The main result of this section says that if κ is any uncountable regular cardinal and X is a compact T2 space containing a free sequence of length κ, thenX also contains a convergent such sequence. Let us note that having a free sequence {pα : α ∈ κ}
converging to p in X also yields a closed set F ⊂ X with χ(p, F) = κ. Indeed, F ={pα :α∈κ}works since the sets{pα :α∈κ\β}are clopen inF for allβ ∈κ, for the sequence{pα :α ∈κ}is free, i.e.,
{pα :α∈β} ∩ {pα :α∈κ\β}=∅
forβ ∈κ.
The proof of the main result is split into two cases according to whetherXcontains a closed set that maps continuously onto 2κ or not. In the first case we get somewhat stronger results and now we turn to their discussion.
Theorem 1.1. Letκ be any uncountable cardinal andf : X → 2κ be an irreducible continuous map of the compactT2 spaceX onto2κ . Then for every point x ∈ X we have
(i) there is a (relatively) discrete subspace D ⊂ X \ {x} with |D| = κ that "con-verges" tox in the strong sense that every neighbourhood of x contains all but countably many elements ofD;
(ii) moreover, ifcf(κ)> ω, then there is a free sequence{xα :α∈κ} ⊂X\ {x}that converges tox.
Proof.We need several simple facts concerning irreducible maps that are probably well known, although we have not found them in the literature. For the sake of completeness we present them here with their proofs. First, the image of a regular closed set (in particular, of a clopen set)A under an irreducible closed map f : X → Z is regular closed. Indeed, if we hady∈f(A)\intf(A)6=∅thenA\f−1
intf(A) , hence G= intA\f−1
intf(A) 6=∅
becauseAis regular closed. Sincef is irreducible we haveintf(G)6=∅, contradicting that bothintf(G)\intf(A)andf(G)∩intf(A) = ∅.
Next we show that iff is as above andA, Bare regular closed sets inX, then intf(A∩B) = intf(A)∩intf(B).
Indeed, assume thatintf(A∩B)$intf(A)∩intf(B). Then
C = (intf(A)∩intf(B))\f(A∩B) 6=∅, henceD= intA∩f−1(C)6= ∅, as well.
But thenD∩B =∅andf(D)\C\f(B); consequently,f(X\D) =Y, contradicting thatf is irreducible.
Let us now return to the proof of the theorem, where, for technical reasons, we first assume thatXis also0-dimensional.
For S ⊂ κ a subset H of 2κ is called S-determined if z ∈ H implies z0 ∈ H whenever z0 ∈ 2κ and z0 S = z S. It is well known (cf. [15]) that every regular closed (or open) subset of2κ isS-determined for some countableS ⊂ κ. Moreover, it is obvious that ifH isS-determined then so areH andintHas well.
Forα ∈κandi∈2we shall put
Uα,i ={z ∈2κ :z(α) = i}
so theUα,iform the canonical subbase for the product topology on2κ.
Now, for any subset S ⊂ κwe let AS be the collection of all those clopen subsets A of X that contain x and whose f-image f(A) is S-determined. According to our preliminary remarks, for every clopen neighbourhood A of xthere is some countable set S ⊂ κ such that A ∈ AS . Moreover, each collection AS is closed under finite
becauseXis compact andASis closed under finite intersections. But for allA ∈ AS we havey∈f(A)andf(A)isS-determined, henceΦS ⊂f(A), i.e.,ΦS ⊂f(FS). On
That Dis discrete follows immediately from the fact that the set {yα : α ∈ κ} is discrete. Next, ifAis any clopen neighbourhood ofxthen there is a countable setS ⊂κ withA∈ A :S. But then for everyα∈κ\S, we have
Clearly {yα : α ∈ κ} is a free sequence in 2κ. Moreover, yα ∈ Φα+1 for each α ∈ κ. Sincef(Fα+1) = Φα+1, we can choose pointsxα ∈ Fα+1 forα ∈ κsuch that f(xα) = yα.
Now {xα : α ∈ κ} is a free sequence in X: if z ∈ {xα :α∈β} then f(z) ∈ {yα :α∈β} by the continuity of f; hence f(z) ∈ {y/ α :α ∈β\β} because {yα} is free; thusz /∈ {xα :α∈β\β}.
Finally to see that {xα : α ∈ κ} converges to x it clearly suffices to show that T{Fα+1 :α∈κ}={x}since theFα+1 are decreasing andxα ∈Fα+1\Fα+2. But for every clopen neighbourhoodAofxthere is a countableS ⊂κwithA ∈ AS, hence by cf(κ)> ωthere is anα∈κsuch thatS⊂α. Consequently,Fα+1 ⊂Fα ⊂FS ⊂A.
Thus, using again thatX is0-dimensional, (ii) has been established.
