[proofplan] The forward direction is a triangle-inequality estimate: [uniform convergence](/page/Uniform%20Convergence) to a [limit](/page/Limit) $f$ gives a uniform Cauchy condition by routing the difference $f_n(x) - f_m(x)$ through $f(x)$. The converse first constructs the limit pointwise using completeness of $\mathbb{R}$, then promotes pointwise convergence to uniform convergence by fixing $n$ in the Cauchy bound and sending $m \to \infty$. [/proofplan] [step:Show that uniform convergence implies the uniform Cauchy condition] Suppose $(f_n)$ converges uniformly on $E$ to some $f: E \to \mathbb{R}$. Fix $\varepsilon > 0$. By uniform convergence, there exists $N \in \mathbb{N}$ such that $\|f_n - f\|_\infty < \varepsilon / 2$ for all $n \geq N$, where $\|h\|_\infty := \sup_{x \in E} |h(x)|$. For any $m, n \geq N$ and any $x \in E$, the triangle inequality gives \begin{align*} |f_n(x) - f_m(x)| &\leq |f_n(x) - f(x)| + |f(x) - f_m(x)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Since the bound holds for all $x \in E$ and all $m, n \geq N$, the [sequence](/page/Sequence) is uniformly Cauchy. [/step] [step:Construct the pointwise limit using completeness of $\mathbb{R}$] Suppose $(f_n)$ is uniformly Cauchy on $E$. Fix $x \in E$. For every $\varepsilon > 0$, the uniformly Cauchy condition provides $N \in \mathbb{N}$ such that $|f_n(x) - f_m(x)| < \varepsilon$ for all $m, n \geq N$. In particular, the numerical [sequence](/page/Sequence) $(f_n(x))_{n=1}^\infty$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in $\mathbb{R}$. Since $\mathbb{R}$ is complete, $\lim_{n \to \infty} f_n(x)$ exists. Define $f: E \to \mathbb{R}$ by $f(x) := \lim_{n \to \infty} f_n(x)$. [/step] [step:Promote pointwise convergence to uniform convergence by passing to the limit in the Cauchy bound] Fix $\varepsilon > 0$. By the uniformly Cauchy hypothesis, there exists $N \in \mathbb{N}$ such that $|f_n(x) - f_m(x)| < \varepsilon$ for all $m, n \geq N$ and all $x \in E$. Fix any $n \geq N$ and any $x \in E$. For every $m \geq N$, the inequality $|f_n(x) - f_m(x)| < \varepsilon$ holds. Taking $m \to \infty$ and using $f(x) = \lim_{m \to \infty} f_m(x)$ together with [continuity](/page/Continuity) of the absolute value: \begin{align*} |f_n(x) - f(x)| &= \lim_{m \to \infty} |f_n(x) - f_m(x)| \leq \varepsilon. \end{align*} Since this bound holds for all $x \in E$, we have $\|f_n - f\|_\infty \leq \varepsilon$ for all $n \geq N$, which is [uniform convergence](/page/Uniform%20Convergence). [guided] We have the pointwise limit $f(x) = \lim_{m \to \infty} f_m(x)$ from the previous step, but we need uniform convergence: a single $N$ such that $|f_n(x) - f(x)| \leq \varepsilon$ for all $n \geq N$ and all $x \in E$ simultaneously. The key idea is to reuse the uniform Cauchy bound. Fix $\varepsilon > 0$ and let $N$ be the index from the uniformly Cauchy condition, so $|f_n(x) - f_m(x)| < \varepsilon$ for all $m, n \geq N$ and all $x \in E$. Now fix $n \geq N$ and $x \in E$. For every $m \geq N$, $|f_n(x) - f_m(x)| < \varepsilon$. Send $m \to \infty$: since $f_m(x) \to f(x)$ and the absolute value is continuous, \begin{align*} |f_n(x) - f(x)| &= \lim_{m \to \infty} |f_n(x) - f_m(x)| \leq \varepsilon. \end{align*} The bound $\leq \varepsilon$ (rather than $< \varepsilon$) arises because the limit of quantities strictly less than $\varepsilon$ may equal $\varepsilon$. Crucially, the bound is independent of $x$ -- it holds for all $x \in E$ -- because the original Cauchy condition was uniform. Therefore $\|f_n - f\|_\infty \leq \varepsilon$ for all $n \geq N$. [/guided] [/step] [step:Combine both directions to conclude the equivalence] By the first step, uniform convergence implies the uniform Cauchy condition. By the second and third steps, the uniform Cauchy condition implies uniform convergence. Therefore $(f_n)$ converges uniformly on $E$ if and only if $(f_n)$ is uniformly Cauchy on $E$. [/step]