RUDIN’S DOWKER SPACE AND THE LINEARLY LINDELOF¨ PROBLEM
MIHAI ALBOIU
1. Introduction In this report, we present Mary Ellen Rudin’s Dowker space and discuss the closely related linearly Lindel¨ofproblem. Definition 1.1. A Dowker space is a normal topological space X, whose product with the unit interval, X × [0, 1], is not normal. Dowker spaces are named after C.H. Dowker, whose name is attributed to the follow- ing theorem. Before stating the theorem, we introduce the relevant notion of countable paracompactness. Definition 1.2. A space X is said to be countably paracompact (metacompact) iff every countable open cover of X has a locally-finite (point-finite) refinement. Theorem 1.3 (Dowker’s Theorem). Let X be a normal space. The following are equiv- alent: T (i) If D0 ⊇ D1 ⊇ · · · is a nested sequence of closed sets in X with {Dn|n < ω} = ∅, T then, for each n < ω, there is an open Un ⊇ Dn such that {Un|n < ω} = ∅; (ii) X is countably paracompact; (iii) X × [0, 1] is normal.
Proof. See [DO]. The existence of Dowker spaces was first brought into question by Dowker in [DO]. The above theorem completely classifies Dowker spaces as spaces which are both normal and not countably paracompact. Dowker spaces have proved to be exceedingly difficult to come by. This sheds some light on the subtlety between normality and countable paracompactness: it is quite hard to come up with a space that is both normal and not countably paracompact. In the next section, we exhibit Rudin’s example of a Dowker space (see [RU]), which is a ∞ subspace of 2 [0, ωi]. To prove that the space is not countably paracompact, we prove i=1 that condition (i) of the above theorem does not hold. This is often the condition that is worked with in practice when dealing with countable paracompactness in normal spaces. Rudin’s space is special because it was the first example of a Dowker space in ZFC. It was discovered some twenty years after Dowker published his paper.
A useful characterization of countable paracompactness (metacompactness) that is independent of normality is due to Ishikawa [IS]. 1 2 MIHAI ALBOIU
Proposition 1.4. Let X be a topological space. (i) X is countably paracompact iff whenever D0 ⊇ D1 ⊇ · · · is a nested sequence of T closed sets in X with {Dn|n < ω} = ∅, there exists, for each n < ω, an open Un ⊇ Dn T such that {Un|n < ω} = ∅; (ii) X is countably metacompact iff whenever D0 ⊇ D1 ⊇ · · · is a nested sequence of T closed sets in X with {Dn|n < ω} = ∅, there exists, for each n < ω, an open Un ⊇ Dn T such that {Un|n < ω} = ∅; Thus, in a normal space, countable paracompactness and countable metacompact- ness are equivalent.
A related problem is the so called linearly Lindel¨ofproblem. Definition 1.5. Let X be a topological space. We say X is Lindel¨of provided every open cover of X has a countable subcover. We say X is linearly Lindel¨of iff every increasing open cover {Uα|α < κ} has a countable subcover. Here, increasing means Uα ⊆ Uβ, whenever α < β. Problem 1.6. Is every normal, linearly Lindel¨ofspace Lindel¨of? In section 3, we present Miˇsˇcenko’s example (see [MI]) of a linearly Lindel¨ofspace ∞ Q that is not Lindel¨of.This example, which is a subspace of [0, ωi], is similar to and i=1 helped inspire Rudin’s Dowker space. In section 4, we discuss Buzyakova and Gru- enhage’s example (see [AR]) of a linearly Lindel¨ofspace that is not Lindel¨of.It is a subspace of {0, 1}ℵω and is slightly easier to describe than Miˇsˇcenko’s space. The following proposition relates the linearly Lindel¨ofproblem and Dowker spaces. Proposition 1.7. Every linearly Lindel¨of,countably metacompact space is Lindel¨of. Proof. Suppose X is linearly Lindel¨ofand countably metacompact. We will inductively show that every open cover of size κ > ω has a smaller subcover. Fix κ > ω and suppose {Oα|α < κ} is an open cover of X. Assume that for all β < κ, any open cover of X having size β has a smaller subcover. S Case one: cf(κ) > ℵ0. In this case, for β < κ, let Vβ = {Oα|α ≤ β}. Then, {Vβ|β < κ} is an increasing open cover of X, which, by linear Lindel¨ofness, has a countable subcover
{Vβn |n < ω}. Since cf(κ) > ℵ0, γ = sup{βn|n < ω} < κ, so that {Vγ}, and thus, {Oα|α ≤ γ} covers X. Case two: cf(κ) = ℵ0. In this case, let {κn|n < ω} be an increasing cofinal subset of S κ. For each n < ω, put Dn = X \ {Oα|α < κn}. Then, D0 ⊇ D1 ⊇ · · · is a nested T sequence of closed sets with {Dn|n < ω} = ∅. By 1.4, for each n < ω, there is an T open Un ⊇ Dn such that {Un|n < ω} = ∅. Then, for each n < ω, {Oα|α < κn}∪{Un} ≤ℵ0 is an open cover of X of size κn. By the inductive hypothesis, the is an An ∈ [κn] S S such that X \ Un ⊆ {Oα|α ∈ An}. Let A = {An|n < ω}. Then, A is countable, and S X ⊆ {Oα|α ∈ A}. The significance of the above proposition is the following. Suppose there is a normal, linearly Lindel¨ofspace X that is not Lindel¨of.By the previous proposition, X would not be countably metacompact, hence, not countably paracompact. But then, since X is normal, Dowker’s theorem gives us that X × [0, 1] is not normal. Hence, X would have to be a Dowker space. RUDIN’S DOWKER SPACE AND THE LINEARLY LINDELOF¨ PROBLEM 3
2. Rudin’s Dowker Space In this section, we describe Rudin’s famous ZFC Dowker space. We will follow her construction in [RU] very closely. For this proof, N = {1, 2,...}. Here’s the setup. Let: ∞ Y F = {f : N → ωω| for all n ∈ N, f(n) ≤ ωn} = [0, ωn]; n=1 X = {f ∈ F | there exists i ∈ N, such that, for all n ∈ N, ω < cf(f(n)) < ωi}. Suppose f, g ∈ F . Say: f < g iff for all n ∈ N, f(n) < g(n); f ≤ g iff for all n ∈ N, f(n) ≤ g(n); f
Uf,g = {h ∈ X|f < h ≤ g}.
