Chebyshev Distance is Limit of P-Product Metric

Theorem
Let $M_{1'} = \struct {A_{1'}, d_{1'} }$ and $M_{2'} = \struct {A_{2'}, d_{2'} }$ be metric spaces.

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

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

Let $d_p: \AA \times \AA \to \R$ be the $p$-product metric on $\AA$:
 * $\map {d_p} {x, y} := \paren {\paren {\map {d_{1'} } {x_1, y_1} }^p + \paren {\map {d_{2'} } {x_2, y_2} }^p}^{1/p}$

and let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
 * $\ds \map {d_\infty} {x, y} := \max \set {\map {d_{1'} } {x_1, y_1}, \map {d_{2'} } {x_2, y_2} }$

where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.

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

in the sense that:
 * $\ds \max \set {\map {d_{1'} } {x_1, y_1}, \map {d_{2'} } {x_2, y_2} } = \lim_{p \mathop \to \infty} \paren {\paren {\map {d_{1'} } {x_1, y_1} }^p + \paren {\map {d_{2'} } {x_2, y_2} }^p}^{1/p}$

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

Let $a = \map {d_{1'} } {x_1, y_1}, b = \map {d_{2'} } {x_2, y_2}$.

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

Then:

and:

The result follows by the Squeeze Theorem.