Proof of A Sequentially Compact Subset of a Metric Space is Totally Bounded

Let $\varepsilon$ be a real number with $\varepsilon>0$, and suppose for contradiction that no finite subset $F\subseteq K$ satisfies $K\subseteq\bigcup_{a\in F}B_d(a,\varepsilon)$.

Step 1 (a relation to iterate). Let $\mathcal{S}$ be the set of all finite subsets of $K$. The empty set is finite and is a subset of $K$, so $\emptyset\in\mathcal{S}$ and $\mathcal{S}$ is nonempty. Define a binary relation $R$ on $\mathcal{S}$, that is a subset of the Cartesian product $\mathcal{S}\times\mathcal{S}$, by declaring $(F,G)\in R$ exactly when there is $y\in K$ with

We check that for every $F\in\mathcal{S}$ there is $G\in\mathcal{S}$ with $(F,G)\in R$. Let $F\in\mathcal{S}$. By the contradiction hypothesis $K\subseteq\bigcup_{a\in F}B_d(a,\varepsilon)$ fails, so there is $y\in K$ with $y\notin\bigcup_{a\in F}B_d(a,\varepsilon)$. For each $a\in F$ the identity-of-indiscernibles axiom of a metric gives $d(a,a)=0<\varepsilon$, so $a\in B_d(a,\varepsilon)$ by Open Ball in a Metric Space; since $y$ lies in none of these balls, $y\ne a$. Hence $y\notin F$, and by claim 2 of Basic Properties of Finite Sets the set $F\cup\{y\}$ is finite: if $F$ is empty it equals $\{y\}$, which has $1$ element, and if $F$ has $k$ elements it has $\sigma(k)$ elements, where $\sigma$ denotes the successor map of Natural Numbers. It is also a subset of $K$. So $G=F\cup\{y\}$ lies in $\mathcal{S}$ and $(F,G)\in R$.

Step 2 (an infinite $\varepsilon$-separated sequence). By Axiom of Dependent Choice, applied to $\mathcal{S}$, $R$, and the starting element $\emptyset$, there is a sequence $(F_m)_{m\in\mathbb{N}}$ in $\mathcal{S}$ with $F_1=\emptyset$ and $(F_m,F_{m+1})\in R$ for every $m\in\mathbb{N}$.

Fix $m\in\mathbb{N}$. Any $y$ witnessing $(F_m,F_{m+1})\in R$ satisfies $y\notin F_m$, by the argument of Step 1, and $F_{m+1}=F_m\cup\{y\}$, so $F_{m+1}\setminus F_m=\{y\}$. Hence $F_{m+1}\setminus F_m$ has exactly one element, and that element is a witness; define $x_m$ to be it. No choice principle is involved, since $x_m$ is determined by $F_m$ and $F_{m+1}$. Thus $(x_m)_{m\in\mathbb{N}}$ is a sequence in $X$ with $x_m\in K$, and for every $m$,

An induction using the principle of induction and the first of these gives $F_j\subseteq F_m$ whenever $j\le m$.

Now let $j,m\in\mathbb{N}$ with $j<m$. By claim 7 of Properties of the Order on the Natural Numbers there is $t\in\mathbb{N}$ with $m=j+t$, and $1\le t$ by claim 4, so $j+1\le j+t=m$ by claim 6. Hence $F_{j+1}\subseteq F_m$ and therefore $x_j\in F_m$. Since $x_m\notin\bigcup_{a\in F_m}B_d(a,\varepsilon)$, in particular $x_m\notin B_d(x_j,\varepsilon)$, so $d(x_j,x_m)<\varepsilon$ fails and comparability of the order on $\mathbb{R}$ gives

Step 3 (contradiction with sequential compactness). By Sequentially Compact Subset of a Metric Space there are $x\in K$ and a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$, in the sense of Subsequence of a Sequence in a Set, such that $(x_{n_k})_{k\in\mathbb{N}}$ converges to $x$ in $(X,d)$. By claim 8 of Elementary Order Arithmetic in an Ordered Field, $0<\varepsilon\cdot 2^{-1}$ and $\varepsilon\cdot 2^{-1}+\varepsilon\cdot 2^{-1}=\varepsilon$.

By Convergent Sequence in a Metric Space there is $N\in\mathbb{N}$ with $d(x_{n_k},x)<\varepsilon\cdot 2^{-1}$ for every $k$ with $N\le k$. Both $k=N$ and $k=N+1$ satisfy $N\le k$, by claims 1 and 6 of Properties of the Order on the Natural Numbers. Since $(n_k)$ is strictly increasing, $n_N<n_{N+1}$, so Step 2 gives $\varepsilon\le d(x_{n_N},x_{n_{N+1}})$.

On the other hand, the symmetry axiom of a metric gives $d(x,x_{n_{N+1}})=d(x_{n_{N+1}},x)$, so adding the two strict inequalities by claim 3 of Elementary Order Arithmetic in an Ordered Field,

and the triangle inequality axiom gives $d(x_{n_N},x_{n_{N+1}})\le d(x_{n_N},x)+d(x,x_{n_{N+1}})$. By claim 2 of Elementary Order Arithmetic in an Ordered Field we conclude $d(x_{n_N},x_{n_{N+1}})<\varepsilon$, and combining with $\varepsilon\le d(x_{n_N},x_{n_{N+1}})$ gives $\varepsilon<\varepsilon$, which is false.

This contradiction shows that some finite subset $F\subseteq K$ satisfies $K\subseteq\bigcup_{a\in F}B_d(a,\varepsilon)$. Since such an $F$ is also a finite subset of $X$, and $\varepsilon>0$ was arbitrary, Totally Bounded Subset of a Metric Space shows that $K$ is totally bounded in $(X,d)$.