Then, {Uf,g|f, g ∈ F } is a basis for a topology on X. The aim of this section will be to prove that X, equipped with the topology generated by the aforementioned basis, is a Dowker space. We start by showing that X is not countably paracompact. For n ≥ 1, let:
Dn = {f ∈ X| there exists an i ≥ n with f(i) = ωi}
Cn = {f ∈ X| for all i < n, f(i) = ωi and, for all i ≥ n, f(i) < ωi}
Claim 2.1. (i) D1 ⊇ D2 ⊇ · · · T (ii) {Dn|n ∈ N} = ∅ (iii) Each Dn is closed. Proof. (i) is clear. To see (ii), note that if f ∈ X, there is an n ∈ N such that cf(f(i)) < ωn for all i ∈ N. In particular, f(i) 6= ωi for all i ≥ n, so that f∈ / Dn. For (iii): if f ∈ X \ Dn, then, for all i ≥ n, f(i) < ωi. Hence, f ∈ U0,f ⊆ X \ Dn, so that Dn is closed.
Now, for each n ∈ N, let Un ⊇ Dn be open. To show that X is not countably T paracompact, it suffices to show, by 1.4 and the above claim, that {Un|n ∈ N} = ∅. We do this through the following three lemmas.
Lemma 2.2. Suppose 1 < n ∈ N, U is an open set, f ∈ Cn+1, and U ⊇ {h ∈ Cn+1|f Let K = {kα|α < λ} be a maximal well-ordered family of terms in Cn \ U such that for all α < β < λ, k 0 0 ∗ 0 that f Then, g ∈ X and, since λ < ωn, g(i) < ωi for all i ≥ n. Whence, g ∈ Cn. Furthermore, if 0 h ∈ Cn \U, satisfies g Proof. For each n > 1, choose fn as in 2.3; i.e., such that Un ⊇ {h ∈ C2|fn <2 h}. For each i > 1, choose an ordinal αi such that cf(αi) = ω1 and such that for all n > 1, fn(i) < αi < ωi. Set α1 = ω1. Define g ∈ X by g(i) = αi for all i ∈ N. Then, g ∈ C2 and, for all n > 1, fn <2 g. By 2.3, it follows that g ∈ Un for all n > 1. Also, g(1) = ω1, T so that g ∈ D1 ⊆ U1. Hence, {Un|n ∈ N} 3 g. We have just shown that X is not countably paracompact. To prove that the X is normal requires a lot more work. Just as Rudin did in [RU], we will show that X is collectionwise normal; that is, for every discrete family of closed sets {Fi|i ∈ I} there is a collection {Ui|i ∈ I} of pairwise disjoint open sets, such that Ui ⊇ Fi. Let’s get started! Suppose H = {Hj|j ∈ J} is a discrete collection of closed sets in X. Let H = S {Hj|j ∈ J}. If U ⊆ F , define tU ∈ F by tU (n) = sup{f(n)|f ∈ U} for all n ∈ N, and note that U ⊇ V implies tU ≥ tV . Goal. For each α < ω1, we will define, by induction, a cover Jα of H consisting of disjoint open sets having the following property: If β < α < ω1 and V ∈ Jα, then there exists a U ∈ Jβ such that: (1) V ⊆ U; i.e., Jα refines Jβ; (2) if V intersects at least two members of H, then tV 6= tU ; (3) if U intersects at most one member of H, then U = V . We will first show that the existence of the Jα, as described in Goal, is enough to RUDIN’S DOWKER SPACE AND THE LINEARLY LINDELOF¨ PROBLEM 5 find a set of pairwise disjoint open sets {Uj|j ∈ J} such that Uj ⊇ Hj, which is what we need in order to show X is collectionwise normal. Suppose f ∈ H. If α < ω1, then, since Jα covers H with disjoint open sets, there is a unique Uα ∈ Jα such that f ∈ Uα. If β < α < ω1, then, by (1), Uα ⊆ Uβ; hence, tUα ≤ tUβ . By (2), if Uα intersects more than one term of H, there is an n such that tUα (n) < tUβ (n). Since there is no infinite descending sequence of ordinals, for each n ∈ N, an inequality of the above form can only hold for finitely many β < α < ω1. So, considering all n ∈ N and all β < α < ω1, an inequality of the form tUα (n) < tUβ (n) can hold at most countably many times. So, since ω1 is uncountable, there must be an αf < ω1 such that Uα intersects at most one term of H. Therefore, by (3), if f S αf < β < ω1, then Uβ = Uαf . Define Uj = {Uαf |f ∈ Hj} for all j ∈ J. Note that for each j ∈ J, Hj ⊆ Uj. We will now show that the Uj’s are pairwise disjoint. It suffices to show that Uαf ∩ Uαg = ∅, whenever f and g belong to different terms of H. To this end, choose γ < ω1 greater than αf and αg. Then, by the above, the term of Jγ which contains f is Uαf , and the term of Jγ which contains g is Uαg . If f and g do not belong to the same term of H, then Uαf 