Open not implies regular open

Statement

It is possible to have an open subset in a topological space that is not a regular open subset, i.e., it is not the interior of its closure.

Proof

Example in the real line

Consider the subset $U=(-1,0)\cup (0,1)$ of the real line, equipped with the Euclidean topology. The closure of $U$ is the closed interval $[-1,1]$, and the interior of this is $(-1,1)$, which is strictly bigger than $U$.

Note that the key reason the subset is not regular open is that it is missing some points that are well on the inside of it, and not just points on the boundary.

In the real line, it is true that any connected open subset is regular open.

Example in the Euclidean plane

We can construct a similar example in the Euclidean plane, this time of a connected open subset that is not regular open. Let $U$ be the punctured open unit disk:

$U=\{(x,y)\mid 0<x^{2}+y^{2}<1\}$

Then, $U$ is an open subset. Its closure is the closed unit disk $\{(x,y)\mid x^{2}+y^{2}\leq 1\}$, and the interior of its closure is the open unit disk $\{(x,y)\mid x^{2}+y^{2}<1\}$. This is strictly bigger than $U$ because it contains the origin.