[proofplan] We first record that the supremum formula defines a norm on the [vector space](/page/Vector%20Space) of bounded $K$-valued functions. To prove completeness, we take an arbitrary [Cauchy sequence](/page/Cauchy%20Sequence) in the supremum norm and use the scalar completeness of $K=\mathbb R$ or $K=\mathbb C$ to define a pointwise limit. The Cauchy estimate is uniform in $x$, so passing one index to the pointwise limit gives [uniform convergence](/page/Uniform%20Convergence) to the limit function. Finally, a uniform limit of bounded functions is bounded by comparing it with one fixed bounded term of the sequence. [/proofplan] [step:Verify that the supremum formula defines a norm on $B(X;K)$] The pointwise sum and scalar multiple of bounded functions are bounded, since for $f,g\in B(X;K)$ and $\lambda\in K$, \begin{align*} |(f+g)(x)|\le |f(x)|+|g(x)|\le \|f\|_\infty+\|g\|_\infty \end{align*} for every $x\in X$, and \begin{align*} |(\lambda f)(x)|=|\lambda|\,|f(x)|\le |\lambda|\,\|f\|_\infty \end{align*} for every $x\in X$. The remaining vector-space axioms follow pointwise from the corresponding field axioms in $K$. Hence $B(X;K)$ is a vector space over $K$ under pointwise operations. The quantity $\|f\|_\infty$ is finite for every $f\in B(X;K)$ by boundedness. Non-negativity follows from $|f(x)|\ge 0$ for every $x\in X$. If $\|f\|_\infty=0$, then $|f(x)|=0$ for every $x\in X$, so $f(x)=0$ for every $x\in X$; if $X=\varnothing$, the unique function $X\to K$ is the zero function. Thus $\|f\|_\infty=0$ holds exactly for the zero function. For $\lambda\in K$ and $f\in B(X;K)$, if $X\ne\varnothing$, then \begin{align*} \|\lambda f\|_\infty=\sup_{x\in X}|\lambda|\,|f(x)|=|\lambda|\sup_{x\in X}|f(x)|=|\lambda|\,\|f\|_\infty. \end{align*} If $X=\varnothing$, both sides are $0$ by the stated convention. Similarly, for $f,g\in B(X;K)$, the scalar triangle inequality gives \begin{align*} |(f+g)(x)|\le |f(x)|+|g(x)|\le \|f\|_\infty+\|g\|_\infty \end{align*} for every $x\in X$, and taking the supremum over $x\in X$ gives \begin{align*} \|f+g\|_\infty\le \|f\|_\infty+\|g\|_\infty. \end{align*} When $X=\varnothing$, this inequality reads $0\le 0$. Therefore $\|\cdot\|_\infty$ is a norm on $B(X;K)$. [/step] [step:Define the pointwise limit of a supremum-norm Cauchy sequence] Let $(f_n)_{n\in\mathbb N}$ be a Cauchy sequence in $(B(X;K),\|\cdot\|_\infty)$. If $X=\varnothing$, then $B(X;K)$ consists only of the unique empty function, so every sequence is constant and hence convergent. We therefore assume $X\ne\varnothing$ for the rest of the proof. Fix $x\in X$. Since $(f_n)_{n\in\mathbb N}$ is Cauchy in the supremum norm, for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that \begin{align*} \|f_n-f_m\|_\infty<\varepsilon \end{align*} whenever $n,m\ge N$. For such $n,m$, the definition of the supremum norm gives \begin{align*} |f_n(x)-f_m(x)|\le \|f_n-f_m\|_\infty<\varepsilon. \end{align*} Thus $(f_n(x))_{n\in\mathbb N}$ is a Cauchy sequence in $K$. By the standard completeness theorem for the scalar fields $\mathbb R$ and $\mathbb C$, the scalar field $K$ is complete. Hence there exists a unique scalar in $K$, denoted $f(x)$, such that \begin{align*} \lim_{n\to\infty} f_n(x)=f(x). \end{align*} This defines a function \begin{align*} f:X\to K. \end{align*} [guided] Let $(f_n)_{n\in\mathbb N}$ be a Cauchy sequence in the normed space $(B(X;K),\|\cdot\|_\infty)$. The first question is: how can a candidate limit function be constructed? Since the elements are functions, the natural construction is pointwise. If $X=\varnothing$, there is exactly one function from $X$ to $K$, namely the empty function. This function is the zero vector in $B(X;K)$, and its norm is $0$ by the convention in the