Graph of Continuous Mapping between Metric Spaces is Closed in Chebyshev Product

Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $\displaystyle \mathcal A = A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.

Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the Chebyshev distance on $\mathcal A$:
 * $\displaystyle d_\infty \left({x, y}\right) = \max \left\{ {d_1 \left({x_1, y_1}\right), d_2 \left({x_2, y_2}\right)}\right\}$

where $x = \left({x_1, x_2}\right), y = \left({y_1, y_2}\right) \in \mathcal A$.

Let $\Gamma_f$ be the graph of $f$.

Then $\Gamma_f$ is a closed set of $\left({\mathcal A, d_\infty}\right)$.