6= Uαg . Thus, since Jγ is a cover of disjoint open sets, Uαf and Uαg must, in fact, be disjoint. So, we need to prove the existence of the Jα as described in Goal. We proceed by induction. Define J0 = {X}. Fix α < ω1 and suppose that Jβ has been defined and satisfies the properties in Goal for all 0 ≤ β < α. We first deal with the case when α is a limit ordinal. For f ∈ H and β < α, let Uf (β) denote the term of Jβ to which f belongs, T and set Uf = {Uf (β)|β < α}. Put Jα = {Uf |f ∈ H}. We need a lemma to ensure that each Uf is open. Lemma 2.5. X is a P-space; that is, the intersection of a countable collection of open sets is open. T Proof. Suppose {On|n ∈ N} is a collection of open sets in X. Suppose h ∈ {On|n ∈ N}. For each n ∈ N, choose a function gn ∈ F such that Ugn,h ⊆ On. Define g ∈ F by g(i) = sup{gn(i)|n ∈ N} for all i ∈ N. Then, g ≤ h. If g(i) is a maximum instead of a true supremum, then g(i) = gn(i) for some n, and hence, g(i) < h(i). Otherwise, cf(g(i)) ≤ ω < ω1 ≤ cf(h(i)), and thus, g(i) < h(i) in this case, too. Therefore, g < h T T and h ∈ Ug,h ⊆ {On|n ∈ N}, so that {On|n ∈ N} is open. We now know that Jα = {Uf |f ∈ H} is a collection of open sets; they clearly cover H. To see that the sets in Jα are pairwise disjoint, suppose Uf 6= Ug and h ∈ Uf ∩ Ug. Then, h ∈ Uf (β) ∩ Ug(β) for all β < α. So, by the inductive hypothesis, Uf (β) = Ug(β) for all β < α, and hence, Uf = Ug, which is a contradiction. Now, fix β < α. It suffices to verify the other three properties listed in Goal with V = Uf and U = Uf (β). Clearly Uf (β) ⊇ Uf , so that (1) is satisfied. If Uf intersects two terms in H, then, since α is a limit, so do Uf (β) and Uf (β + 1). Since Jβ and Jβ+1 consist of disjoint sets, the induc- tive hypothesis gives that Uf (β + 1) ⊆ Uf (β), and hence, tUf (β+1) 6= tUf (β). Therefore, there is an n ∈ N such that tUf (n) ≤ tUf (β+1)(n) < tUf (β)(n), so that (2) holds. To see (3), suppose Uf (β) intersects at most one member of H. Then, for all β < γ < α, Uf (β) = Uf (γ) by condition (3) for Jγ. So, since δ1 < δ2 implies that Uf (δ1) ⊇ Uf (δ2), T T it follows that Uf = {Uf (δ)|δ < α} = {Uf (δ)|δ ≤ β} = Uf (β), which is (3). This 6 MIHAI ALBOIU concludes the case where α is a limit. We now move on to the case when α is a successor. Suppose α = β + 1. We begin with a claim. Claim 2.6. Suppose that for all U ∈ Jβ, there is a collection JU of disjoint open subsets of U covering U ∩ H, such that V ∈ JU implies (2) and (3) of Goal are satisfied with S respect to U. Then, Jα := {JU |U ∈ Jβ} will have the desired properties of Goal. Proof. Let γ < α and let V ∈ Jα. Then, there is a U ∈ Jβ such that V ∈ JU . Choose 0 0 a set U ∈ Jγ containing U. We’ll show that V and U satisfy the properties of Goal. We have V ⊆ U ⊆ U 0, so that (1) is satisfied. If V intersects at least two terms of H, then tV 6= tU by the assumption of the claim. So, tV < tU ≤ tU 0 , which proves that (2) holds. Finally, if U 0 intersects at most one terms of H, then, so does U. Thus, by the assumption of the claim, V = U. If γ = β, then U = U 0. Otherwise, by the inductive 0 0 hypothesis applies to Jβ, U = U . In either case, V = U , which is (3). For the rest of the proof, we hereby fix a U ∈ Jβ and let t := tU . We need to con- struct a collection JU as specified by the above claim. We consider three cases. Case One: U intersects at most one term of H. In this case, define JU = {U}. Then, if V ∈ JU , V = U, so (3) is satisfied. And, (2) holds trivially. Case Two: U intersects at least two members of H and there is an i ∈ N such that cf(t(i)) ≤ ω. In this case, (3) never holds, so we only need to ensure (2). Because U intersects two members of H, U 6= ∅, and hence, t(i) 6= 0. We claim that cf(t(i)) = ω. Other- wise, 0 < cf(t(i)) < ω, and so, t(i) = γ for some ordinal γ. If f ∈ U ⊆ X, then cf(f(i)) > ω > cf(t(i)) and f(i) ≤ t(i). Hence, f(i) < t(i). But then, f(i) ≤ γ, and, since this holds for all f ∈ U, it follows