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. $\endgroup$ – Arturo Magidin Jan 1 '12 at 23:29
  • $\begingroup$ @Arturo You are right, it isn't correct. $\endgroup$ – WLOG Jan 1 '12 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 '12 at 23:36

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. $\endgroup$ – Henno Brandsma Jan 2 '12 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 $X\setminus U$. $\endgroup$ – Brian M. Scott Jan 2 '12 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 $\endgroup$ – Brian M. Scott Jan 2 '12 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 '12 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 '12 at 12:36

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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