Set Interior is Largest Open Set

Theorem

Let $\left({T, \vartheta}\right)$ be a topological space.

Let $H \subseteq T$.

Let $H^\circ$ be the interior of $H$.

Then $H^\circ$ can alternatively be defined as the largest open set contained in $H$, in the sense that:

If $K$ is an open set such that $K \subseteq H$, then $K \subseteq H^\circ$.

Proof

Let $\mathbb K$ be defined as:

$\mathbb K := \left\{{K \in \vartheta: K \subseteq H}\right\}$

That is, let $\mathbb K$ be the set of all open sets of $T$ contained in $H$.

Then from the definition of the interior of $H$, we have:

$\displaystyle H^\circ = \bigcup_{K \in \mathbb K} K$

That is, $H^\circ$ is the union of all the open sets of $T$ contained in $H$.

• Let $K \subseteq T$ such that $K$ is open in $T$ and $K \subseteq H$.

That is, let $K \in \mathbb K$.

Then from Subset of Union it follows directly that $K \subseteq H^\circ$.

So any open set in $T$ contained in $H$ is a subset of $H^\circ$, and so $H^\circ$ is the largest open set of $T$ contained in $H$.

$\Box$

Let $U$ be the largest open set of $T$ contained in $H$.

Let $K$ be open in $T$ such that $K \subseteq H$.

Then by definition $K \subseteq U$.

It follows from Union Smallest: General Result that $U$ is the union of all open sets of $T$ contained in $H$.

That is:

$\displaystyle U = \bigcup_{K \in \mathbb K} K$

where $\mathbb K := \left\{{K \in \vartheta: K \subseteq H}\right\}$.

From the definition of the interior of $H$ it follows that $U = H^\circ$.

$\Box$

The implication has been demonstrated to hold in both directions, so the interior of $H$ can be defined as the largest open set of $T$ contained in $H$.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 1$: Closures and Interiors