Is $\mathbb{R}$ with $\tau_{CF}$ and $\tau_{CN}$ locally compact?

Let $\mathbb{R}$ be the set of real numbers and $\tau_{CF}$ and $\tau_{CN}$ the cofinite topology and the countable complement topology respectively. Study if any if the topologies above make $\mathbb{R}$ locally compact.

In $\tau_{CF}$, all subsets of $\mathbb{R}$ are compact, so let $x\in \mathbb{R}$ and take $A= \mathbb{R}\setminus \{a_1, \dots , a_s\}$ such that $x \not \in \{a_1, \dots , a_s\}$. Then $A$ is an open neighbourhood of $x$ which is compact and hence $(\mathbb{R},\tau_{CF})$ is locally compact.

In $\tau_{CN}$ I'm not sure how to prove it, since I lack some knowledge about countable sets, but here is my try. Take $U$ some neighbourhood of $x\in \mathbb{R}$, then, $\mathbb{R}\setminus U$ is countable. Assume that there is some compact subset of $\mathbb{R}$, $C$, such that $x\in U \subset C$. Then $$\mathbb{R}\setminus C \subset \mathbb{R}\setminus U$$ But on the other hand, the only compact sets of $\mathbb{R}$ with the countable complement topology are finite sets, so $C$ is finite, and since $\mathbb{R}$ is uncountable, $\mathbb{R}\setminus C$ is uncountable, and hence so is $\mathbb{R}\setminus U$, which contradicts the hypothesis that $U$ is an open neighbourhood of $x$. Therefore, there is no compact subset of $(\mathbb{R},\tau_{CN})$ containing an open neighbourhood and thus, $(\mathbb{R},\tau_{CN})$ isn't locally compact.

Is this proof right? If not, I'd like some help to prove it correctly. Thanks in advance!

• To me it's fine. The same in different arrangement: Assume $C$ is a compact neighbourhood of $x$. Then $C$ is finite, Hence any open subset of $C$ is also finite and so cannot be cocountable. Hence the only open subset of $C$ is $\emptyset$ and $C$ fails to be a neighbourhood of $x$. – Hagen von Eitzen Aug 11 '16 at 10:40
• @HagenvonEitzen Thanks! That's precisely what I was unsure of, whether the complement of a inite subset of an uncountable set was uncountable. Intuitively it seems so, but I didn't have a formal proof of that. – user313212 Aug 11 '16 at 11:21