Minimal KC –spaces are countably compact

Minimal KC –spaces are countably compact

T. Vidalis

Abstract. In this paper we show that a minimal space in which compact subsets are closed is countably compact. This answers a question posed in [1].

Keywords: KC-space, weaker topology Classification: 54A10

1. Introduction

A topological space (X, τ) is said to be a KC-space if every compact set is closed. Since everyKC-space isT1 and everyT2 space isKC, theKC-property can be thought of as a separation axiom betweenT1 andT2.

In 1943 E. Hewitt [3] proved that a compact T2 space is minimal T2 and maximal compact, see also [5], [6], [7]. R. Larson [4] asked whether a space is maximal compact iff it is minimal KC. A related question is whether every KC-topology contains a minimalKC-topology. W. Fleissner proved that this is not always true. In [2] he constructed aKC-topology which does not contain a minimalKC-topology.

In a recent paper, [1], the authors proved that every minimalKC-topology on a countable set is compact and posed the question whether minimal KC-spaces are countably compact.

In this paper we answer affirmatively this question by proving thateveryKC- space which is not countably compact has a strictly weaker KC-topology.

2. Preliminaries and notations

A filter over a setX is a collectionF of subsets ofX such that:

(i) ∅∈ F;/

(ii) ifF₁ ∈ F andF₂∈ F thenF₁∩F₂ ∈ F;

(iii) ifA, B⊂X, A∈ F andB⊃AthenB∈ F. A filterF over a setX is an ultrafilter if

∀A⊂X either A∈ F or X−A∈ F.

With|A|we denote the cardinality of a setA, and with A^c the complement of a setA.

Forκan infinite cardinal number, an ultrafilterF overκis uniform if |F|=κ for allF∈ F.

3. MinimalKC-spaces are countably compact

Let (X, τ) be aKC-space which is not countably compact. Then there exists a set {xn : n ∈ ω} ⊂ X which has no accumulation points. We define a new topologyτ^′ onX as follows:

For everyx∈X withx6=x₀ the open neighborhoods ofxinτ^′ coincide with the open neighborhoods ofxin τ.

(NT) An open neighborhood ofx₀ inτ^′ is everyτ-open set containing x₀ and a member ofF, whereF is a uniform ultrafilter defined over the set{x_n: 0< n < ω}.

Remark 3.1. It is clear thatτ^′ is aT₁-topology and thatx₀ is the unique point which can beτ^′-accumulation point for a setK⊂Xwhile it is notτ-accumulation point of it.

Our aim is to show that if (X, τ) is aKC-space, which is not countably com- pact, then the topologyτ^′ defined by (NT) is also aKC-topology.

LetK⊂X beτ^′-compact. Ifx₀∈/K thenKisτ-compact, thusτ-closed, and since{xn : n∈ ω} has no accumulation points we have that {xn :n ∈ω} ∩K is finite. Hencex₀ is not aτ^′-accumulation point ofK and it follows that K is τ^′-closed.

So it remains to prove that if K ⊂ X is τ^′-compact and x₀ ∈ K, then K is τ^′-closed, or equivalently it is τ-closed. Therefore we assume for the rest of the paper thatx₀∈K.

To prove that aτ^′-compact setK isτ^′-closed we consider the following cases for a member of the ultrafilterF in relation withK:

(1) F ⊂K;

(2) F∩K^τ =∅;

(3) F ⊂(K^τ −K).

Lemma 3.2 below refers to case (1), Lemma 3.3 to case (2), while Lemmas 3.4 and 3.5 to case (3).

Lemma 3.2. Let(X, τ) be aKC-space which is not countably compact,{xn : n ∈ω} a set without accumulation points, F a uniform ultrafilter defined over {xn: 0< n < ω},τ^′ the topology defined by(NT)andKaτ^′-compact set. Then there is anF ∈ F, such thatF∩K=∅.

Proof: SinceF is an ultrafilter, either there exists anF ∈ F such that F⊂K, or there is anF ∈ F withF∩K=∅.

In the first case letF =F₁∪F₂ withF₁∩F₂=∅and|F₁|=|F₂|=ω.

Then ifF1∈ F, there exists an open setU(F1) containingF1 with U(F₁)∩F₂ =∅.

Thus there is aτ^′-open neighborhood ofx₀,U^′(x₀), with F₂∩U^′(x₀) =∅,

and F2 will be an infinite subset ofK withoutτ^′-accumulation points, which is impossible. So there must be anF∈ F such that: F∩K=∅.

Lemma 3.3. With the assumptions of Lemma3.2if there exists anF0 ∈ Fsuch thatF0∩K^τ =∅, thenK isτ^′-closed.

Proof: Sincex₀∈K it suffices to show thatK isτ-closed.

Let{U_i :i∈I}, be aτ-open cover ofK and letV₀ be an open set containing F₀ such thatV₀∩K=∅.

Then the collection{U_i∪V₀ :i∈I}, is aτ^′-open cover ofK and thus it has a finite subcover, say,U_i1∪U_i2∪. . .∪U_in∪V₀.

The set S

{U_ik : k = 1,2, . . . , n} coversK, so K is τ-compact and therefore

τ-closed.

It remains to consider the case where there is anF ∈ F such thatF ⊂(K^τ− K). We will show first that in this caseKis countably compact.