that t(i) ≤ γ < γ+1 = t(i), which is impossible. Thus, cf(t(i)) = ω. Now, let {λn|n ∈ N} be a cofinal increasing subset of t(i). Define V1 = {f ∈ U|f(i) ≤ λ1}, and, for 1 < n ∈ N, set Vn = {f ∈ U|λn−1 < f(i) ≤ λn}. Clearly each Vn is open, and the Vn’s are disjoint. Thus, JU = {Vn|n ∈ N} is a set of disjoint open subsets of U, which cover U ∩ H. Furthermore, to see that (2) holds, note that for each n ∈ N, tVn (i) ≤ λn < t(i). Case Three: U intersects at least two members of H and cf(t(n)) > ω for all n ∈ N. This case will require substantially more work than the previous two. Consider the following lemma. Lemma 2.7. There is an f ∈ F such that f < t and {h ∈ U|f < h} intersects at most one term of H. Claim 2.8. If the above lemma holds, then we can define JU with the desired properties of 2.6. Proof. Suppose that the above lemma holds, and let f ∈ F be given as in the lemma. We define JU as follows. For M ⊆ N, set VM = {h ∈ U|h(n) ≤ f(n) for all n ∈ M and h(n) > f(n) for all n ∈ N \ M}. RUDIN’S DOWKER SPACE AND THE LINEARLY LINDELOF¨ PROBLEM 7 Define JU = {VM |M ⊆ N}. Since U intersects at least two members of H, (3) is satisfied trivially, just as in Case Two. To see that (2) holds, we prove the contrapositive. Suppose tVm = t. Then, tVm (n) = t(n) ∀n ∈ N =⇒ tVm (n) = t(n) ∀n ∈ M =⇒ t(n) ≤ f(n) ∀n ∈ M =⇒ M = ∅ since f < t. But, V∅ = {h ∈ U|h > f}, which, by the above lemma, intersects at most one on term of H. This proves (2) via the contrapositive. So, since JU is a set of disjoint open sets, JU satisfies the condition of 2.6. By the above claim, it suffices to prove 2.7 in order to conclude the proof. In order to do this, we will appeal to the following lemma. For each n ∈ N, we define Un = {h ∈ U| for all i ∈ N, cf(h(i)) ≤ ωn}. Lemma 2.9. Suppose n ∈ N. There exists a g ∈ F with g < t such that {h ∈ Un|g < h} intersects at most one term of H. Claim 2.10. If the above lemma holds, then so does 2.7. Proof. Suppose the above lemma holds, and, for each n ∈ N, let gn be given as in the lemma. Define f ∈ F by f(i) = sup{gn(i)|n ∈ N} for each i ∈ N. Then, f ≤ t. In fact, f < t. This follows by an argument similar to the one in the proof of 2.5 since cf(t(n)) > ω for all n ∈ N by the assumption of Case Three. We will now show that {h ∈ U|f < h} intersects at most one term of H. Suppose h0, h1 ∈ {h ∈ U|f < h} ∩ H. We’ll show that h0 and h1 belong to the same term of H. There are i, j ∈ N such that h0 ∈ Ui and h1 ∈ Uj. Let n = i + j. Then, h0, h1 ∈ Un and gn ≤ f < h0, h1. Hence, by 2.9, h0 and h1 belong to the same element of H. Therefore, by the above claim, in order to conclude the proof that X is collectionwise normal, we must prove 2.9. The rest of this section is devoted to this proof. Proof of 2.9. Fix n ∈ N. We shall proceed by contradiction, assuming that 2.9 is false. I.e., assume that for all f ∈ F with f < t, there are terms h, k ∈ Un such that f < h and f < k, but h and k belong to different terms of H. Before proceeding with the proof, we need to introduce some necessary notation. For i ≤ n, define Mi = {j ∈ N|cf(t(j)) = ωi}. Let M = {j ∈ N|cf(t(j)) > ωn}. By S the assumption of Case Three, N = {Mi|i ≤ n} ∪ M. Let R = {r : (1, 2, . . . , n) → ωn|r(i) < ωi for all i ≤ n}. Note that |R| = |ω1 × · · · × ωn| = ωn. Since ωn is regular, we can enumerate R by {rλ|λ < ωn}, such that if λ < ωn and r ∈ R, then there is a γ with λ < γ < ωn and rγ = r. Finally, for all i ≤ n and j ∈ Mi, choose an increasing cofinal subset {sj,σ|σ < ωi} of t(j). In order to help aid with the proof later on, we will define by induction, for each ordinal λ < ωn, an fλ ∈ F and hλ, kλ ∈ Un as follows. Let f0 ∈ F be given by ( s if i ≤ n and j ∈ M f (j) = j,r0(i) i . 