> [!definition] > > Let $X, Y$ be [[Topological Space|topological spaces]], then the **[[Compactness|compact]]-[[Open Set|open]] topology** on $C(X, Y)$ is the topology generated by sets of the form > $ > S(K, U) = \bracs{f \in C(X, Y): f(C) \subset U} > $ > where $K \subset X$ is compact and $U \subset Y$ is open. If $Y$ is a [[Metric Space|metric space]], then the above topology is known as the **topology of uniform convergence on compact sets**, generated by the [[Base|base]] consisting of sets of the form > $ > B_K(f, \eps) = \bracs{g \in C(X, Y): \sup_{x \in K}d(f(x), g(x)) < \eps} > $ > [!theorem] > > Let $X$ be a topological space, $Y$ be a [[Metric Space|metric space]] and $f, g \in C(X, Y)$. If $f \in B_K(g, 1/n)$, then there exists $m \in \nat$ such that $f \in B_K(f, 1/m) \subset B_K(g, 1/n)$. > > *Proof*. Let $\varepsilon = \sup_{x \in K}|g(x) - f(x)|$, then there exists $m \in \nat$ with $\frac{1}{m} < \varepsilon$. Take $B_K(f, 1/m)$, then for any $h \in B_K(f, m)$, > $ > \sup_{x \in K}|h(x) - g(x)| \le \sup_{x \in K}|g(x) - f(x)| + \frac{1}{m} < \frac{1}{n} > $ > [!theorem] > > Let $X$ be a [[Locally Compact Hausdorff Space|LCH]] space, then the topology of uniform convergence on compact sets on $C(X)$ is defined by the [[Seminorm|seminorms]] $p_K(f) = \sup_{x \in K}\abs{f(x)}$, making it a [[Topological Vector Space|TVS]]. If $X$ is $\sigma$-compact, then $\complex^X$ with this topology is a [[Fréchet Space|Fréchet space]].