ω is Fréchet α 3. But we do not know the answer to: Question 2. Does X ω Fréchet imply X is α 3? We also do not know if there are ZFC examples which show that the bottom down arrow, the third one from the top, or the second from the top for n 2, do not reverse. If X is FU fin, then X is σ-fu n for all n, so it follows from the chart that X ω is Fréchet α 2. In fact, we can show that X FU fin is equivalent to X ω FU fin, and the same holds for σ-boundedly FU fin.

18 18 GARY GRUENHAGE AND PAUL J. SZEPTYCKI Theorem 4.5. Let X be a space with exactly one non-isolated point. Then: (1) X is FU fin iff X ω is FU fin ; (2) X is σ-boundedly FU fin iff X ω is σ-boundedly FU fin. Proof. We show (1), with (2) being entirely analogous. To prove the non-trivial direction, suppose X is FU fin. Denote the non-isolated point of X by, and let P be a π-net at,,... X ω consisting of finite subsets of X ω. For each P P and n ω, let π(p, n) = {x(i) : i n, x P }, and let P(n) = {π(p, n) : P P}. It is elementary to check that each P(n) is a π-net at of finite subsets of X. Hence we can find P n P such that π(p n, n). Let us see that P n,,... in X ω. A basic open set containing,,... has the form U k X ω for some k ω. For sufficiently large n, we have π(p n, n) U. If also n k, it follows that P n U k X ω. It follows from the above results, together with the examples of the previous section, that X ω can have any one of the properties we are considering that imply boundedly FU fin without having any stronger property. References [1] A. V. Arhangel skiǐ, Frequency spectrum of a topological space and classification, Dokl. Akad. Nauk. SSSR 206(1972), [2] A. V. Arhangel skiǐ, Classes of topological groups, Russian Math. Surveys 36 (1981), Russian original in: Uspekhy Mat. Nauk 36 (1981), [3] A. Dow, Two classes of Fréchet-Urysohn spaces Proc.Amer. Math.Soc. 108 (1) (1990) [4] A.Dow and J. Steprans Countable Frchet α 1 spaces may be first-countable Arch. Math. Logic 32 (1992), no. 1, [5] G. Gruenhage Infinite games and generalizations of first-countable spaces Gen. Top. Appl. 6(1976), [6] G. Gruenhage A note on the product of Fréchet spaces Topology Proc. 3 (1978), no. 1, [7] G. Gruenhage The story of a topological game Rocky Mountain J. Math., to appear. [8] G. Gruenhage and D.K. Ma Baireness of C k (X) for locally compact X, Topology Appl. 80(1997), [9] J. Gerlits and Zs. Nagy Some properties of C(X), I Top. Appl. 14 (1982) [10] G. Gruenhage and P. Szeptycki, Fréchet Urysohn for finite sets, Topology Appl., to appear. [11] K. Kunen, (κ, λ ) gaps under MA, handwritten notes (August, 1976). [12] R. Laver On the consistency of the Borel conjecture Acta Mathematica 137 (1976) [13] D.K. Ma The Cantor tree, the γ-property, and Baire function spaces Proc. Amer. Math. Soc. 119(1993) [14] A. Miller Special subsets of the real line in: Handbook of Set-theoretic Topology, K. Kunen and J. Vaughan, eds., North-Holland, Amsterdam (1984) [15] A. Miller, On lamda -sets, Topology Proceedings, to appear. [16] T. Nogura, The product of α i -spaces, Topology Appl. 21 (1985), no. 3, [17] P. Nyikos Subsets of ω ω and the Fréchet-Urysohn and α i -properties Top. Appl. 48 (1992) [18] P. Nyikos The Cantor tree and the Fréchet-Urysohn property Ann. New York Acad. Sci. 552 (1989) [19] P. Nyikos Private communication [O] R. C. Olson, Bi-quotient maps, countably bi-sequential spaces, and related topics, General Topology and Appl. 4 (1974), [20] E. Reznichenko and O. Sipacheva Fréchet-Urysohn type properties in topological spaces, groups and locally convex vector spaces Moscow University Mathematics Bulletin 54(3) (1999) 33-38