Proof of A Product of Compact Subsets is Compact in the Product Metric

Throughout, $\mathbb{N}$ denotes the set of natural numbers equipped with the order relations $<$ and $\le$ of that definition.

Step 1 (Each factor is sequentially compact). By hypothesis $K$ is compact in $X$, where $X$ carries the collection of its subsets open in $(X,d_X)$. By Compactness and Sequential Compactness Agree for Subsets of a Metric Space, $K$ is sequentially compact in $(X,d_X)$. The same argument applied to $(Y,d_Y)$ shows that $L$ is sequentially compact in $(Y,d_Y)$.

Step 2 (Two successive extractions). Let $(z_m)_{m\in\mathbb{N}}$ be a sequence in $X\times Y$ with $z_m\in K\times L$ for every $m\in\mathbb{N}$. By the definition of the Cartesian product, each $z_m$ is an ordered pair whose first entry lies in $K$ and whose second entry lies in $L$; write

$$z_m=(x_m,y_m),\qquad x_m\in K\subseteq X,\quad y_m\in L\subseteq Y .$$

Then $(x_m)_{m\in\mathbb{N}}$ is a sequence in $X$ with all terms in $K$, and $(y_m)_{m\in\mathbb{N}}$ is a sequence in $Y$ with all terms in $L$.

Applying sequential compactness of $K$ to $(x_m)_{m\in\mathbb{N}}$ gives a point $x\in K$ and a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ such that $(x_{n_k})_{k\in\mathbb{N}}$ converges to $x$ in $(X,d_X)$.

The sequence $(y_{n_k})_{k\in\mathbb{N}}$ is a sequence in $Y$ with all terms in $L$. Applying sequential compactness of $L$ to it gives a point $y\in L$ and a strictly increasing sequence $(k_j)_{j\in\mathbb{N}}$ in $\mathbb{N}$ such that $(y_{n_{k_j}})_{j\in\mathbb{N}}$ converges to $y$ in $(Y,d_Y)$.

Step 3 (The two extractions compose). By claim 2 of A Subsequence of a Subsequence is a Subsequence, the sequence $(n_{k_j})_{j\in\mathbb{N}}$ is strictly increasing, and by claim 3 of that lemma the sequence $(z_{n_{k_j}})_{j\in\mathbb{N}}$ is a subsequence of $(z_m)_{m\in\mathbb{N}}$. Moreover, applying A Subsequence of a Convergent Sequence Has the Same Limit to the convergent sequence $(x_{n_k})_{k\in\mathbb{N}}$ and the strictly increasing sequence $(k_j)_{j\in\mathbb{N}}$ shows that $(x_{n_{k_j}})_{j\in\mathbb{N}}$ converges to $x$ in $(X,d_X)$.

Step 4 (The composed subsequence converges in the product). Put $z=(x,y)$, so that $z\in K\times L$. Let $\varepsilon$ be a real number with $\varepsilon>0$. Since $(x_{n_{k_j}})_{j\in\mathbb{N}}$ converges to $x$, there is $N_1\in\mathbb{N}$ with

$$d_X\bigl(x_{n_{k_j}},x\bigr)<\varepsilon\quad\text{for every }j\in\mathbb{N}\text{ with }N_1\le j,$$

and since $(y_{n_{k_j}})_{j\in\mathbb{N}}$ converges to $y$, there is $N_2\in\mathbb{N}$ with

$$d_Y\bigl(y_{n_{k_j}},y\bigr)<\varepsilon\quad\text{for every }j\in\mathbb{N}\text{ with }N_2\le j .$$

By claim 3 of Properties of the Order on the Natural Numbers, exactly one of $N_1<N_2$, $N_1=N_2$, $N_2<N_1$ holds. Put $N=N_2$ in the first case and $N=N_1$ in the other two. In either case $N_1\le N$ and $N_2\le N$, by claim 1 of that lemma, which gives both reflexivity of $\le$ and the implication from $<$ to $\le$.

Let $j\in\mathbb{N}$ with $N\le j$. By transitivity of $\le$, claim 1 of Properties of the Order on the Natural Numbers, we get $N_1\le j$ and $N_2\le j$, so both displayed inequalities hold at this $j$. Since $z_{n_{k_j}}=(x_{n_{k_j}},y_{n_{k_j}})$ and $z=(x,y)$, claim 3 of The Product Metric is a Metric yields

$$d_{X\times Y}\bigl(z_{n_{k_j}},z\bigr)<\varepsilon .$$

As $\varepsilon>0$ was arbitrary, $(z_{n_{k_j}})_{j\in\mathbb{N}}$ converges to $z$ in the metric space $(X\times Y,d_{X\times Y})$, which is a metric space by claim 1 of The Product Metric is a Metric.

Step 5 (Conclusion). Steps 2 to 4 show: for every sequence in $X\times Y$ with all terms in $K\times L$ there are a point $z\in K\times L$ and a subsequence converging to $z$ in $(X\times Y,d_{X\times Y})$. Hence $K\times L$ is sequentially compact in $(X\times Y,d_{X\times Y})$. Applying Compactness and Sequential Compactness Agree for Subsets of a Metric Space in the metric space $(X\times Y,d_{X\times Y})$, in the direction from sequential compactness to compactness, we conclude that $K\times L$ is compact in $X\times Y$ equipped with the collection of its subsets open in $(X\times Y,d_{X\times Y})$.

Step 6 (The restatement in terms of $d_{K\times L}$). By the definition of compactness of a subset, the assertion just proved is that the set $K\times L$, equipped with the subspace topology inherited from $X\times Y$, is a compact topological space. By claim 2 of The Restriction of a Metric to a Subset Induces the Subspace Topology, applied to the subset $K\times L$ of $X\times Y$ and the restricted metric $d_{K\times L}$, the collection of subsets of $K\times L$ open in the metric space $(K\times L,d_{K\times L})$ is exactly that subspace topology. So the two formulations concern one and the same topological space, and the second assertion of the statement follows. $\blacksquare$