A map between [[Metric Space|metric spaces]] $f:(X,d_{X})\to (Y, d_{Y})$ is said to be *uniformly continuous* iff for every $\varepsilon>0$, there exists $\delta>0$ such that for all $x_{1}, x_{2} \in X$, $d_{X}(x_{1},x_{2})<\delta \implies d_{Y}(f(x_{1}), f(x_{2}))<\varepsilon.$ The following lemma generalises [[Continuous Real Function on Closed Real Interval is Uniformly Continuous]]. > [!NOTE] Lemma > Any [[Continuity|continuous map]] from a [[Compactness|compact]] metric is uniformly continuous. **Proof**: Let $f: X\to Y$ be continuous and choose $\varepsilon>0$. For $z\in X$, define $U_{z} = f^{-1}\left( B_{Y}\left( f(z), \frac{\varepsilon}{2} \right) \right).$Then the collection $\mathcal{U} = \{ U_{z}: z\in X \}$ is an open cover of $X$. By [[Lebesgue Number Lemma]], there exists $\delta>0$ such that for all $x\in X$, there exists $z$ such that $B(x, \delta) \subset U_{z}$. So if $x, y\in X$ and $d_{X}(x,y)< \delta$ we have $y \in B(x, \delta)\subset f^{-1}\left( B_{Y}\left( f(z), \frac{\varepsilon}{2} \right) \right)$ thus $d_{Y}(f(x), f(y)) \leq d_{Y}(f(x), f(z))+ d_{Y}(f(z), f(y))< \varepsilon. $$\square$