A Banach space is a normed vector space, where a Cauchy sequence converges {{WRT}} the supplied norm. To prove the theorem, we need to show that a Cauchy sequence in $\struct {\ell^\infty, \norm {\,\cdot\,}_\infty}$ converges. We take a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\struct {\ell^\infty, \norm {\,\cdot\,}_\infty}$. Then we consider the $k$th component and show, that a real Cauchy sequence $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$ with the limit $x^{\paren k}$ and denote the entire set as $\mathbf x$. Finally, we show that $\sequence {\mathbf x_n}_{n \in \N}$, composed of components $x_n^{\paren k},$ converges in $\struct {\ell^\infty, \norm {\,\cdot\,}_\infty}$ with the limit $\mathbf x$. Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {\ell^\infty, \norm{\, \cdot \,}_\infty}$. Denote the $k$th component of $\mathbf x_n$ by $x_n^{\paren k}$. === $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$===  Let $\epsilon >0$. Then: :$\displaystyle \exists N \in \N : \forall m,n \in \N : m,n > N : \norm {\mathbf x_n - \mathbf x_m}_\infty < \epsilon$ For same $N, m, n$ consider $\size {x_n^{\paren k} - x_m^{\paren k} } $: {{begin-eqn}} {{eqn | l = \size {x_n^{\paren k} - x_m^{\paren k} }        | o = \le       | r = \sup_{k \mathop \in \N} \size {x_n^{\paren k} - x_m^{\paren k} } }} {{eqn | r = \norm {\mathbf x_n - \mathbf x_m}_\infty       | c = {{defof|Supremum Norm}} }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} Hence, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\R, \size {\, \cdot \,}}$. From Real Number Line is Complete Metric Space, $\R$ is a complete metric space. Consequently, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$. {{qed|lemma}} Denote the limit $\displaystyle \lim_{n \mathop \to \infty} \sequence {x_n^{\paren k}}_{n \mathop \in \N} = x^{\paren k}$. Denote $\sequence {x^{\paren k}}_{k \mathop \in \N} = \mathbf x$. === $\mathbf x$ belongs to $\ell^\infty$ === From previous lemma, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\R, \size {\, \cdot \,}}$. Fix any $n > N$ and $k \in \N$. Then: :$\displaystyle \forall m > N : \size {x_n^{\paren k} - x_m^{\paren k}} < \epsilon$ Take the limit $m \to \infty$: :$\size {x_n^{\paren k} - x^{\paren k}} < \epsilon$ Since $k$ was arbitrary: {{begin-eqn}} {{eqn | ll = \forall k \in \N :       | l = \size {x_n^{\paren k} - x^{\paren k} }       | o = \le       | r =  \sup_{k \mathop \in \N} \size {x_n^{\paren k} - x^{\paren k} }        | c = {{defof|Supremum of Real Sequence}} }} {{eqn | r = \norm {\mathbf x_n - \mathbf x}_\infty       | c = {{defof|Supremum Norm}} }} {{eqn | o = <       | r = \epsilon }} {{eqn | o = <       | r = \infty }} {{end-eqn}} By definition, $\mathbf x_n - \mathbf x \in \ell^{\infty}$. By assumption, $\mathbf x_n \in \ell^\infty$. Hence, $\mathbf x \in \ell^\infty$. {{qed|lemma}} === $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ to $\mathbf x$=== From previous lemma we have that for some $\epsilon > 0$, $N \in \N$ and fixed $n > N$: :$\displaystyle \norm {\mathbf x_n - \mathbf x}_\infty < \epsilon$ Repeat the same argument for all $\epsilon \in \R_{>0}$ and note that $n$ was arbitrary: :$\displaystyle \forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {\mathbf x_n - \mathbf x}_\infty < \epsilon$ Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\ell^\infty, \norm{\, \cdot \,}_\infty}$. {{qed}}