Spaces where compact sets are closed under countable union

Question

Are there non-compact spaces such that a countable union of compact subspaces is always compact?

If so, are there any non-compact spaces (necessarily uncountable) such that a subset is countable iff it is compact?

Answer 1

No, every non-compact spact has a non-compact countable subspace. (Of course a countable space is a countable union of compact sets.)

Suppose $X$ is not compact, and let $\mathcal V$ be an open cover of $X$ having no finite subcover. Choose a sequence $x_0,V_0,x_1,V_1,\dots,x_n,V_n,\dots$ as follows. Having chosen $x_1,V_1,\dots,x_{n-1},V_{n-1}$, choose $x_n\in X\setminus(V_0\cup\cdots\cup V_{n-1})$ (possible since $\{V_0,\dots,V_{n-1}\}$ is not a cover of $X$) and then choose $V_n\in\mathcal V$ so that $x_n\in V_n$ (possible since $\mathcal V$ is a cover of $X$).

Now $\{x_0,x_1,\dots\}$ is a countable subspace of $X$ which is not compact, because $\{V_0,V_1,\dots\}$ is an open cover with no finite subcover. Since $\{x_0,x_1,\dots\}$ is a countable union of compact sets, this answers the first question in the negative.

Second question: Is there any space $X$ in which a subset is compact iff it is countable?

Such a space is necessarily countable: if $X$ is uncountable, either $X$ is compact (so it's an uncountable compact subset of itself), or else $X$ is non-compact, and then it has a countable non-compact subset, as shown above.

On the other hand, there are countable spaces in which every subset is compact. Any finite space has this property; more generally, any space in which every nonempty open set is cofinite; or any finite topological sum of such spaces.

Answer 2

"Are there non-compact spaces such that a countable union of compact subspaces is always compact?" I don't have a proof, but it seems unlikely. In any case, it doesn't seem "natural" to ask about the union being compact, seems more sensible to ask about the closure of the union.

Let $X=\omega_1$, the set of countable ordinals with the order topology. Then $S\subset X$ is countable if and only if $\overline S$ is compact, answering a revised version of the second question. Also any compact subset is countable and hence the closure of a countable union of compact subsets is compact.