---- > [!definition] Definition. ([[topology of compact convergence]]) > Let $X$ be a [[topological space]] and $(Y,d)$ a [[metric space]]. Given $f \in Y^{X}$, $K$ a [[compact]] [[subspace topology|subspace]] of $X$, and $\varepsilon>0$, define $B_{K}(f, \varepsilon):= \{ g \in Y^{X}: \rho |_{K}(f, g) <\varepsilon \},$ where $\rho:Y \times Y \to [0, \infty]$ is the map $\rho(f,g)=\sup_{x \in X} d\big( f(x), g(x) \big)$ and $\rho |_{K}$ denotes its restriction to $K \times K$. The sets $B_{K}(f, \varepsilon)$ form a [[basis for a topology]] on $X$, [[topology generated by a basis|called]] the **topology of compact convergence** or the **topology of uniform convergence on compact sets**. ^definition > [!proposition] On compact convergence. > By construction, a [[sequence]] $(f_{n})$ in $Y^{X}$ converges to $f$ with respect to the topology of compact convergence if and only if for all $K \subset X$ [[compact]], the sequence $(f_{n} |_{K}:K \to Y)$ [[uniform convergence|converges]] to $f$ in the [[uniform metric|topology of uniform convergence]] on $K$. > > > [!proof]+ Proof. > > Indeed, $(f_{n}) \to f$ in the topology of compact convergence if and only if for all pairs $(K, \varepsilon)$ there exists $N$ such that $f_{n} \in B_{K}(f ,\varepsilon)$ for all $n \geq N$, that is, $\sup_{x \in K}d(f_{n}(x),f(x))< \varepsilon \text{ for all }n \geq N.$ > > This is precisely the statement that for all [[compact]] $K \subset X$, $f_{n} \to f$ [[uniform convergence|uniformly]] on $K$. > > > [!justification] > To see that $\{ B_{K}(f, \varepsilon) : \varepsilon>0, f \in Y^{X} , K \text{ compact}\}$ is indeed a [[basis for a topology]] note that [[cover|covering]] is obvious and if $h \in B_{K}(f, \varepsilon) \cap B_{K'}(f', \varepsilon')$ then setting $\begin{align} > \hat{\varepsilon}&:= \min \{ \varepsilon- \rho |_{K}(f,h) , \varepsilon'- \rho |_{K'}(f', h) \} \\ > \hat{K}& := K \cup K' > \end{align}$ > we have $B_{\hat{K}}(h, \hat{\varepsilon}) \subset B_{K}(f, \varepsilon) \cap B_{K'}(f', \varepsilon').$ > What we want to show is that for any $g \in B_{\hat{K}}(h, \hat{\varepsilon})$, i.e. $g$ for which $\rho |_{K \cup K'}(h,g)< \varepsilon- \rho |_{K}(f, h) \text{ and } \rho |_{K \cup K'}(h, g) < \varepsilon'-\rho |_{K'}(f', h),$we have $\rho |_{K}(f,g)< \varepsilon$ and $\rho_{K'}(f', g)<\varepsilon'$. This is clear because $\rho |_{K}(g,f) \leq \rho |_{K}(h, g)+ \rho |_{K}(f, h) \leq \rho_{K \cup K'}(h, g)+ \rho |_{K}(f,h)<\varepsilon$ > and similarly for the primed version. (Here we have used the triangle inequality pointwise and monotonicity of the [[supremum]].) ---- #### ---- #### 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 > ```