If all closed subsets of a set are compact, does it follow that this set is subset of a compact set?

It is well known that closed subsets of compact sets are themselves compact. Now the reverse is not true: A set of which all closed subsets are compact needs not to be compact itself; for example, consider non-closed bounded sets in $$\mathbb R^n$$.

However those sets are themselves subsets of compact sets (as bounded sets, they are subsets of closed balls, which are compact). And it is obvious that the initially quoted theorem also holds for arbitrary subsets of compact sets, since the subset relation is transitive.

However I wonder: Can there exist a set in some topological space, no matter how weird, such that all closed subsets of that set are compact, but the set itself is not the subset of a compact set?

There was a related question that asked about the case where all proper closed subsets of a topological space are compact, and the conclusion was that the space itself is compact. However if this helps with the subset case, then I don't see how.

Clarification: Since it seems to have caused a lot of confusion in the comments: In the context of my post, “closed” is to be understood in the topology of the full space, not in the subspace topology of the subset (those are very different notions of “closed”!)

• A set is one of its closed subset. So… – Bernard Aug 20 '19 at 10:12
• @Bernard: I don't understand. An open ball certainly is not a closed subset of itself, is it? – celtschk Aug 20 '19 at 10:14
• Yes it is: $X$ is always closed in $X$. – Mindlack Aug 20 '19 at 10:16
• Any space is contained in its one-point compactification. – Kavi Rama Murthy Aug 20 '19 at 10:17
• @KaviRamaMurthy: But the closed sets of the one-point compactification are not the same as the closed sets of the original space. – celtschk Aug 20 '19 at 10:21

Let $$X$$ be the space of countable ordinals with the order topology (a locally compact Hausdorff space, completely normal, but not metrizable), and let $$Y$$ be the set of all isolated points of $$X$$.

Every subset of $$Y$$ which is closed in $$X$$ is finite, since every infinite subset of $$X$$ has a limit point in $$X$$. (An infinite set of ordinals contains an increasing sequence; the limit of an increasing sequence of countable ordinals is a countable ordinal, i.e., an element of $$X$$.

$$Y$$ is not contained in any compact subset of $$X$$ because no uncountable subset of $$X$$ is compact or even Lindelöf.

P.S. Here is another example, a first countable, separable, locally compact Hausdorff space $$X$$ with a dense open subset $$Y$$ such that: $$Y$$ is countable and discrete; the only subsets of $$Y$$ which are closed in $$X$$ are the finite sets; and $$Y$$ is not contained in any countably compact subset of $$X$$.

Let $$\mathcal A$$ be an infinite maximal almost disjoint family of infinite subsets of $$\omega$$. $$\mathcal A$$ must be uncountable, as there is no maximal almost disjoint family of cardinality $$\aleph_0$$. Let $$X$$ be the corresponding $$\Psi$$-space, that is, $$X=Y\cup\mathcal A$$ where $$Y=\omega$$, and a set $$U\subseteq X$$ is open if $$A\setminus U$$ is finite for each $$A\in U\cap\mathcal A$$. All the properties claimed above are easily verified; the fact that every infinite subset of $$Y=\omega$$ has a limit point in $$X\setminus Y=\mathcal A$$ follows from the maximality of the almost disjoint family $$\mathcal A$$.

Unlike the previous example, this space $$X$$ is not normal; if $$\mathcal A_0$$ is a countably infinite subset of $$\mathcal A$$, then $$\mathcal A_0$$ and $$\mathcal A\setminus\mathcal A_0$$ are disjoint closed sets which can not be separated by open sets.

• hmm, why does every infinite subset of $X$ contain a limit point in $X$? – Cronus Aug 20 '19 at 11:08
• Probably a good idea. And thanks again :) I'm trying to come up with a metric example/a proof it can't happen. I think that perhaps one can prove such a set $A$ is totally bounded and hence so is its closure, which means its closure is incomplete, which means one can find a Cauchy sequence in $A$ which is not convergent, so its set of points is closed in the whole space but not compact. – Cronus Aug 20 '19 at 11:23
• @Cronus: that’s pretty much it for your proof if $S$ is metric and $A$ is as required. Indeed, $\overline{A}$ can’t be compact, so there is a sequence $a_n \in \overline{A}$ without a limit point. Take $b_n \in A$ at distance $<1/n$ of $a_n$ and consider $B=\{b_n,\,n\}$. $B \subset A$ is closed in $S$ and is not compact. – Mindlack Aug 20 '19 at 11:33
• @mindlack yes, I just wanted to work out the details. Perhaps I overcomplicated things a bit. – Cronus Aug 20 '19 at 11:35
• @bof Do you think it's possible to find a separable (Hausdorff) example, or even second countable? – Cronus Aug 21 '19 at 10:03

Here's a proof this can't happen in a metric space. Suppose $$X$$ is a metric space and that $$A\subseteq X$$ is a subset such that every $$B\subseteq A$$ which is closed in $$X$$ is compact.

Lemma. $$A$$ is totally bounded. proof. Assume it is not. Then there is $$\varepsilon>0$$ such that no finite collection of $$\varepsilon$$-balls cover $$X$$. Therefore we can define recursively a sequence of points $$a_1,a_2,...$$ each two of which are at distance at least $$\varepsilon$$ from one another. This is a subset of $$A$$ which is closed (if $$x$$ is in its closure then by taking an $$\varepsilon/2$$ neighbourhood of it we see we must have $$x=a_n$$ for some $$n$$) but clearly not compact (it's discrete and infinite).

Corollary. The closure of $$A$$ is totally bounded as well.

By assumption the closure of $$A$$ is not compact. Therefore, it's not complete, so it contains a Cauchy sequence which is not convergent. Therefore $$A$$ contains such a sequence as well. The set of points of this sequence is closed, but not compact. Contradiction.

• Or, more directly, given a sequence in $\overline{A}$ with no convergent subsequence, you can pick a sequence in $A$ that approximates it in the limit and so has the same convergence subsequences. So that sequence in $A$ forms a closed set which is not compact. – Eric Wofsey Aug 21 '19 at 4:42
• @EricWofsey Yeah, this was pointed out to me already. When I was trying to come up with a proof it just seemed to me like a good idea to first try to prove total boundedness (since I think this is essentially what's going on here), and then when I was done I didn't realise it wasn't really necessary. – Cronus Aug 21 '19 at 6:58
• @DanielWainfleet No, I just showed that the closure of $A$ contains such a sequence and then deduced $A$ itself does too. – Cronus Aug 21 '19 at 10:00

If you don't mind abandoning all separation axioms, then it's really easy to find an example, because you can easily make there be very few closed subsets of your set. For instance, let $$Y$$ be any non-compact topological space, let $$X=Y\times\{0,1\}$$ where $$\{0,1\}$$ has the indiscrete topology, and let $$A=Y\times\{0\}$$. Then no nonempty subset of $$A$$ is closed in $$X$$, but $$A$$ is not contained in any compact subset of $$X$$.