Open Sets of Cartesian Product of Metric Spaces under Chebyshev Distance

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Theorem

Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.

Let $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.

Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:

$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$

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


For $i \in \set {1, 2, \ldots, n}$, let $U_i$ be open in $M_i$.


Then $\ds \prod_{i \mathop = 1}^n U_i$ is open in $M = \struct {\AA, d_\infty}$.


Proof

A set $U$ is open if and only if it is the neighborhood of each of its points.

That is:

$\forall a \in U: \exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq U$

where $\map {B_\delta} a$ denotes the open $\delta$-ball of $a$.


Let $I = \set {1, 2, \ldots, n}$.

For all $i \in I$, let $U_i$ be open in $M_i$.


Then:

\(\ds \forall i \in I: \, \) \(\ds a_i\) \(\in\) \(\ds U_i\)
\(\ds \leadsto \ \ \) \(\ds \map {B_\delta} {a_i}\) \(\subseteq\) \(\ds U_i\)
\(\ds \leadsto \ \ \) \(\ds \forall x_i \in A: \, \) \(\ds \map d {x_i, a_i}\) \(\in\) \(\ds U_i\)
\(\ds \leadsto \ \ \) \(\ds x_i\) \(\in\) \(\ds U_i\)


For all $i \in I$, let $\map d {x_i, a_i} < \delta$.

Then:

\(\ds \max_{i \mathop \in I} \set {\map d {x_i, a_i} }\) \(<\) \(\ds \delta\)
\(\ds \leadstoandfrom \ \ \) \(\ds \family {x_i}\) \(\in\) \(\ds \map {B_\delta} a\) where $a = \family {a_1}$
\(\ds \leadsto \ \ \) \(\ds x_i\) \(\in\) \(\ds U_i\)
\(\ds \leadstoandfrom \ \ \) \(\ds \forall i \in I: \, \) \(\ds \family {x_i}\) \(\in\) \(\ds \prod_{i \mathop \in I} U_i\)
\(\ds \leadsto \ \ \) \(\ds \map {B_\delta} a\) \(\subseteq\) \(\ds \prod_{i \mathop \in I} U_i\)

Hence $\ds \prod_{i \mathop \in I} U_i$ is open in $M$.

$\blacksquare$


Sources

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