# Statement(s) > [!NOTE] Statement 1 ($C(\mathbb{R}^n)$ forms Banach Space) > Let $C_{b}$ denote the set of [[Bounded Real Function|bounded]], [[Continuous Real Function|continuous real functions]]. Then $(C_{b},\lVert \cdot \rVert_{\infty})$ is a [[Complete metric spaces|complete metric space]]. # Proof(s) **Proof of statement 1:** Notice that a sequence in $C_{b}$ is Cauchy iff it is [[Uniformly Cauchy Sequence of Real Functions|uniformly Cauchy]]. It follows from [[Sequence of Real Functions is Uniformly Convergent iff Uniformly Cauchy]] that the limit of any Cauchy sequence exists. It follows from [[Uniform Limit of Sequence of Continuous Real Function is Continuous]] that this limit function is continuous. To see that this limit function is continuous, notice that for every $x$ in its domain $|f(x)|\leq |f(x)-f(x_{n})|+|f_{n}(x)|$for every $n.$ If $f_{n}\rightrightarrows f,$ then there exist $n$ large enough for $|f_{n}(x)-f(x)|<1.$ For that $n$, since $f_{n}$ is bounded we have $\lVert f_{n}\rVert_{\infty}\leq M$ for some $M>0.$ Thus $\lVert f(x)\rVert_{\infty}\leq M+1,$ that is, $f\in C_{b}$ $\blacksquare$