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This is an exercise from a topology book.

$X$ is a topological space. If every open cover of $X$ has a locally finite subcover, then $X$ is compact.

What I have tried: I have considered the cover composed of all open subsets of $X$; this has a locally finite subcover. So we can conclude that in this space every point is in a finite number of open sets.

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    $\begingroup$ I don't think what you conclude is accurate; you know that you can find a family of open subsets such that each point is in only finitely many of them and which covers the space, but I don't see how it follows that every point is in only finitely many open sets. For example, a compact space certainly satisfies the hypothesis that every cover has a locally finite subcover, but it need not be true that every point is in only a finite number of open sets; e.g., $[0,1]$ with the usual topology. $\endgroup$ Jan 1, 2012 at 23:29
  • $\begingroup$ @Arturo You are right, it isn't correct. $\endgroup$
    – WLOG
    Jan 1, 2012 at 23:33
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    $\begingroup$ While your argument is wrong, as pointed by Arturo, I still think that it is an interesting implication and hopefully by the time I wake up someone would answer this. $\endgroup$
    – Asaf Karagila
    Jan 1, 2012 at 23:36

3 Answers 3

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Suppose first that $X$ is not countably compact, and let $\mathscr{U}=\{U_n:n\in\omega\}$ be a countable open cover of $X$ with no finite subcover. For $n\in\omega$ let $V_n=\bigcup\limits_{k\le n}U_k$, and let $\mathscr{V}=\{V_n:n\in\omega\}$. Clearly $\mathscr{V}$ has no finite subcover. Suppose that $A\subseteq \omega$ is infinite, and let $m=\min A$; then every point of $V_m$ is in every member of $\{V_n:n\in A\}$, which is therefore not even point-finite, let alone locally finite. It follows that $X$ must be countably compact.

On the other hand, $X$ is obviously paracompact and hence metacompact, and it’s a standard result that every metacompact countably compact space is compact.

Thus, the hypothesis is actually stronger than necessary: it’s sufficient to assume that every open cover has a point-finite subcover.

Added: The proof isn’t very hard. Let $\mathscr{U}$ be an open cover of $X$; by hypothesis $\mathscr{U}$ has a point-finite subcover $\mathscr{V}$. Let $\mathfrak{S}$ be the family of subcovers of $\mathscr{V}$, ordered by $\supseteq$. Suppose that $\mathfrak{C}$ is a chain in $\mathfrak{S}$, and let $\mathscr{C}=\bigcap\mathfrak{C}$; clearly $\mathscr{C}\subseteq\mathscr{V}$. If $\mathscr{C}$ isn’t a cover of $X$, let $x\in X\setminus\bigcup\mathscr{C}$. Let $\mathscr{V}_x=\{V\in\mathscr{V}:x\in V\,\}$; for each $V\in\mathscr{V}_x$ there is a $\mathscr{C}_V\in\mathfrak{C}$ such that $V\notin\mathscr{C}_V$. $\mathscr{V}_x$ is finite, so we can enumerate $\mathscr{V}_x=\{V_1,\dots,V_n\}$ in such a way that $\mathscr{C}_{V_1}\supseteq\dots\supseteq \mathscr{C}_{V_n}$. But then $x\notin\bigcup\mathscr{C}_{V_n}$, which is impossible, since $\mathscr{C}_{V_n}$ is a cover of $X$. Thus, $\mathscr{C}$ does cover $X$, so $\mathscr{C}\in\mathfrak{S}$ and is clearly an upper bound for $\mathfrak{C}$ in $\mathfrak{S}$. $\mathfrak{C}$ was an arbitrary chain in $\mathfrak{S}$, so by Zorn’s lemma $\mathfrak{S}$ has a $\supseteq$-maximal element $\mathscr{W}$. In other words, $\mathscr{W}$ is a subcover of $\mathscr{V}$ with no proper subcover: every member of $\mathscr{W}$ contains at least one point of $X$ that is not in any other member of $\mathscr{W}$. Such a cover is said to be irreducible, and we’ve just shown that every point-finite open cover has an irreducible subcover.

Finally, observe that an irreducible open cover $\mathscr{W}$ of a countably compact space must be finite. If not, let $\mathscr{W}_0=\{W_n:n\in\omega\}$ be an infinite subset of $\mathscr{W}$, and let $W=\bigcup\Big(\mathscr{W}\setminus\mathscr{W}_0\Big)$; then $\mathscr{W}_0\cup\{W\}$ is a countable open cover of $X$ with no finite subcover, since each of the sets $W_n$ contains at least one point not covered by any other member of the cover.

