Interior of Open Set

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $U \subseteq T$ be open in $T$.

Then:
 * $U^\circ = U \iff U \in \tau$

where $U^\circ$ is the interior of $U$.

That is, a subset of $X$ is open in $T$ iff it equals its interior.

Proof
Let $U \subseteq T$ be open in $T$.

From Set Interior is Largest Open Set, if $H$ is an open set such that $H \subseteq U$, then $H \subseteq U^\circ$.

So $U \subseteq U^\circ$.

But as by definition of interior $U^\circ \subseteq U$.

Hence $U^\circ = U$

Now suppose $U^\circ = U$.

By definition $U^\circ$ is the union of all subsets of $U$ which are open in $U$.

By the definition of a topology, that means $U$ itself must also be open.