It seems that the answer is no, but I'm struggling to think of a proper example.
Would I be best trying to think of a closed set whose complement is open?
It seems that the answer is no, but I'm struggling to think of a proper example.
Would I be best trying to think of a closed set whose complement is open?
$ S:=\left\{\frac{1}{n}: n\ge 1\right\} $ is neither closed nor open (in the usual topology of $\mathbf{R}$).
$\mathbb{Q}$ is not closed in $\mathbb{R}$ (with the usual metric)