----- > [!proposition] Proposition. ([[continuous iff uniformly continuous on compact spaces]]) > Let $X,Y$ be [[metric space|metric spaces]]; $f:X \to Y$ a function. Certainly if $f$ is [[uniformly continuous]], then $f$ is [[continuous]]. If $X$ is [[compact]], then the converse holds. ^proposition > [!proof]- Proof. ([[continuous iff uniformly continuous on compact spaces]]) > > Fix $\varepsilon>0$. > > By [[continuous|continuity]] of $f$, for any $x \in X$ there exists $\delta_{x}>0$ such that $d\big( f(x), f(y) \big)< \frac{\varepsilon}{2}$ whenever $d(x,y)<\delta_{x}$. The sets $\left\{ U_{x}:=B_{ \frac{\delta_{x}}{2}}(x) \right\}$ form an [[cover|open cover]] of $X$; take $U_{1},\dots,U_{n}$ a finite subcover using compactness. Put $\delta:=\min\left\{ \frac{\delta_{x_{i}}}{2} \right\}$. > > Now suppose $d(x,y)<\delta$. We know $x \in U_{x_{i}}$ for some $i$, so that $d(x, x_{i}) < \frac{\delta_{x_{i}}}{2}$. Also, $d(x,y) < \delta< \frac{\delta_{x_{i}}}{2}$. Summing, we see $\underbrace{ d(x,x_{i})+ d(y, x) }_{ \geq d(x_{i}, y) } \leq \delta_{x_{i}},$ meaning $d(x_{i}, y) \leq \delta_{x_{i}}.$ > Now continuity of $f$ entails $d\big( f(x_{i}), f(y) \big)< \frac{\varepsilon}{2}$. Put together with the fact that $d\big( f(x), f(x_{i}) \big) < \frac{\varepsilon}{2}$ , we get $d\big( f(x), f(x_{i}) \big) + d\big( f(x_{i}), f(y) \big) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}=\varepsilon,$ > and the result follows from the triangle inequality. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```