P-Product Metric Induces Product Topology

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $M_A = \struct{A, d_A}$ and $M_B = \struct{B, d_B}$ be metric spaces.

Let $\tau_A$ and $\tau_B$ be the topologies on $A$ and $B$ induced by $d_A$ and $d_B$, respectively.


Let $p \ge 1$ be an extended real number.

Let $M = \struct{A \times B, d}$ be the $p$-product of $M_A$ and $M_B$.

We have that $M$ is a metric space.

Let $\tau$ be the topology on $A \times B$ induced by $d$.


Then $\struct{A \times B, \tau}$ is the product space of $\struct{A, \tau_A}$ and $\struct{B, \tau_B}$.


Proof

By $p$-Product Metrics are Lipschitz Equivalent and Lipschitz Equivalent Metrics are Topologically Equivalent, it suffices to consider the case $p = \infty$.


Let $\struct{A \times B, \tau'}$ be the product space of $\struct{A, \tau_A}$ and $\struct{B, \tau_B}$.


By the definition of $d$, it follows that an open ball in $M$ is the (Cartesian) product of an open ball in $M_A$ and an open ball in $M_B$.

Since an open ball is an open set, it follows from Equivalence of Definitions of Analytic Basis that $\tau \subseteq \tau'$.


Let $W \in \tau'$, $\tuple {x, y} \in W$.

By the definition of $\tau'$, it follows from Equivalence of Definitions of Analytic Basis that there exists a $U$ open in $M_A$ and a $V$ open in $M_B$ such that $\tuple {x, y} \in U \times V \subseteq W$.

By the definition of an open set:

There exists a strictly positive real number $\alpha \in \R_{>0}$ such that the open $\alpha$-ball of $x$ in $M_A$ is contained in $U$
There exists a strictly positive real number $\beta \in \R_{>0}$ such that the open $\beta$-ball of $y$ in $M_B$ is contained in $V$

Let $\epsilon = \min \set {\alpha, \beta}$.

Then $\epsilon \in \R_{>0}$, and the open $\epsilon$-ball of $\tuple {x, y}$ in $M$ is contained in $U \times V$.

Hence, $W$ is open in $M$.

That is, $\tau' \subseteq \tau$.


By definition of set equality:

$\tau = \tau'$

as desired.

$\blacksquare$


Sources