P-Product Metric Induces Product Topology

Theorem
Let $M_A = \left({A, d_A}\right)$ and $M_B = \left({B, d_B}\right)$ 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 $M = \left({A \times B, d}\right)$ 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 $\left({A \times B, \tau}\right)$ is the topological product of $\left({A, \tau_A}\right)$ and $\left({B, \tau_B}\right)$.

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

Let $\left({A \times B, \tau'}\right)$ be the topological product of $\left({A, \tau_A}\right)$ and $\left({B, \tau_B}\right)$.

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 $W \in \tau'$, $\left({x, y}\right) \in W$.

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 $\left({x, y}\right) \in U \times V \subseteq W$.

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 $\epsilon = \min {\left\{{\alpha, \beta}\right\}}$.

Then $\epsilon \in \R_{>0}$, and the open $\epsilon$-ball of $\left({x, y}\right)$ 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.