Proof of The Product Metric Induces the Product Topology

Throughout, $B_{d_X}$, $B_{d_Y}$ and $B_{d_{X\times Y}}$ denote open balls in the respective metric spaces, and we write $s<t$ for real numbers $s,t$ to mean that $s\le t$ and $s\ne t$.

Step 0: balls of the product metric are products of balls. Let $(x,y)\in X\times Y$ and let $r$ be a real number with $0<r$. For $(x',y')\in X\times Y$, claim 3 of The Product Metric is a Metric says that $d_{X\times Y}((x,y),(x',y'))<r$ holds if and only if $d_X(x,x')<r$ and $d_Y(y,y')<r$. Hence

$$B_{d_{X\times Y}}\bigl((x,y),r\bigr)=B_{d_X}(x,r)\times B_{d_Y}(y,r).$$

Step 1: metric openness implies membership in the product topology. Let $W$ be open in $(X\times Y,d_{X\times Y})$ and let $(x,y)\in W$. By the definition of an open subset of a metric space there is a real number $r>0$ with $B_{d_{X\times Y}}((x,y),r)\subseteq W$. Put $U=B_{d_X}(x,r)$ and $V=B_{d_Y}(y,r)$. By Open Ball in a Metric Space is Open we have $U\in\mathcal{T}_X$ and $V\in\mathcal{T}_Y$, and by Step 0

$$(x,y)\in U\times V=B_{d_{X\times Y}}\bigl((x,y),r\bigr)\subseteq W ,$$

where $(x,y)\in U\times V$ because $d_X(x,x)=0<r$ and $d_Y(y,y)=0<r$ by condition 2 in the definition of a metric. As $(x,y)\in W$ was arbitrary, $W$ belongs to the product topology.

Step 2: membership in the product topology implies metric openness. Let $W$ belong to the product topology and let $(x,y)\in W$. Choose $U\in\mathcal{T}_X$ and $V\in\mathcal{T}_Y$ with $(x,y)\in U\times V\subseteq W$. Since $x\in U$ and $U$ is open in $(X,d_X)$, there is a real number $r_1>0$ with $B_{d_X}(x,r_1)\subseteq U$; likewise there is $r_2>0$ with $B_{d_Y}(y,r_2)\subseteq V$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is a real number $r$ with $r\le r_1$, $r\le r_2$, and $r$ equal to $r_1$ or to $r_2$; in either case $0<r$.

Let $(x',y')\in B_{d_{X\times Y}}((x,y),r)$. By Step 0 we have $d_X(x,x')<r$ and $d_Y(y,y')<r$, so claim 2 of Elementary Order Arithmetic in an Ordered Field gives $d_X(x,x')<r_1$ and $d_Y(y,y')<r_2$, that is $x'\in B_{d_X}(x,r_1)\subseteq U$ and $y'\in B_{d_Y}(y,r_2)\subseteq V$. Hence

$$B_{d_{X\times Y}}\bigl((x,y),r\bigr)\subseteq U\times V\subseteq W .$$

As $(x,y)\in W$ was arbitrary, $W$ is open in $(X\times Y,d_{X\times Y})$.

Steps 1 and 2 give the stated equivalence, and therefore the equality of the two collections of subsets of $X\times Y$.