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Revised: Let $X$ be a noncompact topological space. We need to show $X$ has an open cover with no locally finite subcover. In fact we can do slightly better. Let $p \in X$ be fixed and arbitrary. We will produce an open cover $\mathscr{V}$ of $X$ such that every subcover of $\mathscr{V}$ fails to be point finite at $p$.

Claim. There is an open neighbourhood $U$ of $p$ such that $X \setminus U$ is noncompact.

Proof of claim. Suppose this is false and let $\mathscr{U}$ be an open cover of $X$. There is a $U \in \mathscr{U}$ with $p \in U$. By supposition, $X \setminus U$ is compact, so there's a finite family $\mathscr{F} \subset \mathscr{U}$ covering $X \setminus U$. Then, $\mathscr{F} \cup \{U\}$ is a finite subcover of $\mathscr{U}$. Since $\mathscr{U}$ was arbitrary, we contradict our assumption that $X$ is noncompact. QED.

Let $U$ be as in the above claim. Since $X \setminus U$ is noncompact, it has an open cover $\mathscr{V}_0$ with no finite subcover. Let $\mathscr{V} = \{ V \cup U : V \in \mathscr{V} _0\}$ so that $\mathscr{V}$ is an open cover of $X$. Any subcollection of $\mathscr{V}$ which is locally finite at $p$ is finite (since every set in $\mathscr{V}$ contains $p$) and, therefore, fails to cover $X \setminus U$ (since $\mathscr{V}_0$ has no finite subcover).

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  • $\begingroup$ but you have to show that it does not have a locally finite subcover. $\endgroup$ Jan 2, 2012 at 18:08
  • $\begingroup$ @Henno: Mike should make it explicit, but it’s clear that $\mathscr{V}$ doesn’t even have a subcover that’s point-finite at $p$: any subfamily of $\mathscr{V}$ that’s point-finite at $p$ is finite and therefore fails to cover $X\setminus U$. $\endgroup$ Jan 2, 2012 at 20:04
  • $\begingroup$ The argument is basically sound, but you ought to fill in a few more details to make this completely clear. It’s a nice argument that uses less heavy machinery than mine $\endgroup$ Jan 2, 2012 at 20:06
  • $\begingroup$ Thanks @Henno, Brian - I sort of shot from the hip on this one. I've patched it up now. $\endgroup$
    – Mike F
    Jan 3, 2012 at 4:08
  • $\begingroup$ @Mike and Brain: Very preciate your proof, however, it may be a little complex. The answer will be very simple if you notice this fact: For every locally finite open cover of $X$, if we choose a point from every element of such cover, the set of such points is closed discrete. $\endgroup$
    – Paul
    Jul 29, 2012 at 12:36
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I think it is worth converting Mike F's answer into a direct proof.

Given a nonempty topological space $X$ with the property described in the question and any open cover $\mathscr{U}$, we will construct a finite subcover. Pick a point $p\in X$, and pick some $U\in\mathscr{U}$ containing $p$. Then $\mathscr{U}':=\{U\cup V:V\in\mathscr{U}\}$ is an open cover of $X$ with a point-finite subcover $\mathscr{V}'$. Since every element of $\mathscr{U}'$ contains $p$, $\mathscr{V}'$ is finite, so the set of $V$s that give rise to $\mathscr{V}'$ together with $U$ itself forms a finite subcover of $X$.

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  • $\begingroup$ $\mathscr{V}'$ is finite, but that doesn't imply that "the set of Vs that give rise to $\mathscr{V}'$" is finite. You need to choose one $V$ for each distinct $U \cup V$. $\endgroup$ Mar 30, 2021 at 8:58
  • $\begingroup$ @DavidHartley This depends on whether you define an open cover to come with an indexing. If you define an open cover to just be a set of open subsets whose union is the whole space, then you are right, but if an open cover comes with an indexing (and the indexing on my construction is induced by the original indexing) then the set of Vs that give rise to $\mathscr{V}$ is finite. I think the usual convention is that covers come with indexings (to allow repeated items in a cover), but I can't seem to find a reference either way. $\endgroup$ Apr 1, 2021 at 2:39

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