ZFC examples

The key to Tkachuk’s above mentioned CH results in [65] was his Theorem 3.1 which says that if X has countable tightness and π-character, moreover d(X) ≤ ω₁ then πsw(X) ≤ ω. In his list of problems (Problem 4.11), Tkachuk asked if the assump-tion of countable tightness could be omitted here. It is immediate from our next result that this question has an affirmative answer.

Theorem 3.13. LetXbe any topological space withd(X)≤πχ(X)⁺. Thenπsw(X)≤ πχ(X).

Proof. Let us set πχ(X) = κ. If d(X) ≤ κ then we even have π(X) = κ. So we may assume d(X) = κ⁺ and, as is well-known, we may then fix a dense set D = {x_α : α < κ⁺} that is left-separated in this well-ordering. This means that for everyα < κ⁺there is a neighbourhoodU_α ofx_αwith

{x_β :β < α} ∩U_α =∅.

Let us now fix a local π-base B_α of the point x_α such that |B_α| ≤ κ and B ⊂ U_α wheneverB ∈ B_α. Then B = S

{B_α : α < κ⁺}is a π-base ofX such that for every xβ ∈Dwe have

ord(x_β,B) =|{B ∈ B:x_β ∈B}| ≤κ.

We claim that then we have

ord(B) = sup{ord(x,B) :x∈X} ≤κ

as well. Assume, on the contrary, that ord(x,B) = κ⁺ for some pointx ∈ X.Since πχ(x, X) ≤ κ, this implies that there areκ⁺-many members of B(containing x) that include a fixed non-empty open setV. This, however, is impossible becauseD∩V 6=∅.

We may now turn to our first aim that is to produce, in ZFC, first countable spaces without point-countableπ-bases.

Theorem 3.14. There is a first countable, 0-dimensional Hausdorff (hence Tychonov) spaceX withπsw(X)≥ ℵ_ω.

Proof.The underlying set of our space isX =Q

{ω_n:n < ω}. Forf, g∈X we write f ≤g to denote thatf(n)≤g(n)for alln < ω.The topologyτ that we shall consider onX will be generated by all sets of the formU_n(f)(withf ∈X andn < ω), where

U_n(f) = {g ∈X :f ≤gandf n=g n}.

Note that ifg ∈U_n(f)thenU_n(g)⊂U_n(f), and ifg /∈U_n(f)then there isk < ωsuch thatU_k(g)∩U_n(f) = ∅. It follows that, for anyf ∈ X, the family{U_n(f) : n < ω}

forms a clopen neighbourhood base off with respect to the topology τ, consequently the spacehX, τiis indeed first countable, 0-dimensional, and Hausdorff.

It is also easy to see from the definitions that if{U_nα(f_α) :α < κ}is aπ-base ofτ then{f_α :α < κ}must be cofinal in the partial orderhX,≤i. But it is well-known that the cofinality ofhX,≤iis greater thanℵ_ω, consequently we haveπ(X)>ℵ_ω.(Actually, it is easy to see thatπ(X) = cf hX,≤i

but we shall not need this.)

Next we claim that, for anyk < ω, the pair(ℵ_ω+1,ℵ_k)is apair caliberof the space X, i.e. among any ℵ_ω+1 open sets one can find ℵ_k whose intersection is non-empty.

Without any loss of generality, it suffices to check this for a family of basic open sets of the form {U_n(f) : f ∈ F} whereF ∈ [X]^ℵω+1 and n > k is fixed. We may also

Putting together the previous two paragraphs we conclude that the order of any π-base ofhX, τimust be at leastℵ_ω, that is we haveπsw(X)≥ ℵ_ω.

It is clear that if we replace in the above proof the sequence of cardinalshω_n:n < ωi with any other strictly increasing ω-sequence of regular cardinals, say hκ_n:n < ωi, then we obtain a first countable, 0-dimensional space in which the order of anyπ-base is at leastP

n<ωκ_n.

The referee has pointed out that the method of constructing such spaces was pub-lished by TodorˇcevicŽ in [66], Theorem 0.5 (of course, the fact that they do not have a point-countableπ-base is not mentioned there).

The cardinality of our above example is ℵ^ℵ_ω⁰ that is much larger than the optimal value ℵ₂ permitted by Theorem 3.13. So it is natural to raise the question if we could find other examples of smaller cardinality. As it turns out, we can do slightly better by choosing an appropriate subspaceY of the spaceXfrom Theorem3.14. First, however, we need to fix some notation. Forf, g ∈ X = Q

{ω_n : n < ω} we write f <^∗ g to denote that|{n < ω :f(n)≥g(n)}|is finite, i.e.f is belowgmodulo finite. Similarly, we write f =^∗ g to denote that |{n < ω : f(n) 6= g(n)}|is finite. Finally, it is well-known that there is inXa transfinite sequence of order typeω_ω+1that is increasing with respect to<^∗.

