# 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 $S$ is open in $T$ if and only if it equals its interior.

## Proof

### Necessary Condition

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

$U \subseteq U^\circ$

But by definition of interior:

$U^\circ \subseteq U$

Hence $U^\circ = U$.

$\Box$

### Sufficient Condition

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.

$\blacksquare$