> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. Then > $ > BC(X, \real|\complex) = B(X, \real|\complex) \cap C(X, \real|\complex) > $ > is the space of **[[Space of Bounded Functions|bounded]] [[Space of Continuous Functions|continuous]]** functions. > [!theorem] > > Let $(X, \topo)$ be a topological space, then $BC(X) \subset B(X)$ is [[Closed Set|closed]] with respect to the [[Uniform Norm|uniform metric]]. Moreover, $BC(X)$ is [[Complete Metric Space|complete]]. > > *Proof*. Let $\seq{f_n} \subset BC(X)$ be a [[Cauchy Sequence|Cauchy sequence]], $f \in B(X)$ be its [[Limit|limit]], and $\varepsilon > 0$. Let $x \in X$, then since $f_n$ is continuous at $x$. Take $N \in \nat$ such that $\norm{f - f_n} < \varepsilon/3$ for all $n \ge N$, and $U \in \cn(x)$ be a [[Neighbourhood|neighbourhood]] such that $|f_n(y) - f_n(x)| < \varepsilon/3$ for all $y \in U$, then > $ > \begin{align*} > |f(x) - f(y)| &\le |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| \\ > &< \varepsilon \quad \forall n \ge N, y \in U > \end{align*} > $ > and $f$ is also [[Continuity|continuous]] at $x$. Since $f$ is continuous at all points $x \in X$, $f$ is continuous and $f \in BC(X)$. Therefore $BC(X)$ is complete.