Proof of A Closed Totally Bounded Subset of a Complete Metric Space is Compact

Each result cited below is universally quantified over the data in its own statement and is applied with the data named at the point of citation.

Claim 1. Let $(x_{m})_{m\in\mathbb{N}}$ be a sequence in $X$ with $x_{m}\in K$ for every $m\in\mathbb{N}$.

Step 1 (finite nets). For each $k\in\mathbb{N}$ the number $\tfrac{1}{k}$ is positive, so by Totally Bounded Subset of a Metric Space there is a finite set $F_{k}\subseteq X$ with $K\subseteq\bigcup_{a\in F_{k}}B_{d}(a,\tfrac{1}{k})$, where $B_{d}(a,r)=\{y\in X:d(a,y)<r\}$ is the open ball of Open Ball in a Metric Space. Fix such a set $F_{k}$ for every $k$.

Step 2 (nested unbounded index sets). Call a set $I\subseteq\mathbb{N}$ unbounded if for every $N\in\mathbb{N}$ there is $m\in I$ with $N\le m$. We choose recursively unbounded sets $I_{1}\supseteq I_{2}\supseteq\cdots$ and points $a_{k}\in F_{k}$ such that $d(a_{k},x_{m})<\tfrac{1}{k}$ for every $k\in\mathbb{N}$ and every $m\in I_{k}$. Put $I_{0}=\mathbb{N}$, which is unbounded. Let $k\in\mathbb{N}$ and suppose the unbounded set $I_{k-1}$ has been chosen. For $a\in F_{k}$ let $J_{a}=\{m\in I_{k-1}:d(a,x_{m})<\tfrac{1}{k}\}$. Every $m\in I_{k-1}$ satisfies $x_{m}\in K$, hence lies in $J_{a}$ for some $a\in F_{k}$ by the choice of $F_{k}$; thus $I_{k-1}=\bigcup_{a\in F_{k}}J_{a}$, and $F_{k}$ is nonempty because $I_{k-1}$ is. Suppose that no $J_{a}$ were unbounded: then for each $a\in F_{k}$ there would be $N_{a}\in\mathbb{N}$ with $m<N_{a}$ for every $m\in J_{a}$, and the largest $N^{*}$ of the finitely many numbers $N_{a}$, $a\in F_{k}$, would satisfy $m<N^{*}$ for every $m\in I_{k-1}$, contradicting the unboundedness of $I_{k-1}$. Hence we may choose $a_{k}\in F_{k}$ with $J_{a_{k}}$ unbounded, and we put $I_{k}=J_{a_{k}}\subseteq I_{k-1}$.

Step 3 (the indices). Choose $n_{1}\in I_{1}$, and, recursively, once $n_{k}$ has been chosen, choose $n_{k+1}\in I_{k+1}$ with $n_{k}+1\le n_{k+1}$, which is possible because $I_{k+1}$ is unbounded. Then $(n_{k})_{k\in\mathbb{N}}$ is strictly increasing in the sense of Subsequence of a Sequence in a Set, and since the sets $I_{k}$ are nested, $n_{j}\in I_{j}\subseteq I_{k}$ whenever $k\le j$.

Step 4 (a Cauchy subsequence). Put $y_{j}=x_{n_{j}}$. Let $k\in\mathbb{N}$ and let $j,l\in\mathbb{N}$ satisfy $k\le j$ and $k\le l$. By Step 3, $n_{j}$ and $n_{l}$ belong to $I_{k}$, so $d(a_{k},y_{j})<\tfrac{1}{k}$ and $d(a_{k},y_{l})<\tfrac{1}{k}$ by Step 2. By the symmetry and the triangle inequality, conditions 3 and 4 of Metric Space,

$$d(y_{j},y_{l})\le d(y_{j},a_{k})+d(a_{k},y_{l})<\tfrac{2}{k}.$$

Now let $\varepsilon$ be a positive real number. By claim 3 of The Archimedean Property of the Real Numbers, applied to $\tfrac{\varepsilon}{2}$, there is $k\in\mathbb{N}$ with $\tfrac{1}{k}<\tfrac{\varepsilon}{2}$; for all $j,l\ge k$ the display gives $d(y_{j},y_{l})<\tfrac{2}{k}<\varepsilon$. Hence $(y_{j})_{j\in\mathbb{N}}$ is a Cauchy sequence in $(X,d)$ in the sense of Cauchy Sequence in a Metric Space.

Step 5 (the limit lies in $K$). Since $(X,d)$ is complete, Complete Metric Space gives a point $x\in X$ such that $(y_{j})_{j\in\mathbb{N}}$ converges to $x$ in $(X,d)$. Every $y_{j}$ lies in $K$, and $K$ is closed in $(X,\mathcal{T}_{d})$, so $x\in K$ by Sequential Characterization of Closed Subsets of a Metric Space, applied to the set $K$, the sequence $(y_{j})_{j\in\mathbb{N}}$ and the point $x$. Thus the point $x\in K$ and the strictly increasing sequence $(n_{j})_{j\in\mathbb{N}}$ witness the condition of Sequentially Compact Subset of a Metric Space for the sequence $(x_{m})_{m\in\mathbb{N}}$. As this sequence in $K$ was arbitrary, $K$ is sequentially compact in $(X,d)$.

Claim 2. By claim 1, $K$ is sequentially compact in $(X,d)$, so $K$ is compact in $(X,\mathcal{T}_{d})$ by A Sequentially Compact Subset of a Metric Space is Compact, applied to the metric space $(X,d)$ and the set $K$.