Proof of Coordinatewise Convergence, Sequential Compactness and Density in a Product Metric Space

Throughout, for $z=(x,y)$ and $z'=(x',y')$ in $X\times Y$ the product metric is $d_{X\times Y}(z,z')=\max\{d_X(x,x'),d_Y(y,y')\}$. By the definition of the maximum of two elements of a totally ordered set, for real numbers $a,b$ the value $\max\{a,b\}$ is one of $a$ and $b$, and $a\le\max\{a,b\}$ and $b\le\max\{a,b\}$. We use two consequences, for a real number $\varepsilon$:

(M1) If $a<\varepsilon$ and $b<\varepsilon$ then $\max\{a,b\}<\varepsilon$, because $\max\{a,b\}$ is one of $a$ and $b$.

(M2) If $\max\{a,b\}<\varepsilon$ then $a<\varepsilon$ and $b<\varepsilon$, by claim 2 of Elementary Order Arithmetic in an Ordered Field applied to $a\le\max\{a,b\}$ together with $\max\{a,b\}<\varepsilon$, and likewise for $b$.

Claim 1. Write $\delta_m=d_{X\times Y}(z_m,z)=\max\{d_X(x_m,x),d_Y(y_m,y)\}$.

Suppose first that $(z_m)_{m\in\mathbb{N}}$ converges to $z$. Let $\varepsilon$ be a real number with $0<\varepsilon$. By the definition of convergence there is $N\in\mathbb{N}$ with $\delta_m<\varepsilon$ for every $m\ge N$. By (M2), $d_X(x_m,x)<\varepsilon$ and $d_Y(y_m,y)<\varepsilon$ for every $m\ge N$. As $\varepsilon$ was arbitrary, $(x_m)_{m\in\mathbb{N}}$ converges to $x$ in $(X,d_X)$ and $(y_m)_{m\in\mathbb{N}}$ converges to $y$ in $(Y,d_Y)$.

Conversely, suppose both coordinate sequences converge. Let $\varepsilon$ be a real number with $0<\varepsilon$. There are $N_1,N_2\in\mathbb{N}$ with $d_X(x_m,x)<\varepsilon$ for every $m\ge N_1$ and $d_Y(y_m,y)<\varepsilon$ for every $m\ge N_2$. 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. In the first case $N_1\le N_2$ and in the third $N_2\le N_1$, since $m<n$ means $m\le n$ together with $m\ne n$; in the second case both $N_1\le N_2$ and $N_2\le N_1$ hold by reflexivity of the order, which is part of claim 1 of that lemma. So in every case one of $N_1\le N_2$ and $N_2\le N_1$ holds; let $N$ denote whichever of $N_1,N_2$ satisfies $N_1\le N$ and $N_2\le N$. Let $m\ge N$. By claim 1 of Properties of the Order on the Natural Numbers the order on $\mathbb{N}$ is transitive, so $m\ge N_1$ and $m\ge N_2$, whence $d_X(x_m,x)<\varepsilon$ and $d_Y(y_m,y)<\varepsilon$, and therefore $\delta_m<\varepsilon$ by (M1). As $\varepsilon$ was arbitrary, $(z_m)_{m\in\mathbb{N}}$ converges to $z$ in $(X\times Y,d_{X\times Y})$.

Claim 2. Let $(z_m)_{m\in\mathbb{N}}$ be a sequence in $X\times Y$ with $z_m\in K\times M$ for every $m\in\mathbb{N}$, and write $z_m=(x_m,y_m)$, so that $x_m\in K$ and $y_m\in M$ for every $m$.

Then $(x_m)_{m\in\mathbb{N}}$ is a sequence in $X$ all of whose terms lie in $K$, so by sequential compactness of $K$ there are $x\in K$ 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 $(X,d_X)$. Next, $(y_{n_k})_{k\in\mathbb{N}}$ is a sequence in $Y$ all of whose terms lie in $M$, so by sequential compactness of $M$ there are $y\in M$ 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 $(Y,d_Y)$.

By claim 2 of A Subsequence of a Subsequence is a Subsequence the sequence $(n_{k_j})_{j\in\mathbb{N}}$ is strictly increasing in $\mathbb{N}$, and by claim 3 of that lemma $(x_{n_{k_j}})_{j\in\mathbb{N}}$ is a subsequence of $(x_{n_k})_{k\in\mathbb{N}}$. Since $(x_{n_k})_{k\in\mathbb{N}}$ converges to $x$, A Subsequence of a Convergent Sequence Has the Same Limit gives that $(x_{n_{k_j}})_{j\in\mathbb{N}}$ converges to $x$ in $(X,d_X)$.

So the two coordinate sequences of $(z_{n_{k_j}})_{j\in\mathbb{N}}$ converge to $x$ and to $y$ respectively, and claim 1 gives that $(z_{n_{k_j}})_{j\in\mathbb{N}}$ converges to $(x,y)$ in $(X\times Y,d_{X\times Y})$. Moreover $(x,y)\in K\times M$, and $(n_{k_j})_{j\in\mathbb{N}}$ is strictly increasing, so $(z_{n_{k_j}})_{j\in\mathbb{N}}$ is a subsequence of $(z_m)_{m\in\mathbb{N}}$ converging to a point of $K\times M$. As $(z_m)_{m\in\mathbb{N}}$ was an arbitrary sequence with terms in $K\times M$, the set $K\times M$ is sequentially compact in $(X\times Y,d_{X\times Y})$.

Claim 3. Let $(x,y)\in X\times Y$ and let $\varepsilon$ be a real number with $0<\varepsilon$. Since $D$ is dense in $X$ for $\mathcal{T}_X$, the closure of $D$ in $X$ is $X$, so $x$ belongs to it; by Characterization of the Closure in a Metric Space by Open Balls, condition 1 there implies condition 3, so there is $q\in D$ with $d_X(x,q)<\varepsilon$. In the same way there is $q'\in E$ with $d_Y(y,q')<\varepsilon$. Then $(q,q')\in D\times E$ and, by (M1),

As $\varepsilon$ was arbitrary, condition 3 of Characterization of the Closure in a Metric Space by Open Balls holds for the subset $D\times E$ of $X\times Y$ at the point $(x,y)$, so by that theorem $(x,y)$ lies in the closure of $D\times E$ in $X\times Y$. Since $(x,y)$ was arbitrary, $X\times Y$ is contained in that closure; the closure is a subset of $X\times Y$ by definition, so the two sets are equal and $D\times E$ is dense in $X\times Y$ for $\mathcal{T}_{X\times Y}$.

Finally, if $D$ and $E$ are countable then $D\times E$ is countable by claim 1 of Products and Powers of Countable Sets.