# how to prove that countably compact D-space is compact

In the paper A survey of D-spaces by Gary Gruenhage it is written that it is easily seen that any countably compact D-space is compact. However I'm not able to show it. Here is my attempt to prove this claim: Let $(X, \tau)$ be a countably compact D-space with topology $\tau$. Consider a function $N: X \rightarrow \tau$ with $x \in N(x)$ for every $x \in X$ (neighborhood assignment function). I think that to prove compactness of $X$ it is enough to show that open cover $\{N(x):x \in X\}$ of $X$ has finite subcover. Because $X$ is a D-space, there is a closed discrete subset $D$ of $X$ such that $\{N(x):x \in D\}$ covers $X$. It seem to me that to complete the proof (using countable compactness of $X$) I need to use the fact that $D$ is countable. However it is not true in general that close discrete set is countable. Thank you for your help! Michal Čihák

Note 1. The considered spaces are regular and $T_1$.

Note 2. A space $X$ is a D-space if whenever one is given a neighborhood $N(x)$ of $x$ for each $x \in X$, then there is a closed discrete subset $D$ of $X$ such that $\{N(x):x \in D\}$ covers X.

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This looks like a duplicate of this question... is anyone for or against closing? – Zev Chonoles Mar 30 '12 at 19:00
@Zev: of course, they are related and the ideas are similar, but $\sigma$-compactness and countable compactness are quite distinct properties, so I'm against closing this question. – t.b. Mar 30 '12 at 19:17
@t.b.: Ah, I didn't read closely enough - nor did the person who flagged the question :) Thanks for pointing out the difference. – Zev Chonoles Mar 30 '12 at 19:19
@ZevChonoles Also, the implication goes the other direction in the linked question. – Alex Becker Mar 30 '12 at 20:47

Suppose that your set $D$ is infinite. Pick a countably infinite subset of it, say $D_0$. For each $x\in D_0$ let $V(x)$ be a nbhd of $x$ such that $V(x)\cap D_0=\{x\}$; you can do this because $D_0$ is discrete. Moreover, $D_0$ is closed, so $X\setminus D_0$ is open, and $\{X\setminus D_0\}\cup\{V(x):x\in D_0\}$ is then a countable open cover of $X$ that clearly has no finite subcover: each $x\in D_0$ is covered only by the set $V(x)$. This contradicts the countable compactness of $X$, showing that in fact $D$ must be finite.

You are correct that it’s sufficient to look at open covers of the form $\{N(x):x\in X\}$ for nbhd assignments $N$. To see this, let $\mathscr{U}$ be any open cover of $X$. Then for each $x\in X$ let $N(x)$ be any element of $\mathscr{U}$ that contains $x$. Clearly $\{N(x):x\in X\}$ is a subcover of $\mathscr{U}$, so any finite subcover of $\{N(x):x\in X\}$ will automatically be a finite subcover of the original $\mathscr{U}$ as well.

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Thank you very much, your proof is very clear for me. – mcihak Mar 31 '12 at 6:58

It follows from the following characterization: a space $X$ countably compact if and only if every infinite subset of $X$ has an accumulation point. So in particular any discrete subspace is finite. By the way the characterization makes for a nice exercise.

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Note that any countable compact space $X$ is limit point compact. Now let $D \subset X$ be a closed discrete subset of $X$. Then $D$ does not have any limit points, since such a limit point would have to be in $X$, but $D$ is closed. Finally since $X$ is limit point compact it follows that $D$ is finite.

Let $\mathcal U$ be an open cover of $X$. Then for each $x \in X$ pick $U \in \mathcal U$ such that $x \in U$ and set $N(x)=U$. Now if $D$ is a closed discrete subspace of $X$ such that for $x \in D$ we have $N(X)$ covers $D$, then by the previous argument $D$ is finite, so this is a finite subcover of $\mathcal U$.

I am curious though if you can manage to do this without using the axiom of choice, since I seem to be invoking it in order to create the neighborhood function.

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We can do even better than making your set $D$ countable; $D$ is finite! Suppose your set $D$ is infinite, and choose some (necessarily closed) countably infinite discrete subset $D'$ of $D$. Define a neighborhood assignment function $N':D'\to \tau$ such that $x\neq y\implies N'(x)\cap N'(y)\neq \emptyset$, which is possible because $D'$ is discrete. Let $\mathcal O = \{N'(x)\cup (X\setminus D'): x\in D'\}$ and note that $\mathcal O$ is a countable open cover of $X$. Yet if we remove any open set $N'(x)\cup (X\setminus D')$ from $\mathcal O$, then $x\notin \bigcup \mathcal O$. Hence by countable compactness, $D'$ is finite, contradicting our choice of $D'$. Thus $D$ must be finite.

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