Mdthbs

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”!)

Let $X$ be the space of countable ordinals with the order topology (a locally compact Hausdorff space, 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 the only closed set containing $Y$ is the whole space $X$, which is neither Lindelöf nor countably compact, since $Xsetminus Y$ is an uncountable discrete closed subspace.

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, $X=Ycupmathcal A$ where $Y=omega$, and a set $Usubseteq X$ is open if $Asetminus U$ is finite for each $Ain Ucapmathcal A$. All the properties claimed above are easily verified; the fact that every infinite subset of $Y=omega$ has a limit point in $Xsetminus Y=mathcal A$ follows from the maximality of the almost disjoint family $mathcal A$.

Here's a proof this can't happen in a metric space. Suppose $X$ is a metric space and that $Asubseteq X$ is a subset such that every $Bsubseteq A$ 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.

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 $X=Ytimes{0,1}$ where ${0,1}$ has the indiscrete topology, and let $A=Ytimes{0}$. Then no nonempty subset of $A$ is closed in $X$, but $A$ is not contained in any compact subset of $X$.

Let the space S = (-r,r) $cup$ Q where r is an irrational.

Let the set A = (-r,r).

Every closed subset of A is compact.

I dare you to find a compact subset of S that contains A.