Now let us consider the general case with an arbitrary compact T2 space X. By Alexandrov’s well-known theorem (see, e.g., [15, 3.2.2]), there is a0- dimensional com-pactT2spaceZadmitting an irreducible mapg :Z →XontoX. Then the composition h =f ◦g : Z → 2κ is also irreducible, hence the above considerations can be applied toh and a fixed pointz ∈ Z with g(z) = x. This gives us points{zα : α ∈ κ}with h(zα) = yα and{zα : α ∈ κwithh(zα) = yα. Now it is straightforward to check that the setD ={xα =g(zα) :α ∈ κ}and the free sequence{xα =g(zα) : α∈ κ}are as required by (i) and (ii), respectively.
Now we turn to the formulation of our main result.
Theorem 1.2. Let κ be an uncountable regular cardinal. If a compact T2 space X contains a free sequenceS ={xα :α ∈κ}of lengthκthen it also contains one that is convergent.
Proof. Let us first note that the closure S of S can be mapped continuously onto the spaceκ+ 1taken with its usual order topology. Indeed, if for any limit ordinalλ < κ we set
Aλ =\
{xβ :β ∈λ\α}, or equivalently, sinceS is free,
Aλ ={x∈X :λ= min{α:x∈ {xβ :β ∈α}0}}, then the mapf :S →κ+ 1(defined below) is clearly continuous:
f(x) =
n x=xn, n < ω α+ 1 x=xα, α∈κ\ω
λ x∈Aλ, λ≤κlimit
Thus we may assume without any loss of generality that X admits an irreducible mapfontoκ+ 1. Moreover, using1.1(ii), we may also assume that no closed subspace
ofX can be mapped onto2κ. This, however, by Sapirovskil’s celebrated result (cf. [25, 3.18]) implies that every nonempty closed subsetF ofX has a pointpwithπχ(p, F)<
κ.
For every ordinalα < κlet us defineZα =f−1((κ+ 1)\α). In particular, then the setZκ =f−1({κ})is nowhere dense inXbecausef is irreducible.
Letp ∈ Zκ be any point havingπ-characterπχ(p, Zκ) = π less thanκ inZκ. We are going to show that there is a free sequence of lengthκ converging top. This will clearly establish our result.
First we claim that for every open subset GofX ifG∩Zκ 6= ∅thenf(G)∩κis cofinal inκ. Indeed, otherwise we had anα ∈ κsuch thatG\Zκ ⊂ f−l[α], hence as Zκ is nowhere dense we had
G⊂G\Zκ ⊂f−1[α]⊂f−1[α+ 1], contradicting thatG∩Zκ 6=∅.
Usingπχ(p, Zκ) =π(< κ), we may choose open sets{Gγ :γ ∈π}inX such that {Gγ∩Zκ : γ ∈ π}forms a localπ-base atpinZκ. Let us pick a pointqγ ∈ Gγ ∩Zκ for eachγ ∈ π and then choose the open setBγ inX such thatqγ ∈ Bγ ⊂ Bγ ⊂ Gγ. By the above claim, then
Cγ =κ∩f(Bγ)
is a closed unbounded subset ofκ, hence byπ < κso isC = T{Cγ : γ ∈ π}. (Here we use the regularity ofκ.)
Now, for everyα ∈ κwe letVα be the collection of all those open neighbourhoods V ofpfor which Bγ∩Zκ ⊂ V impliesBγ∩Zα for allγ ∈ π. Clearly V1, V2 ∈ Vα impliesV1 ∩V2 ∈ Vκ. MoreoverVα ⊂ Vβ ifα < β. Let us put
Fα =\
{V :V ∈ Vα}
Then we haveFα ⊃Fβ forα < β and f(Fα) =\
{f(V) :V ∈ Vα} by our above remarks.
But for every V ∈ Vα, there is a γ ∈ π such that Bγ ∩ Zα ⊂ Gγ ∩ Zα ⊂ V. Consequently,
C\α⊂Cγ\α⊂f(Bγ∩Zα)⊂f(V).
Thus we obtainC\α⊂f(Fα)and, in particular,Fα6={p}.
Next we show that for every neighbourhood V ofpthere is someα ∈ κ withV ∈ Vα. Indeed, ifBγ∩Zκ ⊂V then
Zκ =\
{Zα :α∈κ},
andZα ⊂ Zβ forβ < αimply that, asX is compact, there is someαγ ∈ κ such that Bγ ∩Zαγ ⊂ V. Thus if a α ∈ κ is chosen with αγ ≤ α for each γ ∈ π, and this is possible sinceκis regular andπ < κ, thenV ∈ Vαindeed.
Putting all this together we obtain from the setsFαa strictly decreasing sequence of closed sets whose intersection is{p} . By suitably thinning out, we may assume that Fα+1 $ Fα for all α ∈ κ; hence ifpα ∈ Fα+1 \Fα then the sequence{pα : α ∈ κ}
converges top. Finally, by passing to a subsequence we may assume that ifα < βthen f(palpha) < f(pβ), and then the subsequence{pα+1 : α ∈ κ}is also free because its f-image{f(pα+1) :α∈κ}is.