0 0 if j ∈ M 8 MIHAI ALBOIU Then, f0 < t, so, since we are assuming the negation of 2.9, we can choose h0, k0 ∈ Un with f0 < h0 and f0 < k0 such that h0 and k0 belong to different terms of H. Now, fix 0 < λ < ωn and suppose hγ, kγ ∈ Un have been chosen for all γ < λ. Define fλ ∈ F by ( sj,rλ(i) if i ≤ n and j ∈ Mi fλ(j) = S . sup {h(j)|h ∈ {{hγ, kγ}|γ < λ}} if j ∈ M Note that if h = hγ or h = kγ for some γ < λ, then h ∈ Un by our inductive hypoth- esis. Hence, for all j ∈ M, cf(h(j)) ≤ ωn < cf(t(j)), so, h(j) < t(j) because h ∈ U. Therefore, if j ∈ M, fλ(j) < t(j) since cf(t(j)) > ωn and λ < ωn. Thus, fλ < t. So, since we are assuming 2.9 is false, we can pick hλ, kλ ∈ Un with fλ < hλ, kλ, such that hλ and kλ belong to different members of H. We need to remember three important facts about the functions just defined: (a) If λ < ωn, i ≤ n, and j ∈ Mi, then fλ(j) = sj,rλ(i) < hλ(j) ≤ t(j) and fλ(j) = sj,rλ(i) < kλ(j) ≤ t(j); (b) If γ < λ < ωn and j ∈ M, then hγ(j) < kλ(j) < hλ+1(j) < t(j) and kγ(j) < hλ(j) < kλ+1(j) < t(j); (c) hλ and kλ belong to different terms of H. Now, let g ∈ F be given by ( t(j) if j ∈ \ M g(j) = N . sup{hλ(j)|λ < ωn} if j ∈ M If j ∈ N \ M, then, by the assumption of Case Three and the definition of M, ω < cf(g(j)) = cf(t(j)) ≤ ωn. If j ∈ M, then, by (b), cf(g(j)) = ωn since ωn is regular. It follows that g ∈ X, and, by (a), g ≤ t. Since H is a discrete family of closed sets, and since g ∈ X, there is an f ∈ F with f < g, such that the basic open set Uf,g intersects at most one term of H. Note, too, that f < t, since f < g and g ≤ t. Now, given an i ≤ n, recall that for j ∈ Mi, {sj,σ|σ < ωi} is an increasing cofinal subset of t(j). So, for each j ∈ Mi, given that f(j) < t(j), there is a σj < ωi such that f(j) < sj,σj . Put µi = sup{σj|j ∈ Mi} < ωi. Define r ∈ R by r(i) = µi for all i ≤ n. Then, for all j ∈ Mi, we have f(j) < sj,σj ≤ sj,µi = sj,r(i). Furthermore, since f < g, the definition of g gives, for each j ∈ M, a δj < ωn such that f(j) < hδj (j). Putting δ = sup{δj|j ∈ M} < ωn, (b) gives us that kγ(j), hγ(j) > hδ(j) > f(j) for all γ > δ and j ∈ M. Finally, by the way we enumerated R, there is a γ such that δ < γ < ωn and rγ = r. We claim that f < hγ, kγ ≤ g. If j ∈ M, then, by the above, f(j) < hγ(j) < g(j) since γ > δ. If instead j ∈ Mi for some i ≤ n, then, by (a), f(j) < sj,r(i) = sj,rγ (i) = fγ(j) < hγ(j) ≤ t(j) = g(j). Therefore, f < hγ ≤ g and a symmetrical argument shows that f < kγ ≤ g. Thus, by (c), the basic open set Uf,g intersects at least two terms of H. This contradicts that assumption that Uf,g is contained in at most one term of H, which completes the proof. RUDIN’S DOWKER SPACE AND THE LINEARLY LINDELOF¨ PROBLEM 9 3. Miˇscenko’sˇ Finally Compact Space Before describing Miˇsˇcenko’s space, we introduce a few notions and preliminary results. Among these is the notion of complete accumulation points and their relation to the linear Lindel¨ofproperty. Definition 3.1. Let X be a topological space and let M ⊆ X. A point x ∈ X is said to be a complete accumulation point of M iff for every neighbourhood U of x, |U ∩ M| = |M|. Definition 3.2. Suppose a and b are infinite cardinals with a ≤ b. (a) We say a space X is [a, b]AC -compact iff for all regular cardinals c with a ≤ c ≤ b, every M ⊆ X of size c has a complete accumulation point. We write [a, ∞]AC -compact iff X is [a, b]AC -compact for all b ≥ a. (b) We say a space X is [a, b]-compact iff for each cardinal c with a ≤ c ≤ b, every open cover of X of size c has a subcover of size < a.[a, ∞]-compact is defined similarly as in (a). The following result, due to Alexandrov and Urysohn (see [AL]), relates the notions of complete accumulation points and open covers in the case of regular cardinals. Theorem 3.3. Let X be a space and suppose κ is a regular cardinal. Then, every set A ⊆ X of cardinality κ has a complete accumulation point iff every open cover of size κ has a subcover of strictly smaller size. Proof. Fix a regular cardinal κ. Suppose first that every open cover of X of size κ has a subcover of strictly smaller size. Assume to the contrary that X contains a subset A = {xα|α < κ} of size κ without a complete accumulation point. For