Interior of Cartesian Product is Product of Interiors

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.

Let $H \subseteq T_1$ and $K \subseteq T_2$.

Then:
 * $\Int {H \times K} = \Int H \times \Int K$

where $\Int H$, for example, denotes the interior of $H$.

Proof
By definition of interior, both $\Int H$ and $\Int K$ are open in $T_1$ and $T_2$ respectively.

By Projection from Product Topology is Continuous, it follows that $\Int {H \times K}$ is an open set of $T$.

It remains to be shown that $\Int {H \times K}$ is the largest open subset of $H \times K$.

Let $H' \times K'$ be an open set of $T$ such that $H' \times K' \subseteq H \times K$.

Then from Projection from Product Topology is Open, $H'$ and $K'$ are open sets of $T_1$ and $T_2$ respectively such that $H' \subseteq H$ and $K' \subseteq K$.

By definition of interior, $\Int H$ and $\Int K$ are the largest open sets of $T_1$ and $T_2$ respectively such that $\Int H \subseteq H$ and $\Int K \subseteq K$.

Hence $H' \subseteq \Int H$ and $K' \subseteq \Int K$.

Hence by Cartesian Product of Subsets:
 * $H' \times K' \subseteq \Int H \times \Int K$

That is, any open subset of $H \times K$ is also a subset of $\Int {H \times K}$.

Hence $\Int {H \times K}$ is the largest open subset of $H \times K$.