Let $F$ be a non-empty countable subset of $\mathbb{R}$. $F$ may or may not be closed in usual topology depending on the choice of $F$

Question

I believe I need to construct examples for both cases.

$\mathbb{Q}$ is a countable subset of $\mathbb{R}$ and $\mathbb{Q}$ is not closed in $\mathbb{R}$ since $\mathbb{R}-\mathbb{Q}$ is not open in $\mathbb{R}$.

The set of all primes $\mathbb{P}$ is a countable subset of $\mathbb{R}$ which is closed in $\mathbb{R}$.

Are my examples correct?

Answer 1

Yes, they are correct.

Another example:

$\left\{\frac1n : n \in \mathbb{N}\right\}$ is not closed in $\mathbb{R}$, but $\left\{\frac1n : n \in \mathbb{N}\right\} \cup \{0\}$ is closed in $\mathbb{R}$.