each x ∈ X, choose an open set N(x) 3 x such that |N(x) ∩ A| < κ. Let β(x) = min{ξ < S κ|N(x) ∩ A ⊆ {xα|α < ξ}}, and, for α < κ, put Vα = {N(x)|β(x) ≤ α}. Then, {Vα|α < κ} is an increasing open cover of X. By assumption, there is smaller subcover {Vαξ |ξ < λ}. Let δ = sup{αξ|ξ < λ}. Then, δ < κ since κ is regular, and Vδ = X. Hence, |A| = |A ∩ Vδ| ≤ |{xα|α < δ}| ≤ δ < κ, which is a contradiction. Conversely, suppose that every subset of X of size κ has a complete accumulation point, S and let {Uα|α < κ} be an open cover of X. For β < κ, let Vβ = {Uα|α ≤ β}. Since κ is regular, it suffices to show that {Vβ|β < κ} has a smaller subcover. So, suppose on contrary that {Vβ|β < κ} does not have a smaller subcover. We may assume, without loss of generality, that Vα ( Vβ whenever α < β. For each α < κ, choose xα ∈ Vα+1 \Vα and let A = {xα|α < κ}. By assumption, A has a complete accumulation point x. There is β < κ such that x ∈ Vβ. But then, |Vβ ∩ A| = |{xα|α < β}| ≤ β < κ, contradicting that x is a complete accumulation point of A. Corollary 3.4. A space is [a, b]AC -compact whenever it is [a, b]-compact. This motivates the question of whether these two notions of compactness are in fact the same. Definition 3.5. A topological space X is said to be finally compact in the sense of accumulation points (FCAP) iff every M ⊆ X of uncountable regular cardinality has a complete accumulation point. In the previous terminology, this amounts to saying X AC is [ℵ1, ∞] -compact. 10 MIHAI ALBOIU AC Miˇsˇcenko’s example is that of a space which is [ℵ1, ∞] -compact, but not [ℵ1, ∞]- compact. His example, therefore, proves that the converse of 3.4 does not hold. Fur- thermore, as the next Corollary shows, Miˇsˇcenko’s example provides us with a space that is linearly Lindel¨ofbut not Lindel¨of. Corollary 3.6. X is linearly Lindel¨ofiff X is FCAP. Proof. Suppose first that X is linearly Lindel¨of.If κ is an uncountable regular cardinal, {Uα|α < κ} is an open cover of X, and {Vβ|β < κ} is the corresponding increasing open cover, then {Vβ|β < κ} has a countable subcover. Hence, since κ is regular, {Uα|α < κ} must have a subcover of smaller size. Since κ was arbitrary, it follows, by 3.3, that X is FCAP. Conversely, suppose that X is FCAP. Then, by 3.3, every increasing open cover of regular cardinality has a strictly smaller subcover. Also, every increasing open cover of singular cardinality has a strictly smaller subcover. Since there is no infinite decreas- ing sequence of ordinals, it follows that every increasing open cover has a coutnable subcover. We are almost ready to describe Miˇsˇcenko’s example, but we need to introduce one more notion first. Definition 3.7. Let X be a space and let m be an infinite cardinal. Then, X is said to be m-bounded iff for every A ⊆ X with |A| ≤ m, there is a compact C ⊆ X with A ⊆ C. Lemma 3.8. Every m-bounded space is [ℵ0, m]-compact. Proof. Suppose X is an m-bounded space. Assume to the contrary that X is not [ℵ0, m]- compact. Let O be an open cover of X with |O| = κ ∈ [ℵ0, m] having no finite subcover. <ω S <ω For each N ∈ [O] , choose xN ∈ X \ N. Then, {xN |N ∈ [O] } is a set of size at most κ. Since X is m-bounded, there is a compact C ⊆ X containing A. Since O <ω covers C, there is a finite subcover U ∈ [O] covering C, too. But then, xU ∈/ A, S S since xU ∈/ U and A ⊆ C ⊆ U. This contradiction shows that X must be [ℵ0, m]- compact. Lemma 3.9. Let m be an infinite cardinal. If {Xα|α < λ} are a collection of m-bounded Q topological spaces, then X = {Xα|α < λ} is m-bounded. Proof. Let πα : X → Xα denote the projection map onto the α-coordinate. Suppose A ⊆ X is of size |A| ≤ m. For each α < λ, |πα(A)| ≤ m, so, since each Xα is m- bounded, there is a compact Cα ⊆ Xα containing πα(A). By the Tychonoff theorem, Q C = α Cα ⊆ X is compact and A ⊆ C. Thus, X is m-bounded. Example 3.10 (Miˇsˇcenko’s Space). The space will be a subset of the product space ∞ Y R = [0, ωk]. k=1 For 1 ≤ k < ω, let k ∞ Y Y Rk = [0, ωi] × [0, ωi), i=1 i=k+1 RUDIN’S DOWKER SPACE AND THE LINEARLY LINDELOF¨ PROBLEM 11 and set ∞ ∗ [ R = Rk. k=1 ∗ AC ∗ We will show that R is [ℵ1, ∞] -compact. The topology on R will be the subspace topology of the ordinary product topology on R with the usual Tychonoff basis B. ∗ AC Note that |B| = ℵω. Hence, R is [ℵω+1, ∞] -compact. To see this, take a subset A ⊆ X with |A| regular and |A| ≥ ℵω+1. If A has no complete accumulation point in X, then, for each x ∈ X, we can find a basic open set N(x) ∈ B containing x, such that |A∩N(x)| < |A|. But then, |A| = |A∩S{N(x)|x ∈ X}| = | S{A∩N(x)|x ∈ X}| < |A|, since |B| = ℵω and |A| ≥ ℵω+1 is regular, which is a contradiction. Now, fix 1 ≤ n < ω and suppose k ≥ n. Each factor in Rk is ℵn-bounded, thus, by 3.9, Rk is ℵn-bounded, too. Applying 3.8 then gives that Rk is [ℵ0, ℵn]-compact. ∞ ∗ ∗ S ∗ Since R is an increasing union, R = Rk. So, R is [ℵ1, ℵn]-compact, and hence, k=n AC ∗ AC [ℵ1, ℵω) -compact by 3.4. Thus, R is [ℵ1, ∞] -compact. ∗ We now show that R is not [ℵ1, ∞]-compact. To do this, we will exhibit an open cover ∗ of R of size ℵω having no subcover of smaller size. Here’s the setup. For 1 ≤ k < ω and ∗ for each α < ωk, let Γα,k = {x = {x(i)|i < ω} ∈ R |x(k) < α}. Let πk = {Γα,k|α < ωk}, S ∗ and set π = πk. Then, π is an open cover of R of size ℵω. Suppose π had a subcover 1≤k<ω π∗ of smaller size. Then, there must be an n < ω with the property that for all k > n, ∗ there is an αk < ωk, such that for all α > αk,Γα,k ∈/ π . Define x = {x(i)} as follows: ( ω if 1 ≤ i ≤ n x(i) = i . αi + 1 if i > n ∗ S ∗ ∗ Then, x ∈ Rn ⊆ R , but x∈ / π , which is a contradiction. Thus, R is not [ℵ1, ∞]- compact. Of course, at this point, one might very well ask whether Miˇsˇcenko’s example provides us with a counterexample to settle the famous linearly Lindel¨ofproblem. Note that R∗ is completely regular, being the subspace of a completely regular space, but is it normal? The answer is no. Claim 3.11. R∗ is not normal. k ∞ Q Q Proof. For 1 ≤ k < ω, let pk = {ωi} × {0}. Let A = {pk|1 ≤ k < ω} and i=1 i=k+1 B = {x = {x(i)} ∈ R∗| for all i, x(i) 6= 0}. It is easy to see that A and B are disjoint closed subsets of R∗. Suppose we are given open sets U ⊇ A and V ⊇ B. We’ll show k k that U ∩ V 6= ∅. For each 1 ≤ k < ω, there exist α1, . . . , αk such that k ∞ Y k Y pk ∈ (αi , ωi] × {0} ⊆ U. i=1 i=k+1 j For 1 ≤ k < ω, let θk = sup{αk|j ≥ k} < ωk. Then, for each 1 ≤ k < ω, k ∞ Y Y Hk := (θi, ωi] × {0} ⊆ U. i=1 i=k+1 12 MIHAI ALBOIU ∞ Q Now, consider z = {θi + 1}. Clearly, z ∈ B ⊆ V . But, we also have that z ∈ i=1 S Hk ⊆ U. Hence, U ∩ V 6= ∅, which is what we set out to show. 1≤k<ω 4. Buzyakova and Gruenhage’s Space In this section, we present another interesting example of a linearly Lindel¨ofspace that is not Lindel¨of.This example, attributed independently to Buzyakova and Gru- enhage, is somewhat simpler and more self-contained than Miˇsˇcenko’s. Let’s get right to it. Example 4.1 (Buzyakova and Gruenhage’s Space). Consider the two-point discrete space D = {0, 1}, and the corresponding product Dℵω . The space in question will be ℵω X = {x = x(α) ∈ D || supp(x)| < ℵω}, where supp(x) = {α < ℵω|x(α) = 1}. To help define the topology on X, we introduce the following definition. Definition 4.2. Given a function f, we let [f] = {g ∈ X| for all j ∈ dom(f), g(j) = f(j)}. We declare sets of the form [f] with dom(f) ⊆ ωω and |dom(f)| < ω to be elementary open sets. The topology on X will be generated by the collection B consisting of all such elementary open sets. We first show that X is linearly Lindel¨of.Note that B has cardinality ℵω. Thus, AC X is [ℵω+1, ∞] -compact, just as in Miˇsˇcenko’s example. So, by 3.6, it suffices to AC show that X is [ℵ1, ℵω) -compact. To this end, let A ⊆ X and suppose |A| = ℵk for some 1 ≤ k < ω. We want to show A has a complete accumulation point in X. For 1 ≤ n < ℵω, let Xn = {x ∈ X|| supp(x)| = ℵn} and let