Chebyshev Distance is Limit of P-Product Metric

Theorem
Let $M_{1'} = \left({A_{1'}, d_{1'}}\right)$ and $M_{2'} = \left({A_{2'}, d_{2'}}\right)$ be metric spaces.

Let $\mathcal A = A_{1'} \times A_{2'}$ be the cartesian product of $A_{1'}$ and $A_{2'}$.

Let $p \in \R_{\ge 1}$.

Let $d_p: \mathcal A \times \mathcal A \to \R$ be the $p$-product metric on $\mathcal A$
 * $d_p \left({x, y}\right) := \left({\left({d_{1'} \left({x_1, y_1}\right)}\right)^p + \left({d_{2'} \left({x_2, y_2}\right)}\right)^p}\right)^{1/p}$

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

Then:
 * $\displaystyle d_\infty = \lim_{p \mathop \to \infty} d_p$

in the sense that:
 * $\displaystyle \max \left\{ {d_{1'} \left({x_1, y_1}\right), d_{2'} \left({x_2, y_2}\right)}\right\} = \lim_{p \mathop \to \infty} \left({\left({d_{1'} \left({x_1, y_1}\right)}\right)^p + \left({d_{2'} \left({x_2, y_2}\right)}\right)^p}\right)^{1/p}$

Proof
Let $x$ and $y$ be arbitrary.

Let $a = d_{1'} \left({x_1, y_1}\right), b = d_{2'} \left({x_2, y_2}\right)$.

, suppose that $\max \left\{ {a, b}\right\} = a$.

Then:

and:

The result follows by the Squeeze Theorem.