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There is an open problem in this paper: Classes defined by stars and neighbourhood assignments by van Mill and others.

Problem 4.8. Is a regular (Tychonoff) star compact space metrizable if it has a $G_\delta$-diagonal?

A topological space $X$ is said to be star compact if whenever $\mathscr{U}$ is an open cover of $X$, there is a compact subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$. More on star compactness see here.

My question is this: Is always the cardinality of such regular (Tychonoff) star compact space less than $2^{\omega_0}$? See the related link here.

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Do you mean less-than-or-equal-to, rather than less-than? –  Joel David Hamkins Feb 6 '13 at 2:22
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Please don't write $2^{\omega_0}$ for cardinality. This is very misleading as $\omega_0$ usually denotes the first infinite ordinal, and hence $2^{\omega_0}$ would be an ordinal. It is better to write $2^{\aleph_0}$. –  Andrej Bauer Feb 6 '13 at 9:25
    
@Joel David Hamkins: Yes, I mean that less than or equal to. –  Paul Feb 6 '13 at 10:42
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@Andrej: For me it is a pain to write $\aleph$ in a blackboard, so I often use $\omega$, $\omega_1$ and $2^\omega$ for cardinalities; I'm almost certain I never confused anyone by doing so. Also the "cardinals=initial ordinals" disclaimer is frequent in papers related with set theory. –  Ramiro de la Vega Feb 7 '13 at 8:57

1 Answer 1

We shall show that every metrizable space of cardinality greater than $2^{\aleph_{0}}$ is not star-compact. Assume that $(X,d)$ is a metric space with $|X|>2^{\aleph_{0}}$. Color the collection of pairs $\{\{x,y\}|x,y\in X,x\neq y\}$ where $\{x,y\}$ has color $n$ for some natural number $n$ with $d(x,y)>\frac{1}{n}$. Then by the Erdos-Rado theorem there is a uncountable subset $A$ and some $n$ such that if $a,b\in A,a\neq b$, then $\{a,b\}$ has color $n$ (i.e. $A$ is homogeneous). Therefore $d(a,b)>\frac{1}{n}$ for distinct $a,b\in A$. We may therefore assume that $d(a,b)>8$ for $a,b\in A,a\neq b$. Now consider the open cover $\mathcal{U}=\{B_{2}(a)|a\in A\}\cup(\bigcup_{a\in A}\overline{B_{1}(a)})^{c}$. Let $K\subseteq X$ be compact. Then since $K$ is totally bounded, the set $K$ can intersect only finitely many balls $B_{2}(a)$ for $a\in A$. Therefore $St(K,\mathcal{U})$ can only contain finitely many points in $A$, so $St(K,\mathcal{U})\neq X$. Therefore $X$ is not star-compact.

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Then can you rewrite the question so it is clear what you are asking? –  Joseph Van Name Feb 6 '13 at 14:52
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@Joseph: Since any star-compact metrizable space must have size at most $2^{\aleph_0}$ (as you showed in your answer), a way to relax van Mill´s question is to ask: "Is the cardinality of a regular (Tychonoff) star-compact space at most $2^{\aleph_0}$ if it has a $G_\delta$ diagonal?". And I think that is Peter´s question. –  Ramiro de la Vega Feb 6 '13 at 17:30
    
Anyway, +1 for the Erdos-Rado argument. –  Ramiro de la Vega Feb 6 '13 at 17:54

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