Proof of A Totally Bounded Subset of a Nonempty Metric Space is Bounded

Throughout, $\mathbb{R}$ denotes the real numbers, an ordered field whose order $\le$ is a total order, with the addition of the underlying field, additive identity $0$ and multiplicative identity $1$; for real numbers $s,t$ we write $s<t$ to mean $s\le t$ and $s\ne t$. For $a\in X$ and a real number $r>0$, $B_d(a,r)$ denotes the open ball in $(X,d)$ with center $a$ and radius $r$. Let $\mathbb{N}$ be the natural numbers.

Step 1: a $1$-net. Statement 6 of Elementary Order Arithmetic in an Ordered Field gives $0<1$ in $\mathbb{R}$. Applying the definition of total boundedness with $\varepsilon=1$ yields a finite subset $F\subseteq X$ such that

$$K\subseteq\bigcup_{a\in F}B_d(a,1).$$

Step 2: the case $F=\emptyset$. By the convention recorded in the definition of total boundedness, the union over the empty index set is empty, so $K\subseteq\emptyset$ and therefore $K=\emptyset$. Since $X$ is nonempty, fix a point $x\in X$ and put $R=1$. Then $0<R$, and the requirement that $d(x,y)\le R$ for every $y\in K$ holds vacuously because $K$ has no elements. Hence $K$ is bounded in $(X,d)$.

Step 3: the case $F\ne\emptyset$, a bound on the centers. Since $F$ is finite and nonempty, the definition of a finite set gives a natural number $n$ such that $F$ has $n$ elements, and hence a bijection $\beta:[n]\to F$, where $[n]$ is the initial segment of $\mathbb{N}$ determined by $n$. Fix a point $x_0\in F$.

Let $c$ be the $n$-tuple in $\mathbb{R}$ whose components are

$$c_k=d\bigl(x_0,\beta(k)\bigr)\qquad (k\in[n]).$$

By Greatest Element of a Finite Family in a Totally Ordered Set, applied to the totally ordered set $\mathbb{R}$, there is $j\in[n]$ with $c_k\le c_j$ for every $k\in[n]$. Put $M=c_j$. Because $\beta$ is a bijection onto $F$, every $a\in F$ equals $\beta(k)$ for some $k\in[n]$, so

$$d(x_0,a)\le M\qquad\text{for every } a\in F.$$

Step 4: the radius. Put $R=1+M$. Condition 1 in the definition of a metric gives $0\le d(x_0,\beta(j))=M$. Applying statement 3 of Elementary Order Arithmetic in an Ordered Field to the pair of inequalities $0<1$ and $0\le M$ gives $0+0<1+M$, and $0+0=0$ by Additive Cancellation and Elementary Additive Identities in a Field. Hence $0<R$.

Step 5: the bound on $K$. Let $y\in K$. By Step 1 there is $a\in F$ with $y\in B_d(a,1)$, that is, $d(a,y)<1$ by the definition of the open ball. From $d(a,y)<1$ and $d(x_0,a)\le M$, statement 3 of Elementary Order Arithmetic in an Ordered Field gives

$$d(a,y)+d(x_0,a)<1+M=R,$$

and addition in a field is commutative, so $d(x_0,a)+d(a,y)<R$. Condition 4 in the definition of a metric gives $d(x_0,y)\le d(x_0,a)+d(a,y)$, so mixed transitivity, statement 2 of Elementary Order Arithmetic in an Ordered Field, yields $d(x_0,y)<R$ and in particular $d(x_0,y)\le R$.

Thus $x_0\in F\subseteq X$ and the real number $R$ satisfy $0<R$ and $d(x_0,y)\le R$ for every $y\in K$, so $K$ is bounded in $(X,d)$.