
> 
> **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".



A topology on $\Bbb R$ where the compact sets are precisely the countable sets
There is no such topology.


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.


