Proof of Heine-Cantor Theorem: Continuity on a Compact Subset Implies Uniform Continuity

Let $\mathbb{R}$ denote the real numbers with the order and arithmetic of their ordered field structure, and write $s<t$ to mean $s\le t$ and $s\ne t$. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$; we must produce $\delta\in\mathbb{R}$ with $0<\delta$ such that all $x,y\in K$ with $d_X(x,y)<\delta$ satisfy $d_Y(f(x),f(y))<\varepsilon$.

The degenerate case. If $K=\varnothing$, take $\delta=1$, which satisfies $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field. There are no points $x,y\in K$, so the required condition holds vacuously. Assume from now on that $K$ is nonempty.

A cover with no choices. By claim 8 of Elementary Order Arithmetic in an Ordered Field the element $\eta=\varepsilon\cdot 2^{-1}$ satisfies $0<\eta$ and $\eta+\eta=\varepsilon$. Let $P$ be the set of all pairs $(x,r)$ with $x\in K$ and $r\in\mathbb{R}$ such that $0<r$ and such that every $z\in K$ with $d_X(x,z)<r+r$ satisfies $d_Y(f(z),f(x))<\eta$. For $p=(x,r)\in P$ put $U_p=B_{d_X}(x,r)$, the open ball of radius $r$ about $x$; this assignment is a family of subsets of $X$ indexed by $P$, and each $U_p$ lies in $\mathcal{T}_{d_X}$ by Open Ball in a Metric Space is Open. Note that $P$ is specified outright by a condition, so no selection of a radius for each point is made.

This family is an open cover of $K$ in $X$. Indeed, let $x\in K$. By continuity of $f$ at $x$ relative to $K$, applied with the tolerance $\eta$, there is $\delta_0\in\mathbb{R}$ with $0<\delta_0$ such that every $z\in K$ with $d_X(x,z)<\delta_0$ satisfies $d_Y(f(z),f(x))<\eta$. Put $r=\delta_0\cdot 2^{-1}$; by claim 8 of Elementary Order Arithmetic in an Ordered Field we have $0<r$ and $r+r=\delta_0$, so $(x,r)\in P$. Moreover $d_X(x,x)=0<r$ by condition 2 of the definition of a metric, so $x\in U_{(x,r)}$.

Extracting a radius. Since $K$ is compact in $(X,\mathcal{T}_{d_X})$, the implication from condition 1 to condition 2 of Compact Subset Criterion via Open Covers in the Ambient Space gives a finite subset $J\subseteq P$ with $K$ contained in the union of the sets $U_p$ for $p\in J$. As $K$ is nonempty, some point of $K$ lies in some $U_p$ with $p\in J$, so $J$ is nonempty. Hence $J$ has $n$ elements for some natural number $n$; fix a bijection from the initial segment $[n]$ onto $J$ and write $(x_k,r_k)$ for the image of $k\in[n]$, so that $J$ consists exactly of the pairs $(x_k,r_k)$ with $k\in[n]$.

Let $c$ be the $n$-tuple in $\mathbb{R}$ with components $c_k=-r_k$. By Greatest Element of a Finite Family in a Totally Ordered Set, applied to $\mathbb{R}$ with its total order, there is $j\in[n]$ with $c_k\le c_j$ for every $k\in[n]$, that is $-r_k\le -r_j$; by claim 4 of Elementary Order Arithmetic in an Ordered Field this says $r_j\le r_k$ for every $k\in[n]$. Put $\delta=r_j$, so $0<\delta$ because $(x_j,r_j)\in P$.

The estimate. Let $x,y\in K$ satisfy $d_X(x,y)<\delta$. Since the sets $U_p$ with $p\in J$ cover $K$, there is $k\in[n]$ with $x\in B_{d_X}(x_k,r_k)$, that is $d_X(x_k,x)<r_k$.

First, $0<r_k$ gives $r_k+0<r_k+r_k$ by claim 1 of Elementary Order Arithmetic in an Ordered Field, so $r_k<r_k+r_k$; with $d_X(x_k,x)<r_k$ and claim 2 of that lemma we get $d_X(x_k,x)<r_k+r_k$. Since $x\in K$ and $(x_k,r_k)\in P$, this yields $d_Y(f(x),f(x_k))<\eta$.

Next, $d_X(x,y)<\delta=r_j$ and $r_j\le r_k$ give $d_X(x,y)<r_k$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, hence in particular $d_X(x,y)\le r_k$. Combining with $d_X(x_k,x)<r_k$ by claim 3 of that lemma gives $d_X(x_k,x)+d_X(x,y)<r_k+r_k$, and condition 4 of the definition of a metric gives $d_X(x_k,y)\le d_X(x_k,x)+d_X(x,y)$; so $d_X(x_k,y)<r_k+r_k$ by claim 2. Since $y\in K$ and $(x_k,r_k)\in P$, this yields $d_Y(f(y),f(x_k))<\eta$.

Finally, condition 4 of the definition of a metric gives

and $d_Y(f(x_k),f(y))=d_Y(f(y),f(x_k))<\eta$ by condition 3 of that definition. From $d_Y(f(x),f(x_k))<\eta$ and $d_Y(f(x_k),f(y))\le\eta$, claim 3 of Elementary Order Arithmetic in an Ordered Field gives $d_Y(f(x),f(x_k))+d_Y(f(x_k),f(y))<\eta+\eta=\varepsilon$, and claim 2 of that lemma gives $d_Y(f(x),f(y))<\varepsilon$.

Since $\varepsilon$ was an arbitrary positive real number, $f$ is uniformly continuous on $K$.