Lemma 3.4. Let (X, τ) be a KC-space which is not countably compact, τ^′ the topology defined by (NT), K a τ^′-compact set, x0 ∈ K and F0 ∈ F with F0 ⊂(K^τ−K). ThenKisτ-countably compact.

Proof: LetF₀∈ F be such thatF₀ ⊂(K^τ −K), with F₀ ={xnk :k∈ω} and suppose for a contradiction thatKis notτ-countably compact.

Then there exists a set{yn:n∈ω} ⊂K withoutτ-accumulation points inK and sincex₀∈K, there is aτ-open neighborhoodU(x₀) ofx₀ with

U(x₀)∩ {yn:n∈ω}=∅.

We claim that for every infinite subset{ynk:k∈ω}of{yn:n∈ω}and for every z∈F₀ there is aτ-open neighborhood ofz,U(z), such that

|U(z)^c∩ {ynk :k∈ω}|=ω.

Actually, for otherwise{y_nk:k∈ω} →zand sinceτ is aKC-topology,zwill be the uniqueτ-accumulation point of{y_nk :k∈ω}.

But, there is anF ∈ F withz /∈F, thus there is an open setW(F) containing F with z /∈ W(F). So z /∈ U(x₀)∪W(F), and consequently x₀ is not a τ^′- accumulation point of{ynk :k∈ω}.

It follows that{ynk :k∈ω}is an infinite subset ofKwith noτ^′-accumulation points inK which is impossible, sinceK isτ^′-compact.

So, letU(xn₁) be an open neighborhood ofxn₁ such that

|U(x_n1)^c∩ {y_n:n∈ω}|=ω,

and let

z₁ ∈U(xn₁)^c∩ {yn:n∈ω}.

LetU(xn₂) be an open neighborhood ofxn₂ with

|U(xn₂)^c∩U(xn₁)^c∩ {yn:n∈ω}|=ω, and let

z₂∈U(xn₂)^c∩U(xn₁)^c∩ {yn:n∈ω},

withz26=z1 and inductively, letU(xnk) be an open neighborhood ofxnk with

|U(xn₁)^c∩U(xn₂)^c∩. . .∩U(xnk)^c∩ {y_n:n∈ω}|=ω, and let

z_k∈U(xn₁)^c∩U(xn₂)^c∩. . .∩U(xnk)^c∩ {y_n:n∈ω}, with

z_k∈ {z/ ₁, z₂, . . . , z_k−1}.

The so defined sequence{zn:n∈ω}is a subset ofK and since {zn:n∈ω} ∩[U(x₀)∪[

{U(xnk) :k∈ω}] =∅,

it follows that it has noτ^′-accumulation points inK, contrary to the hypothesis.

Lemma 3.5. Let(X, τ)be aKC-space which is not countably compact. Then X can be condensed onto a weakerKC-topology.

Proof: Letτ^′ be the topology defined by (NT). We will prove that (X, τ^′) is a KC-space.

For this we will show that there is anF ∈ F withF∩K^τ =∅ and the proof will be a consequence of Lemma 3.3.

Indeed, suppose for a contradiction that there isF₀ ∈ F such that F₀ ⊂K^τ. LetF₁,F₂ be subsets ofF₀ with|F₁|=|F₂|=ω,F₁∪F₂=F₀, andF₁∩F₂ =∅.

Suppose thatF₁ ∈ F. We claim thatF₁∪K isτ-compact.

Actually let{U_i :i∈I} be a τ-open cover ofF₁∪K. Then countably many of theU_i^′s, say,{U_in:n∈ω}, cover the countable setF₁, and if we write

U^′(x₀) =U(x₀)∪[

{U_in:n∈ω},

where U(x₀) is a member of {U_i : i ∈ I} which contains x₀ then U^′(x₀) is a τ^′-open neighborhood ofx₀, and we will have

[{U_i:i∈I}=U^′(x0)∪[

{V_j :j∈J},

where {V_j : j ∈J} is a subcollection of {U_i : i∈I} which covers U^′(x₀)^c∩K.

But{Ui :i∈I}is also a τ^′-open cover ofK. So it contains a finite subcover.

It turns out that finitely manyV_j^′s, say,Vj₁,Vj₂,. . .,Vjk, cover the set K∩(U(x₀)∪[

{U_in :n∈ω})^c =K∩U^′(x₀)^c.

Now [

{Vjm:m= 1,2, . . . , k} ∪[

{Uin :n∈ω} ∪U(x₀)

is a countableτ-open cover ofKand in view of Lemma 3.4 it has a finite subcover.

So K∪F₁ is τ-compact and therefore τ-closed. But this is impossible since everyx∈F₂ is aτ-accumulation point ofK.

So there must be anF ∈ F with

F∩K^τ =∅

and Lemma 3.3 implies thatK isτ-closed. Now from Remark 3.1 it follows that

K isτ^′-closed.

The following theorem answers a question posed in [1]. Its proof is an imme- diate consequence of Lemma 3.5.

Theorem 3.6. Every minimalKC-space is countably compact.

References

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Math. Soc.3(1963), 167–177.

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Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece E-mail: [email protected]

(Received November 10, 2003,revised March 1, 2004)