[proofplan]
We prove that the usual quantifier definition of [uniform continuity](/page/Uniform%20Continuity) is equivalent to packaging the admissible radii into a single positive function of $\varepsilon$. In the forward direction, for each $\varepsilon>0$ we consider the set of all radii that work for that $\varepsilon$ and use its supremum to define a canonical positive radius. In the reverse direction, the given function $\delta$ immediately supplies the radius required in the definition of uniform continuity.
[/proofplan]
[step:Construct a positive radius function from uniform continuity]
Assume that $f$ is uniformly continuous. Thus, for every $\varepsilon>0$, there exists $\eta>0$ such that, for all $x,x'\in X$,
\begin{align*}d_X(x,x')<\eta \implies d_Y(f(x),f(x'))<\varepsilon.\end{align*}
For each $\varepsilon>0$, define the set $A_\varepsilon\subset(0,1]$ by
\begin{align*}A_\varepsilon:=\{r\in(0,1]:\text{ for all }x,x'\in X,\ d_X(x,x')<r\implies d_Y(f(x),f(x'))<\varepsilon\}.\end{align*}
By uniform continuity, $A_\varepsilon$ is nonempty: if $\eta>0$ is admissible for $\varepsilon$, then
\begin{align*}\frac{\min\{\eta,1\}}{2}\in A_\varepsilon\end{align*}
Since $A_\varepsilon\subset(0,1]$, the number
\begin{align*}s(\varepsilon):=\sup A_\varepsilon\end{align*}
is defined and satisfies $0<s(\varepsilon)\le 1$.
Define the function $\delta:(0,\infty)\to(0,\infty)$ by
\begin{align*}\delta(\varepsilon):=\frac{s(\varepsilon)}{2}.\end{align*}
Fix $\varepsilon>0$. Since $0<\delta(\varepsilon)<s(\varepsilon)$, the defining property of the supremum gives some $r_\varepsilon\in A_\varepsilon$ such that $\delta(\varepsilon)<r_\varepsilon$. If $x,x'\in X$ satisfy $d_X(x,x')<\delta(\varepsilon)$, then $d_X(x,x')<r_\varepsilon$, and the definition of $A_\varepsilon$ gives
\begin{align*}d_Y(f(x),f(x'))<\varepsilon.\end{align*}
Therefore $\delta$ has the required modulus property.
[guided]
Assume that $f$ is uniformly continuous. By definition, this means that for every tolerance $\varepsilon>0$ there is at least one positive radius $\eta>0$ such that the implication
\begin{align*}d_X(x,x')<\eta \implies d_Y(f(x),f(x'))<\varepsilon\end{align*}
holds for all $x,x'\in X$.
The point of the theorem is to turn these separate existence statements into one function of $\varepsilon$. To avoid making an arbitrary choice of one radius for every $\varepsilon$, we collect all radii that work. For each $\varepsilon>0$, define
\begin{align*}A_\varepsilon:=\{r\in(0,1]:\text{ for all }x,x'\in X,\ d_X(x,x')<r\implies d_Y(f(x),f(x'))<\varepsilon\}.\end{align*}
The restriction $r\le 1$ is harmless; it only makes the set bounded above. We verify that $A_\varepsilon$ is nonempty. Uniform continuity gives an admissible $\eta>0$ for this $\varepsilon$. Then
\begin{align*}\frac{\min\{\eta,1\}}{2} \in (0,1]\end{align*}
and whenever
\begin{align*}d_X(x,x')<\frac{\min\{\eta,1\}}{2}\end{align*}
we also have $d_X(x,x')<\eta$, so the uniform continuity implication gives $d_Y(f(x),f(x'))<\varepsilon$. Hence
\begin{align*}\frac{\min\{\eta,1\}}{2}\in A_\varepsilon\end{align*}.
Because $A_\varepsilon$ is a nonempty subset of $(0,1]$, its supremum exists as a real number. Define
\begin{align*}
s(\varepsilon):=\sup A_\varepsilon.
\end{align*}
The previous paragraph shows $s(\varepsilon)>0$, and the inclusion $A_\varepsilon\subset(0,1]$ gives $s(\varepsilon)\le 1$. Now define the candidate modulus $\delta:(0,\infty)\to(0,\infty)$ by
\begin{align*}
\delta(\varepsilon):=\frac{s(\varepsilon)}{2}.
\end{align*}
This is positive because $s(\varepsilon)>0$.
It remains to check that this particular value actually works. Fix $\varepsilon>0$. Since $\delta(\varepsilon)<s(\varepsilon)=\sup A_\varepsilon$, the definition of supremum gives some $r_\varepsilon\in A_\varepsilon$ with
\begin{align*}
\delta(\varepsilon)<r_\varepsilon.
\end{align*}
Now take arbitrary points $x,x'\in X$ satisfying $d_X(x,x')<\delta(\varepsilon)$. Then
\begin{align*}
d_X(x,x')<r_\varepsilon.
\end{align*}
Since $r_\varepsilon\in A_\varepsilon$, its defining property applies and yields
\begin{align*}
d_Y(f(x),f(x'))<\varepsilon.
\end{align*}
Thus the function $\delta$ satisfies the required epsilon-delta modulus condition.
[/guided]
[/step]
[step:Recover uniform continuity from a modulus]
Conversely, suppose there exists a function $\delta:(0,\infty)\to(0,\infty)$ such that, for every $\varepsilon>0$ and all $x,x'\in X$,
\begin{align*}
d_X(x,x')<\delta(\varepsilon) \implies d_Y(f(x),f(x'))<\varepsilon.
\end{align*}
Let $\varepsilon>0$ be arbitrary, and define $\eta:=\delta(\varepsilon)$. Since $\delta$ maps into $(0,\infty)$, we have $\eta>0$. For all $x,x'\in X$, the assumed property of $\delta$ gives
\begin{align*}
d_X(x,x')<\eta \implies d_Y(f(x),f(x'))<\varepsilon.
\end{align*}
This is exactly the epsilon-delta definition of uniform continuity. Hence $f$ is uniformly continuous.
[/step]
[step:Identify the modulus condition]
The two implications prove the equivalence. The final sentence is terminology: any function $\delta:(0,\infty)\to(0,\infty)$ satisfying the displayed implication for every $\varepsilon>0$ and all $x,x'\in X$ is, by definition, an epsilon-delta modulus of uniform continuity for $f$.
[/step]
