> [!NOTE] Theorem (Uniform Continuity of Continuous Function on Closed Real Interval) >Let $[a,b]$ be a [[Closed Real Interval|closed real interval]]. > > Let $f:[a,b]\to \mathbb{R}$ be a [[Continuous Real Function|continuous real function]]. Then $f$ is [[Uniformly Continuous Real Function|uniformly continuous]] on $[a,b].$ **Proof**: BWOC, suppose $f$ is not uniformly continuous. Then there exists $\varepsilon>0,$ such that for all $n\in \mathbb{N}^{+},$ there exists $x_{n},y_{n}\in[a,b]$ with $|x_{n}-y_{n}|< \frac{1}{n} \quad \land \quad |f(x_{n})-f(y_{n})|\geq \varepsilon.$ By [[Bolzano-Weierstrass Theorem (Sequential Compactness of The Reals)|sequential compactness of the reals]], there exists a [[Real Subsequence|subsequence]] $x_{n_{k}}$ which converges to some $x\in \mathbb{R}.$ By [[Limits of Real Sequence Preserve Weak Inequalities]], $x\in[a,b].$ By the first condition, $y_{n_{k}}\to x$ as well. By continuity $f(x_{n_{k}})\to f(x) \quad \text{and} \quad f(y_{n_{k}})\to f(x)$But this means that $f(x_{n_{k}})-f(y_{n_{k}})\to 0$which contradicts the fact that $|f(x_{n})-f(y_{n})|\geq \varepsilon$ for all $n\geq 1.$ $\square$ **Proof**: Follows from [[Continuous map from a compact metric space to another metric space is uniformly continuous]].