[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.[/step]

[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: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]