# A topology on $\Bbb R$ where the compact sets are precisely the countable sets

QUESTION.

In there a topology on $$\Bbb R$$ where the compact subsets are precisely the countable subsets?

I am trying to create a counterexample to a certain claim, and I found that what I need is a topology of this kind. I thought hard about this and did quite a lot of searching, but could not find something relevant. Thank you in advance.

Note: "countable" includes "finite".

• Just to narrow the search a bit, let me point out that if every countable subset of a space is compact, then the space is either finite or it fails to be Hausdorff. – Will Brian Nov 21 '16 at 20:19
• @WillBrian Because every infinite Hausdorff space contains a countably infinite discrete subspace, right? – Forever Mozart Nov 21 '16 at 20:24
• A decreasing sequence of countable sets may have empty intersection – Pietro Majer Nov 21 '16 at 20:30
• Although, in $\omega$ with the cofinite topology every subset is compact, but clearly there is a decreasing sequence of sets with empty intersection. So it seems that you need to use the fact that $\mathbb R$ is uncountable... – Forever Mozart Nov 21 '16 at 20:41
• Will Brian's argument used (a weak version of) the Axiom of Choice. Is it possible to prove the non-existence in ZF without using AC? – Lajos Soukup Nov 22 '16 at 20:56

Suppose there were. Then $\mathbb R$ itself is not compact, so there is an open cover $\mathcal U$ of $\mathbb R$ with no finite subcover. Using recursion, we can construct a non-compact countable subset of $\mathbb R$.
To begin, let $x_0 \in \mathbb R$ and let $U_0$ be any member of $\mathcal U$ containing $x_0$. Since $\{U_0\}$ does not cover $\mathbb R$, we may choose $x_1 \in \mathbb R - U_0$, and then we may also choose some $U_1 \in \mathcal U$ containing $x_1$. Continuing in this way, suppose $x_0, \dots, x_n$ are already chosen, and so are sets $U_0, \dots, U_n$ from $\mathcal U$, with $x_i \in U_i$ for each $i$. Since $\{U_0, \dots, U_n\}$ does not cover $\mathbb R$, we may choose $x_{n+1} \in \mathbb R - \bigcup_{i \leq n}U_i$, and then we may also choose $U_{n+1}$ from $\mathcal U$ containing $x_{n+1}$.
In the end, we get a countable set $\{x_0,x_1,x_2,\dots\}$ and a countable open cover $\{U_0,U_1,U_2,\dots\}$ of this set. But it is clear from our choices of the $x_i$ and the $U_i$ that this open cover has no finite subcover.