All sets contain their interior

up:: Interior (Topology)

Let $(X,\tau )$ be a Topological Space, and $S\subseteq X$.

Then, by the definition of Interior (Topology), we know that $\mathring{S}$ is the union of open sets which are contained in $S$.

Since all of them are contained in $S$, their union will also be inside $S$. The “lower bound” of its interior is $\varnothing$, since $\varnothing$ is closed; the “upper bound” is $S$ itself, since all open sets involved are $\subseteq S$; thus, $\varnothing \subseteq \mathring{S}\subseteq S$.

Ad absurdum, suppose $\exists x\in \mathring{S}\mid x\notin S$.
Then $\exists A\subseteq S\mid x\in A⟹x\in S$, absurd.