Let $Y$ be a subset of a locally compact Hausdorff topological space $X$ and consider the following properties.

  1. $\overline{Y}$ is compact.
  2. Every open cover of $X$ has a finite subcover of $Y$.

Certainly 1. implies 2. Does 2. imply 1.?

If $X$ were also second countable, $X$ would metrizable and the answer would be yes.

If $X$ did not have to be locally compact, the answer would be no. To see this, extend the usual subspace topology on the closed unit ball $B$ of $\mathbb{R}^2$ by adding in sets of the form $\{x\}\cup(N\cap B^\circ)$, where $N$ is an open neighbourhood of $x$ in $\mathbb{R}^2$ and $x\in B\setminus B^\circ$. Then $B^\circ$ satisfies 2. but not 1.

Note: In my original question, 1. was instead "every net in $Y$ has a cluster point in $X$". As Nik pointed out, proving 2. implies 1. is then a simple exercise in topology.

  • $\begingroup$ Maybe I'm overlooking something but isn't 2. not just "Y is compact with the subspace topology" ? $\endgroup$ Sep 26, 2014 at 18:38
  • $\begingroup$ @Johannes: no, take $Y = (0,1)$ and $X = \mathbb{R}$. $\endgroup$
    – Nik Weaver
    Sep 26, 2014 at 18:40
  • 1
    $\begingroup$ Dear @Tristan Bice, please revert the question to its original formulation, and start a new thread for your updated question. As it is now, Nik's answer is for a question which no longer exists except in a small footnote. $\endgroup$ Sep 27, 2014 at 22:58
  • 1
    $\begingroup$ I agree that it is good MO form to start new threads rather than change the original question. I do observe that Tristan made a note of the form of the original question at the end, so that the relevance of Nik's answer can still be discerned, but if not now, let's please follow Ricardo's advice in the future. $\endgroup$
    – Todd Trimble
    Sep 28, 2014 at 22:29

1 Answer 1


Tristan pointed out in the comments that my argument showing (2) $\Rightarrow$ (1) in general is faulty. Still, I think it's true for locally compact spaces. Suppose $X$ is locally compact, $Y \subseteq X$, and every open cover of $X$ has a finite subcover of $Y$. Consider the covering of $X$ by all open subsets of whose closure is compact. Since finitely many of them cover $Y$, this implies that $\overline{Y}$ is compact.

My answer to the original question follows.

Let $(x_\alpha)$ be a net in $Y$ and suppose every open cover of $X$ has a finite subcover of $Y$. For each $\alpha$ let $F_\alpha$ be the closure, in $X$, of $\{x_\beta: \beta \geq \alpha\}$, and let $U_\alpha = X \setminus F_\alpha$. If $\{U_\alpha\}$ were an open cover of $X$ then by hypothesis there would be a finite subcover of $Y$, and then by directedness there would be a single $U_\alpha$ containing $Y$, which is absurd. So $\{U_\alpha\}$ cannot be an open cover of $X$, hence there exists $x \in X$ not in any $U_\alpha$, i.e., $x \in F_\alpha$ for all $\alpha$. So $x$ is a cluster point of the net.

  • $\begingroup$ Ah, so my guess was wrong. Great answer - thanks Nik! $\endgroup$ Sep 26, 2014 at 21:17
  • $\begingroup$ Incidentally, the same argument shows that you don't even need $T_1$ to show the equivalence of "every sequence in Y has a cluster point in X" and "every countable open cover of X has a finite subcover of Y" right? Do you still need $T_1$ to show these are equivalent to "every infinite subset of Y has a limit point in X"? $\endgroup$ Sep 26, 2014 at 21:21
  • $\begingroup$ Just one last question - is 1./2. equivalent to relative compactness, i.e. $\overline{Y}$ is compact, in a locally compact Hausdorff space? $\endgroup$ Sep 26, 2014 at 22:43
  • $\begingroup$ Tristan, these exercises are left to the reader! $\endgroup$
    – Nik Weaver
    Sep 26, 2014 at 22:52
  • $\begingroup$ Certainly the answer to my original question was a lot easier than I thought it would be. Sorry if I'm missing something obvious again, but the answer to my last question on locally compact Hausdorff spaces is still not clear to me. Would you care to illuminate on this 'exercise'? Actually this was the question I was really interested in - I was expecting an answer to my original question would also answer this. $\endgroup$ Sep 27, 2014 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.