Theorem 3.15. Let {f_α : α < ω_ω+1} ⊂ X be an increasing sequence with respect to

<^∗and set

Y ={f ∈X : ∃α < ω_ω+1withf =^∗ f_α}.

Then the subspaceY of X, with the subspace topology inherited fromτ, also satisfies πsw(Y)≥ ℵ_ω.

Proof.The proof is very similar to that of Theorem3.14. First we note that, trivially, again we haveπ(Y)>ℵ_ω.Next, we show that(ℵ_ω+1,ℵ_k)is a pair caliber ofY for each k < ω.To see this, we again consider a family {U_n(f) : f ∈ F} whereF ∈ [Y]^ℵω+1

andn > k > 0, moreoverf n =σ for a fixedσ ∈ Q

i<nω_i wheneverf ∈F. Let us choose any subset G⊂ F with|G| = ℵ_k, then there is an ordinalα < ω_ω+1 such that g <^∗ f_α for allg ∈G. We may find an integerm≥nsuch that the set

G^∗ ={g ∈G: ∀i≥m g(i)< f_α(i) } also has cardinalityℵk.

Note that if n ≤ j < m then {g(j) : g ∈ G^∗} is bounded in ω_j, hence we may find a function f ∈ Y such that f n = σ, if n ≤ j < m then g(j) < f(j) for all g ∈ G^∗, moreover f(i) = fα(i) whenever m ≤ i < ω. Clearly, then we have f ∈T{U_n(g) :g ∈G^∗} ∩Y.

We were unable to produce a ZFC example of a first countable space without a point-countableπ-base of cardinality less thanℵω+1. This leads us to the following intriguing open question.

Problem 3.16. Is there, in ZFC, a first countable (Tychonov) space of cardinality less thanℵ_ω that has no point-countableπ-base?

Actually, at this point we do not even have such an example of cardinalityℵ_ω. We conjecture, however, that having such an example is equivalent to having one of size

<ℵ_ω. In fact, we could verify this conjecture under the assumption2^ℵ1 <ℵ_ω.

Theorem 3.17. Assume that 2^ℵ1 < ℵ_ω and X is a first countable space of cardinality ℵ_ω. If every subspace of X of cardinality < ℵ_ω has a point-countable π-base then so doesX.

Proof.Let us start by giving a (very natural) definition. A familyBof non-empty open sets in X is said to be anouter π-base of a subspaceY ⊂ X if for every open set U withU ∩Y 6=∅there is a memberB ∈ Bsuch thatB ⊂ U. We claim that, under the assumptions of our theorem, every subspace ofX of cardinality<ℵ_ωeven has a point-countable outerπ-base. Thus if we haveX =S

n<ωY_n where|Y_n|<ℵ_ω for alln < ω andB_nis a point-countable outerπ-base ofY_ninXthenS

n<ωB_nis a point-countable π-base ofX.

To prove the above claim let us consider anω₁-closed elementary submodelM of a

"universe"H(θ)with |M| < ℵ_ω. (As usual, here θ is a large enough regular cardinal, H(θ) is the collection of all sets of hereditary cardinality < θ, and for M to be ω₁ -closed means that [M]^≤ω1 ⊂ M.) The regular cardinalθ is chosen so large thatH(θ) (and alsoM) containsX and everything else that is relevant, e.g. a mapV that assigns to every pointx∈Xa countable open neighbourhood baseV_x. Now,2^ℵ1 <ℵ_ωimplies that for every Y ∈ [X]^<ℵω there is such an elementary submodel M with Y ⊂ M.

Consequently, our claim will be proven if we show thatX∩M has a point-countable outerπ-base inXwheneverM is like above.

To see this, note first that for every point x ∈ X ∩ M we have V_x ∈ M and hence V_x ⊂ M as well. Consequently V_M = ∪{V_x : x ∈ X ∩ M} ⊂ M is an outer base of X ∩M in X, hence we may choose a subfamily B ⊂ V_M such that B M ={B∩M :B ∈ B}is a point-countableπ-base of the subspaceX∩M.

It suffices to show now that B is a point-countable outer π-base of X ∩M in X.

Indeed,Bis point-countable for ifU ∈[B]^ω1 thenU ∈M becauseM isω₁-closed, and thus∩U 6=∅would imply ∩U ∩M 6= ∅, contradicting thatB M is point-countable.

(Here we used the fact that, by elementarity, the correspondanceB 7→ B∩M is one-to-one onB ⊂ M.) Finally,B is an outerπ-base ofX∩M inX because ifU is open withx ∈ U ∩M 6= ∅then there is V ∈ V_x ⊂ M withV ⊂ U, hence ifB ∈ B with B∩M ⊂V ∈M then we also haveB ⊂V ⊂U.