Let $X$ be a topological space. It is a well-known result that, if $X$ is compact, then every proper closed subset $Y \subset X$ is compact. Out of curiosity, I would like to explore the converse of this statement:

Let $X$ be a topological space with the property that every proper closed subset is compact. Is $X$ itself compact?

Of course, this question is only interesting when $X$ is infinite. We can further restrict ourselves to spaces that have infinitely many open sets since every closed set, and indeed the space itself, will be compact otherwise. If this result is false in general, what hypotheses can be added to make it so?

Should this not hold, I'm interested in finding concrete examples of when it does. For instance, we can take any infinite set endowed with the cofinite topology, which works because every closed set is finite and thus compact, and the space itself would be compact. Another example is an interval $[a, b] \subset \mathbb{R}$, wherein closed subsets of this are closed and bounded in $\mathbb{R}$; as such, they are compact per Heine-Borel.

Update: Zardo's answer reveals that such a topological space is indeed compact. For further discussion on the finite intersection property and its relevance to compactness, see Munkres' Topology: Chapter $3$, theorem $26.9$.

  • $\begingroup$ Any Disconnected topological space will work just fine. The proof in this particular case is pretty obviuos aswell. $\endgroup$ – TheOscillator Jun 8 '15 at 23:07
  • $\begingroup$ I wonder if you can write a proof by noting that it's trivial if there are two disjoint, non-empty sets (since then the space is the union of two compact sets - the complements of those open sets). The condition of "all pairs of open sets intersect" is fairly restrictive - so the proof in that case might be interesting. (This would decidedly a novel proof - Zardo's proof below seems a lot more reasonable) $\endgroup$ – Milo Brandt Jun 9 '15 at 0:22
  • $\begingroup$ @Meelo My guess is that irreducibility isn't restrictive enough of a condition to lead to a different proof. For example, spaces with this property include all open subspaces of the spectra of integral domains. See the addition I just made to my post for an illustration of this case, which I think already possesses the subtlety of the general case. $\endgroup$ – Slade Jun 9 '15 at 1:24

Assume that $X$ is not compact. Then there is a family $\{A_i\}_i$ of closed subsets with the finite intersection property such that $\bigcap_i A_i = \emptyset$. Thus, there exists an $j$ with $A_j \neq X$. Then $A_j$ is a proper closed subset of $X$ but it is not compact, since the family $\{A_i \cap A_j \}_i$ has the finite intersection property and $\bigcap (A_j \cap A_i) = \emptyset$.

  • $\begingroup$ That works! Awesome! $\endgroup$ – Kaj Hansen Jun 8 '15 at 22:52
  • 21
    $\begingroup$ The same proof, translated into open sets: If $X$ has an open cover $\mathcal{C}$, take any nonempty $U\in\mathcal{C}$. $X\setminus U$ is compact by assumption, so a finite subset of $\mathcal{C}$ covers $X\setminus U$, and adding $U$ gives a finite subset covering $X$. $\endgroup$ – Slade Jun 9 '15 at 1:26

I think that the most interesting examples of this are infinite topological spaces $X$ such that every proper closed subset is finite. Such a space trivially has the property you mention, but is compact for less trivial reasons.

One nice example is $X=\mathbb{N}$, with the nonempty open sets $U_n = \{k \mid k\geq n\}$.

Note that this space is an inverse limit of finite topological spaces. In general, an inverse limit of finite topological spaces is compact.

Here is another nice example! Let $R$ be a commutative ring with unity, and $I\subset R$ an ideal. For simplicity, assume that $R$ is reduced. Consider the topological space $X = \operatorname{Spec} R \setminus V(I)$ of prime ideals of $R$ not containing $I$.

Suppose that every proper closed subspace of $X$ is compact. This is equivalent to the condition that $V(J)\setminus V(I)$ is compact for each ideal $J$ with $IJ\neq 0$.

Unraveling this, our condition is that $\overline{I}$ is finitely generated in $R/J$ for all ideals $J$ with $IJ\neq 0$. And the conclusion is that $I$ is itself finitely generated.

Indeed, we can take any nonzero $x\in I$. Then $I\cdot(x)\neq 0$, and if $\overline{I}$ is finitely generated in $R/(x)$, we can lift generators to $R$, and throw in $x$ to get a set of generators for $I$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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