Closure of Cartesian Product is Product of Closures

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:
 * $\map \cl {H \times K} = \map \cl H \times \map \cl K$

where $\map \cl H$, for example, denotes the closure of $H$.

Proof
Consider the relative complements of $H$ and $K$ in $T_1$ and $T_2$ respectively:


 * $\overline H = \relcomp {S_1} H$
 * $\overline K = \relcomp {S_2} K$

Then from Interior of Cartesian Product is Product of Interiors:
 * $\Int {\overline H \times \overline K} = \Int {\overline H} \times \Int {\overline K}$

From Complement of Interior equals Closure of Complement:


 * $\map \cl H = \Int {\overline H}$
 * $\map \cl K = \Int {\overline K}$
 * $\map \cl {H \times K} = \Int {\overline H \times \overline K}$

Hence the result.