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If every closed and proper subset of a topological space is compact, then is the whole space necessarily compact?

The "converse" of this question is well-known, of course, but I'm having difficulty establishing a proof of this. Also, no counterexamples spring to mind either.

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Wouldn't the whole space be compact by the simple fact that, trivially the whole space is a subset of itself (assuming that the whole space is closed)? –  rocinante Feb 23 at 16:02
    
@rocinante OP specified proper subspaces. –  Eric Auld Feb 23 at 16:02
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Note that it would suffice to find two proper closed subsets (not necessarily disjoint) whose union is the whole space. –  Eric Auld Feb 23 at 16:04
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If $\cal O$ is a cover of the space and $O\in\cal O$, then $\cal O$ is a cover of the closed set $O^C$. –  David Mitra Feb 23 at 16:05
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Yes. Choose $O$ so that $O^C$ is proper (if no such $O$ exists, any $O$ covers $X$). Then finitely many elements of $\cal O$ cover $O^C$. Toss in $O$ and you have a cover of $X$. –  David Mitra Feb 23 at 16:15

2 Answers 2

up vote 4 down vote accepted

An equivalent definition of compactness is the following:

A space $X$ is compact if and only if every family of closed subsets of $X$ with the finite intersection property has non-empty intersection.

We say that a family $\mathcal F$ of sets has the finite intersection property if $F_1\cap\cdots\cap F_n\ne\varnothing$, for every $n$ and $F_1,\ldots,F_n\in\mathcal F$. (See also here.)

Assume now that $\mathcal C$ is a collection of closed subsets of $X$ with the finite intersection property, and $F\in\mathcal C$, with $F\ne X$. It is already given that $F$ is compact. Then clearly the family $$ \tilde{\mathcal C}=\big\{F\cap C: C\in\mathcal C\big\}, $$ is another family of closed subsets of $X$ with the finite intersection property, and as they are also closed subsets of $F$, which is assumed compact, the $\tilde{\mathcal C}$ has non-empty intersection, and so does ${\mathcal C}$.

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My suggestion:

Let $T$ be the topological space. Let $U_{i}\neq\emptyset, i\in I$ be an open cover.

Let $i_{0}\in I$ and consider $T^{'}:=T\setminus U_{i_{0}}$. $T^{'}$ is a proper subspace and thus there is a finite subset $I_{F}\subset I$ such that $\left\{U_{i}\right\}_{i\in I_{F}}$ is an open cover for $T^{'}$ but this means that $\left\{U_{i}\right\}_{i\in I_{F}\cup\left\{i_{0}\right\}}$ is a finite open cover for $T$.

Editet it twice, in this version I don't need Zorn's lemma. Check if I forgot something plx :-)

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"w.l.o.g. such that you cannot remove any set without losing the covering property" - Can you explain why this is possible? –  Yiorgos S. Smyrlis Feb 23 at 16:11
    
hah, already done before you asked! I knew someone would state that :-) –  Max Feb 23 at 16:13
    
if you still don't believe me say so. –  Max Feb 23 at 16:20
    
It is true, but not trivial. You need Zorn's Lemma. –  Yiorgos S. Smyrlis Feb 23 at 16:37
    
interesting, thanks! Puzzling with that one is always fun :-) –  Max Feb 23 at 16:41

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