Proof of The Product Metric is a Metric

Write $s=d_X(x_1,x_2)$ and $t=d_Y(y_1,y_2)$, so that $d_{X\times Y}(p,q)=\max\{s,t\}$ by the definition of the product metric. Points of the Cartesian product $X\times Y$ are ordered pairs, so $(x_1,y_1)=(x_2,y_2)$ holds if and only if $x_1=x_2$ and $y_1=y_2$. We use the numbering of the conditions in the definition of a metric, the properties of the maximum recorded in Elementary Properties of the Maximum of Two Elements, and the order arithmetic of Elementary Order Arithmetic in an Ordered Field and Elementary Arithmetic in an Ordered Field. We also use repeatedly that if $a\le b$ and $c\le e$ then $a+c\le b+e$; this follows from claims 2 and 3 of Elementary Arithmetic in an Ordered Field together with the associativity and commutativity of addition in the underlying field.

Claim 2. This is claim 1 of Elementary Properties of the Maximum of Two Elements applied to $s$ and $t$.

Claim 3. Suppose $\max\{s,t\}<r$. By claim 2 we have $s\le\max\{s,t\}$ and $t\le\max\{s,t\}$, so claim 2 of Elementary Order Arithmetic in an Ordered Field gives $s<r$ and $t<r$. Conversely, suppose $s<r$ and $t<r$. By claim 2 of Elementary Properties of the Maximum of Two Elements the element $\max\{s,t\}$ equals $s$ or $t$, and in either case $\max\{s,t\}<r$.

Claim 1. We verify the four conditions in the definition of a metric.

Condition 1. Condition 1 for $d_X$ gives $0\le s$, and claim 2 of the present theorem gives $s\le d_{X\times Y}(p,q)$; transitivity of $\le$ gives $0\le d_{X\times Y}(p,q)$.

Condition 2. Suppose $d_{X\times Y}(p,q)=0$. By claim 2 of the present theorem, $s\le 0$ and $t\le 0$; condition 1 for $d_X$ and $d_Y$ gives $0\le s$ and $0\le t$, so antisymmetry gives $s=0$ and $t=0$. Condition 2 for $d_X$ and $d_Y$ then gives $x_1=x_2$ and $y_1=y_2$, that is $p=q$. Conversely, suppose $p=q$. Then $s=d_X(x_1,x_1)=0$ and $t=d_Y(y_1,y_1)=0$ by condition 2 for $d_X$ and $d_Y$, and $\max\{0,0\}=0$ because $0\le 0$ by reflexivity, so $d_{X\times Y}(p,q)=0$.

Condition 3. Condition 3 for $d_X$ and $d_Y$ gives $d_X(x_2,x_1)=s$ and $d_Y(y_2,y_1)=t$, so $d_{X\times Y}(q,p)=\max\{s,t\}=d_{X\times Y}(p,q)$.

Condition 4. Let $w=(x_3,y_3)\in X\times Y$. Condition 4 for $d_X$ gives

$$d_X(x_1,x_3)\le d_X(x_1,x_2)+d_X(x_2,x_3),$$

and claim 2 of the present theorem, applied to the pairs $p,q$ and $q,w$, gives $d_X(x_1,x_2)\le d_{X\times Y}(p,q)$ and $d_X(x_2,x_3)\le d_{X\times Y}(q,w)$. Adding these two inequalities and using transitivity,

$$d_X(x_1,x_3)\le d_{X\times Y}(p,q)+d_{X\times Y}(q,w).$$

The same argument with $d_Y$ gives $d_Y(y_1,y_3)\le d_{X\times Y}(p,q)+d_{X\times Y}(q,w)$. By claim 3 of Elementary Properties of the Maximum of Two Elements,

$$d_{X\times Y}(p,w)=\max\{d_X(x_1,x_3),d_Y(y_1,y_3)\}\le d_{X\times Y}(p,q)+d_{X\times Y}(q,w).$$

Hence $d_{X\times Y}$ satisfies all four conditions and is a metric on $X\times Y$.