> [!definition] > > Let $X$ be a [[Topological Space|topological space]], then the **topology of [[Uniform Convergence|uniform convergence]] on [[Compactness|compact]] sets** is the topology generated by sets of the form > $ > B_{K}(f, n) = \bracs{g \in \complex^X: \sup_{x \in K}\abs{g(x) - f(x)} < \frac{1}{n}} > $ > where $f \in \complex^X$, $K$ compact, and $n \in \nat$. > [!theorem] > > Let $X$ be a topological space and $f, g \in \complex^X$. If $f \in B_K(g, n)$, then there exists $m \in \nat$ such that $f \in B_K(f, m) \subset B_K(g, n)$. > > *Proof*. Let $\varepsilon = \sup_{x \in K}|g(x) - f(x)|$, then there exists $m \in \nat$ with $\frac{1}{m} < \varepsilon$. Take $B_K(f, m)$, 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 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]].