[Math] If the interior of a closed set is non empty then the set is the closure of its interior

Question

I see that if the interior were empty this would not be true (unless the set itself were empty) – here for example: https://math.stackexchange.com/q/1229080

I suspect it is not true in all cases for a set with a non-empty interior, but can't think of an example. Any help would be appreciated.

Answer 1

You can just take an example with empty interior and add on a disjoint set which is the closure of an open set. For instance, you can take $\{0\}\cup[1,2]\subset\mathbb{R}$.