Why does a Hausdorff but not countably compact space have an infinite closed discrete subset?

In the proof of Theorem 2.1.5 in

  • E. K. van Douwen, G. M. Reed, A. W. Roscoe, and I. J. Tree, MR 1103993 Star covering properties, Topology Appl. 39 (1991), no. 1, 71--103.

the authors prove that a star compact Hausdorff space is countably compact. At the start of the proof they write

Suppose that $X$ is a Hausdorff space that is not countably compact. Then there exists $D = \{ x_n : n \in \mathbb N \} \subseteq X$, an infinite closed discrete subset.

I do not understand why if $X$ is Hausdorff but not countably compact, then there is an infinite closed discrete $D$.

  • $\begingroup$ I deleted an answer because I realized I had overlooked "closed" so it was at least incomplete, possibly worse (e.g. misleading). I'm pretty sure the way I got a discrete infinite set could give a nonclosed one. $\endgroup$ Commented Feb 9, 2017 at 8:21

1 Answer 1


In Hausdorff (in fact T1) spaces, countable compactness is equivalent to limit point compactness. Recall the definition:

  • A topological space $X$ is limit point compact if every infinite $A \subseteq X$ has a limit point, that is, there is an $x \in X$ such that each open neighborhood of $x$ contains a point of $A$ different from $x$ itself.

(For a proof see the answers to this question.)

If $X$ is a Hausdorff space which is not countably compact, there is an infinite subset $D \subseteq X$ with no limit points. It is easy to show that such a set is closed and discrete:

  • For each $x \in D$, since $x$ is not a limit point of $X$, then there is an open neighborhood $U$ of $x$ which contains so other points of $D$, i.e., $U \cap D = \{ x \}$. So $D$ is discrete.
  • Recall that $\overline{D} = D \cup \{ x \in X : x \text{ is a limit point of } D \}$. Since $D$ has no limit points it follows that $D = \overline{D}$, and so it is closed.

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