Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If every closed and proper subset of a topological space $X$ 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.

share|cite|improve this question
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 '14 at 16:02
@rocinante OP specified proper subspaces. – Eric Auld Feb 23 '14 at 16:02
Note that it would suffice to find two proper closed subsets (not necessarily disjoint) whose union is the whole space. – Eric Auld Feb 23 '14 at 16:04
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 '14 at 16:05
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 '14 at 16:15
up vote 6 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 family 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 family $\tilde{\mathcal C}$ has non-empty intersection, and so does family ${\mathcal C}$.

share|cite|improve this answer

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 :-)

share|cite|improve this answer
"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 '14 at 16:11
hah, already done before you asked! I knew someone would state that :-) – Max Feb 23 '14 at 16:13
if you still don't believe me say so. – Max Feb 23 '14 at 16:20
It is true, but not trivial. You need Zorn's Lemma. – Yiorgos S. Smyrlis Feb 23 '14 at 16:37
interesting, thanks! Puzzling with that one is always fun :-) – Max Feb 23 '14 at 16:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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