Interior is open

Statement

Let $X$ be a topological space and let $S \subseteq X$. Then the interior $\operatorname{int} S$ is an open set of $X$.

Remarks

Proof: $\operatorname{int} S$ is a union of open sets, and each union of open sets is open by the axioms of a topology. It follows that $\operatorname{int}(\operatorname{int} S) = \operatorname{int} S$.

Explanations

L1 human

The interior is made by putting together open sets, and open sets put together give an open set.