An = Xn ∩ A. Since A is S uncountable, there is an 1 ≤ n < ω with |An| = |A|. Let S = supp(x), and set x∈An H = {x ∈ X| supp(x) ⊆ S}. Note that H is a closed subset of the compact space Dℵω , and hence, is compact. Thus, every infinite subset of H has a complete accumulation point in H. In particular, there is a complete accumulation point a ∈ H of An. But, H ⊆ X. Indeed, |S| ≤ max{|An|, sup{| supp(x)||x ∈ An}} = max{ℵk, ℵn} < ℵω. Hence, a ∈ X, so that An has a complete accumulation point in X. Since |An| = |A|, a is a complete accumulation point of A, too. Thus, X is linearly Lindel¨of. To see that X is not Lindel¨of,let πα : X → D be the projection onto the α- ← coordinate, and consider the open cover {πα {0}|α < ℵω} of X. If X were Linedl¨of, ← there would be a countable subcover {παn {0}|n < ω} that would cover X. But this cannot be, for the point x = x(β) given by x(β) = 1 iff β = αn for some n < ω is missed by the subcover, albeit still in X. Note that just like Miˇsˇcenko’s space, X is completely regular. Unfortunately, though, just as with Miˇsˇcenko’s space, X fails to be normal, as we shall now see. Claim 4.3. X is not normal. Proof. Let πα : X → D denote the projection onto the α-coordinate. Let H = T ← {πωn {1}|n ≥ 0}, and let K = {pn|n < ω}, where pn(j) = 1 iff j < ωn. It is easy to see that H and K are closed subsets of X, which are disjoint. Suppose U ⊇ H and RUDIN’S DOWKER SPACE AND THE LINEARLY LINDELOF¨ PROBLEM 13 V ⊇ K are open. We’ll show that U ∩ V 6= ∅. For each n < ω, let [fn] be an elementary open set such that pn ∈ [fn] ⊆ V , and define yn ∈ X by 1 if f (j) = 1 for some 0 ≤ k ≤ n k yn(j) = 1 if j = ωk for some k < n . 0 otherwise Then, each yn ∈ [fn]. Indeed, since pk ∈ [fk] for all k < ω, it follows that fk(j) 6= 0 for all j < ωk. Note, too, that the yn’s build upon each other; i.e., if k < l and yk(j) = 1, then yl(j) = 1. Now, let y ∈ X be defined by y(j) = max{yn(j)|n < ω}. By construction, y ∈ H ⊆ U. If [f] is any elementary open set containing y, then, for each α ∈ dom(f) with f(α) = 1, there is an nα < ω such that ynα (α) = y(α) = f(α). Since |dom(f)| < ω and since the yn’s build upon each other, we can let N = max{nα|α ∈ S∞ dom(f) and f(α) = 1}, so that yN ∈ [f]. Thus, y ∈ n=1[fn] ⊆ V . Hence, U ∩ V 6= ∅, as desired. 5. Concluding Remarks It is worth pointing out that in both Miˇsˇcenko’s and Buzyakova and Gruenhage’s example, we proved something slightly stronger than non-normality. In both cases, one of the closed sets witnessing non-normality was countable. Thus, neither of the two spaces is even pseudonormal; a space is pseudonormal iff for every two disjoint closed sets, one of which is countable, there are disjoint open sets separating them. An example of a linearly Lindel¨ofspace that is not Lindel¨of,but that is pseudonormal would, therefore, be quite interesting. As far as I know, there is no such example of a space in the literature. References [AL] P.S. Alexandroff, P.S. Urysohn, M´emoire sur les espaces topologiques compacts, Verh. Kon. Akad. Wetensch. Amsterdam 14 (1929)m no. 1, 96 pp [AR] A.V. Arhangel’skii, R.Z. Buzyakova, On Linearly Lindel¨ofand Strongly Discretely Lindel¨of Spaces, Proc. Amer. Math. Soc. 127 (1999), no. 8, 2449-2458. [DO] C.H. Dowker, On Countably Paracompact Spaces, Canad. J. Math. 3 (1951), 219-224. [GU] S.I. Gulden, W.F. Fleischman, J.H. Weston, Linearly Ordered Topological Spaces, Proc. Am. Math Soc. 24 (1970), 197-203. [HOD] R.E. Hodel, J.E. Vaughan, A note on [a, b]-Compactness Gen. Top. and its App.4 (1974), 179-189. [HOW] N.R. Howes, A Note on Transfinite Sequences, Fund. Math. 106 (1980), 213-216. [IS] F. Ishikawa, On Countably Paracompact Spaces, Proc. Japan Acad. 39: 95-97. [MI] A. Miˇsˇcenko, On Finally Compact Spaces, Dokl. Akad. Nauk SSSR 145 (1962), 1224-1227, Translation: Soviet Math. Dokl. 3 (1962) 1199-1202. [RU] M.E. Rudin, A normal space X for which X × I is not normal, Fund. Math. 73 (1971), 179-186.