Proof of The Product Topology is a Topology

Throughout, we use the description of the product topology: a subset $W\subseteq X\times Y$ belongs to $\mathcal{T}_{X\times Y}$ if and only if for every $(x,y)\in W$ there exist $U\in\mathcal{T}_X$ and $V\in\mathcal{T}_Y$ with $(x,y)\in U\times V\subseteq W$. We also use the fact that, for subsets $U\subseteq X$ and $V\subseteq Y$, a point $(x,y)$ of the Cartesian product $X\times Y$ lies in $U\times V$ exactly when $x\in U$ and $y\in V$. We verify the three conditions of Topological Space.

Condition 1. The set $\varnothing$ has no elements, so the defining condition holds vacuously and $\varnothing\in\mathcal{T}_{X\times Y}$. For $X\times Y$, let $(x,y)\in X\times Y$. By condition 1 of Topological Space applied to $(X,\mathcal{T}_X)$ and to $(Y,\mathcal{T}_Y)$ we have $X\in\mathcal{T}_X$ and $Y\in\mathcal{T}_Y$, and $(x,y)\in X\times Y\subseteq X\times Y$. Hence $X\times Y\in\mathcal{T}_{X\times Y}$.

Condition 2. Let $A$ be a set and let $(W_a)_{a\in A}$ be a family of subsets of $X\times Y$ with $W_a\in\mathcal{T}_{X\times Y}$ for every $a\in A$, and put $W=\bigcup_{a\in A}W_a$. Let $(x,y)\in W$. Then $(x,y)\in W_a$ for some $a\in A$, so there are $U\in\mathcal{T}_X$ and $V\in\mathcal{T}_Y$ with $(x,y)\in U\times V\subseteq W_a$. Since $W_a\subseteq W$, this gives $(x,y)\in U\times V\subseteq W$. Hence $W\in\mathcal{T}_{X\times Y}$.

Condition 3. Let $n$ be a natural number, let $W_1,\dots,W_n\in\mathcal{T}_{X\times Y}$, and put $W=\bigcap_{i=1}^{n}W_i$. Let $(x,y)\in W$, so that $(x,y)\in W_i$ for every $i$ in the initial segment $[n]$.

Let $S=\mathcal{T}_X\times\mathcal{T}_Y$, and for $i\in[n]$ let

Because $W_i\in\mathcal{T}_{X\times Y}$ and $(x,y)\in W_i$, the set $A_i$ is nonempty for every $i\in[n]$. The index set $[n]$ is finite by claim 1 of Basic Properties of Finite Sets, so by Choice for a Family Indexed by a Finite Set there is a function $a:[n]\to S$ with $a(i)\in A_i$ for every $i\in[n]$. For $i\in[n]$ let $U_i\in\mathcal{T}_X$ and $V_i\in\mathcal{T}_Y$ be the first and second components of the pair $a(i)$, so that

Put $U=\bigcap_{i=1}^{n}U_i$ and $V=\bigcap_{i=1}^{n}V_i$. By condition 3 of Topological Space applied in $(X,\mathcal{T}_X)$ and in $(Y,\mathcal{T}_Y)$ we have $U\in\mathcal{T}_X$ and $V\in\mathcal{T}_Y$. From $(x,y)\in U_i\times V_i$ we get $x\in U_i$ and $y\in V_i$ for every $i\in[n]$, hence $x\in U$ and $y\in V$, that is, $(x,y)\in U\times V$.

Finally, let $(x',y')\in U\times V$ and let $i\in[n]$. Then $x'\in U\subseteq U_i$ and $y'\in V\subseteq V_i$, so $(x',y')\in U_i\times V_i\subseteq W_i$. As $i\in[n]$ was arbitrary, $(x',y')\in W$. Therefore $U\times V\subseteq W$, and we have produced $U\in\mathcal{T}_X$ and $V\in\mathcal{T}_Y$ with $(x,y)\in U\times V\subseteq W$. Since $(x,y)\in W$ was arbitrary, $W\in\mathcal{T}_{X\times Y}$.

All three conditions hold, so $(X\times Y,\mathcal{T}_{X\times Y})$ is a topological space.