----- > [!proposition] Proposition. ([[totally bounded under uniform metric implies equicontinuous]]) > Let $X$ be a [[topological space]] and $(Y,d)$ a [[metric space]]. [[continuous|If]] $F \subset C(X,Y):=\text{Hom}_{\mathsf{Top}}(X,Y)$ is [[totally bounded]] under the [[uniform metric]] corresponding to $d$, then $F$ is [[equicontinuous]] under $d$. ^proposition > [!proof]- Proof. ([[totally bounded under uniform metric implies equicontinuous]]) > Assume $F \subset C(X,Y) \subset Y^{X}$ is [[totally bounded]]. Fix $0<\varepsilon<1$ and $x_{0} \in X$. Set $\delta=\frac{\varepsilon}{3}$; [[cover]] $F$ by finitely many open $\delta$-[[metric topology|balls]] $B_{\delta}(f_{1}),\dots, B_{\delta}(f_{n})$ in $C(X,Y)$. Each $f_{i}$ is [[continuous]]; therefore, we can choose a [[neighborhood]] $U$ of $x_{0}$ such that $x \in U \implies d\big( f_{i}(x), f_{i}(x_{0}) \big)<\delta$ for all $i \in [n]$. > > Now let $f \in F$ be arbitrary. Then $f$ belongs to at least one of the above $\delta$-balls, say to $B(f_{i}, \delta)$. Then for $x \in U$, we have $\begin{align} > \overline{d}\big( f(x), f_{i}(x)\big) < \delta \\ > d\big( f_{i}(x) , f_{i}(x_{0}) \big) < \delta \\ > \overline{d}\big( f_{i}(x_{0}), f(x_{0}) \big) < \delta, > \end{align}$ > where the first and third inequalities hold because $\overline{\rho}(f,f_{i})<\delta$. Since $\delta<1$, the second holds with $d$ replaces by $\overline{d}$, now triangle inequality finishes. ----- #### ---- #### 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 > ```