statement. Therefore every sequence in $B(X;K)$ is constant and converges to the empty function. We now assume $X\ne\varnothing$. Fix an arbitrary point $x\in X$. The Cauchy condition in the supremum norm says that for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that \begin{align*} \|f_n-f_m\|_\infty<\varepsilon \end{align*} whenever $n,m\ge N$. The supremum norm controls every individual point evaluation, because for every $n,m\in\mathbb N$, \begin{align*} |f_n(x)-f_m(x)|\le \sup_{y\in X}|f_n(y)-f_m(y)|=\|f_n-f_m\|_\infty. \end{align*} Therefore, whenever $n,m\ge N$, we have \begin{align*} |f_n(x)-f_m(x)|<\varepsilon. \end{align*} This proves that $(f_n(x))_{n\in\mathbb N}$ is a Cauchy sequence in the scalar field $K$. Now we use the standard completeness theorem for the scalar fields $\mathbb R$ and $\mathbb C$: every Cauchy sequence in $K$ converges to an element of $K$. Hence there exists a unique scalar, which we call $f(x)\in K$, such that \begin{align*} \lim_{n\to\infty} f_n(x)=f(x). \end{align*} Because this construction works for each $x\in X$, it defines a function \begin{align*} f:X\to K. \end{align*} At this stage we only know pointwise convergence. The next step upgrades this to convergence in the supremum norm. [/guided] [/step] [step:Pass the uniform Cauchy estimate to the pointwise limit] Let $\varepsilon>0$. Since $(f_n)_{n\in\mathbb N}$ is Cauchy in $\|\cdot\|_\infty$, there exists $N\in\mathbb N$ such that \begin{align*} \|f_n-f_m\|_\infty<\varepsilon \end{align*} whenever $n,m\ge N$. Fix $n\ge N$ and $x\in X$. For every $m\ge N$, \begin{align*} |f_n(x)-f_m(x)|<\varepsilon. \end{align*} Passing to the limit as $m\to\infty$ and using $f_m(x)\to f(x)$ in $K$, the standard continuity of the scalar absolute value map $z\mapsto |z|$ on $K$ gives \begin{align*} |f_n(x)-f(x)|\le \varepsilon. \end{align*} Since this holds for every $x\in X$, taking the supremum over $x\in X$ gives \begin{align*} \|f_n-f\|_\infty\le \varepsilon \end{align*} for every $n\ge N$, provided $f_n-f$ is bounded. The next step verifies that $f\in B(X;K)$, and hence this supremum norm is well-defined in $B(X;K)$. [/step] [step:Show that the pointwise limit is bounded] Choose $N_1\in\mathbb N$ such that \begin{align*} \|f_n-f_m\|_\infty<1 \end{align*} whenever $n,m\ge N_1$. Fix $x\in X$. For every $m\ge N_1$, \begin{align*} |f_{N_1}(x)-f_m(x)|<1. \end{align*} Letting $m\to\infty$ gives \begin{align*} |f_{N_1}(x)-f(x)|\le 1. \end{align*} By the scalar triangle inequality, \begin{align*} |f(x)|\le |f_{N_1}(x)|+|f(x)-f_{N_1}(x)|\le \|f_{N_1}\|_\infty+1. \end{align*} The bound on the right is independent of $x$. Therefore $f$ is bounded, so $f\in B(X;K)$. [/step] [step:Conclude convergence in the supremum norm] Let $\varepsilon>0$. Apply the Cauchy property with $\varepsilon$ in place of the tolerance: there exists $N\in\mathbb N$ such that \begin{align*} \|f_n-f_m\|_\infty<\varepsilon \end{align*} whenever $n,m\ge N$. The pointwise limiting argument above gives \begin{align*} |f_n(x)-f(x)|\le \varepsilon \end{align*} for every $n\ge N$ and every $x\in X$. Since $f_n-f\in B(X;K)$, taking suprema gives \begin{align*} \|f_n-f\|_\infty\le \varepsilon \end{align*} for every $n\ge N$. To obtain the strict convergence criterion, let $\delta>0$ be arbitrary and repeat the same argument with tolerance $\delta/2$. Then there exists $N_\delta\in\mathbb N$ such that \begin{align*} \|f_n-f\|_\infty\le \frac{\delta}{2}<\delta \end{align*} for every $n\ge N_\delta$. Hence $f_n\to f$ in the supremum norm. Since every Cauchy sequence in $B(X;K)$ converges to an element of $B(X;K)$, the normed space $(B(X;K),\|\cdot\|_\infty)$ is complete. Therefore it is a [Banach space](/page/Banach%20Space) over $K$. [/step]