[proofplan] We prove completeness directly. Starting from a [Cauchy sequence](/page/Cauchy%20Sequence) in the [uniform norm](/page/Uniform%20Norm), we first construct a pointwise scalar limit using completeness of $\mathbb{F}$. The same [Cauchy estimates](/theorems/2571) then pass to the limit and give [uniform convergence](/page/Uniform%20Convergence) to the resulting function. Finally, we prove the pointwise limit is continuous by the standard $\varepsilon/3$ argument and verify that continuous scalar-valued functions on compact spaces have finite uniform norm. [/proofplan] [step:Show continuous scalar-valued functions on $K$ have finite uniform norm] Let $h: K \to \mathbb{F}$ be continuous. If $K = \varnothing$, then the convention in the theorem statement gives $\|h\|_\infty = 0$, so the uniform norm is finite. Assume now that $K \neq \varnothing$. For each $m \in \mathbb{N}$, define the open scalar ball $V_m := \{z \in \mathbb{F} : |z| < m\}$. The family $(V_m)_{m=1}^\infty$ covers $\mathbb{F}$, so the family $(h^{-1}(V_m))_{m=1}^\infty$ is an open cover of $K$. Since $K$ is compact and nonempty, there exist $r \in \mathbb{N}$ and $m_1,\dots,m_r \in \mathbb{N}$ such that \begin{align*} K \subset \bigcup_{j=1}^{r} h^{-1}(V_{m_j}). \end{align*} Define $M := \max\{m_1,\dots,m_r\}$. Then for every $x \in K$, there exists $j \in \{1,\dots,r\}$ such that $h(x) \in V_{m_j}$, and hence \begin{align*} |h(x)| < m_j \le M. \end{align*} Therefore $\|h\|_\infty \le M < \infty$. Thus the uniform norm is finite on every element of $C(K;\mathbb{F})$. [/step] [step:Construct the pointwise limit of a uniformly Cauchy sequence] Let $(f_n)_{n=1}^{\infty}$ be a Cauchy sequence in $(C(K;\mathbb{F}),\|\cdot\|_\infty)$. Fix $x \in K$. For every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n,m \ge N$, \begin{align*} \|f_n - f_m\|_\infty < \varepsilon. \end{align*} By the definition of the uniform norm, this implies \begin{align*} |f_n(x)-f_m(x)| \le \|f_n-f_m\|_\infty < \varepsilon. \end{align*} Thus $(f_n(x))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{F}$. Since $\mathbb{F}$ is complete, it converges. Define the function $f: K \to \mathbb{F}$ by \begin{align*} f(x) := \lim_{n \to \infty} f_n(x) \end{align*} for every $x \in K$. [/step] [step:Upgrade the pointwise limit to uniform convergence] We prove that $f_n \to f$ in the uniform norm. Let $\varepsilon > 0$. Since $(f_n)_{n=1}^{\infty}$ is Cauchy in $\|\cdot\|_\infty$, there exists $N \in \mathbb{N}$ such that for all $n,m \ge N$, \begin{align*} \|f_n-f_m\|_\infty < \varepsilon. \end{align*} Fix $n \ge N$ and $x \in K$. Then for every $m \ge N$, \begin{align*} |f_n(x)-f_m(x)| \le \|f_n-f_m\|_\infty < \varepsilon. \end{align*} Taking the limit as $m \to \infty$ in $\mathbb{F}$ and using $f_m(x) \to f(x)$ gives \begin{align*} |f_n(x)-f(x)| \le \varepsilon. \end{align*} Since this bound holds for every $x \in K$, we have \begin{align*} \|f_n-f\|_\infty \le \varepsilon \end{align*} for every $n \ge N$. Hence $\|f_n-f\|_\infty \to 0$. [guided] The pointwise limit was constructed one point at a time, so we must still prove that the convergence is uniform. The Cauchy hypothesis is exactly the uniform information needed for this. Let $\varepsilon > 0$. Because $(f_n)_{n=1}^{\infty}$ is Cauchy with respect to $\|\cdot\|_\infty$, there exists $N \in \mathbb{N}$ such that for all $n,m \ge N$, \begin{align*} \|f_n-f_m\|_\infty < \varepsilon. \end{align*} By the definition of $\|\cdot\|_\infty$, this means that for every $x \in K$ and every $n,m \ge N$, \begin{align*} |f_n(x)-f_m(x)| \le \|f_n-f_m\|_\infty < \varepsilon. \end{align*} Now fix $n \ge N$ and $x \in K$. The sequence $(f_m(x))_{m=1}^{\infty}$ converges in $\mathbb{F}$ to $f(x)$ by the definition of $f$. Since the absolute value map $z \mapsto |z|$ is continuous on $\mathbb{F}$, taking $m \to \infty$ in the inequality \begin{align*} |f_n(x)-f_m(x)| < \varepsilon \end{align*} gives \begin{align*} |f_n(x)-f(x)| \le \varepsilon. \end{align*} This estimate is independent of $x$. Therefore \begin{align*} \|f_n-f\|_\infty = \sup_{x \in K} |f_n(x)-f(x)| \le \varepsilon \end{align*} for every $n \ge N$. Since every positive $\varepsilon$ is allowed, this proves $\|f_n-f\|_\infty \to 0$. [/guided] [/step] [step:Prove the uniform limit is continuous] We prove that $f: K \to \mathbb{F}$ is continuous. Let $x_0 \in K$ and let $\varepsilon > 0$. Since $\|f_n-f\|_\infty \to 0$, choose $N \in \mathbb{N}$ such that \begin{align*} \|f_N-f\|_\infty < \frac{\varepsilon}{3}. \end{align*} The function $f_N: K \to \mathbb{F}$ is continuous at $x_0$, so the set \begin{align*} U := f_N^{-1}\left(\{z \in \mathbb{F} : |z-f_N(x_0)| < \frac{\varepsilon}{3}\}\right) \end{align*} is an open neighbourhood of $x_0$ in $K$. For every $x \in U$, the triangle inequality gives \begin{align*} |f(x)-f(x_0)| \le |f(x)-f_N(x)| + |f_N(x)-f_N(x_0)| + |f_N(x_0)-f(x_0)|. \end{align*} The first and third terms are bounded by $\|f_N-f\|_\infty$, and the middle term is less than $\varepsilon/3$ by the definition of $U$. Therefore \begin{align*} |f(x)-f(x_0)| < \varepsilon. \end{align*} Thus $f$ is continuous at $x_0$. Since $x_0 \in K$ was arbitrary, $f \in C(K;\mathbb{F})$. [/step] [step:Conclude that every Cauchy sequence converges in $C(K;\mathbb{F})$] First verify that $\|\cdot\|_\infty$ is a norm on $C(K;\mathbb{F})$. The first step shows that $\|h\|_\infty < \infty$ for every $h \in C(K;\mathbb{F})$. For $h \in C(K;\mathbb{F})$, nonnegativity follows from $|h(x)| \ge 0$ for all $x \in K$. If $\|h\|_\infty = 0$, then $|h(x)| \le 0$ for every $x \in K$, so $h(x)=0$ for every $x \in K$; conversely the zero function has uniform norm $0$ by the definition and by the empty-set convention when $K=\varnothing$. For $\lambda \in \mathbb{F}$ and $h \in C(K;\mathbb{F})$, the identity $|\lambda h(x)| = |\lambda|\,|h(x)|$ for every $x \in K$ gives $\|\lambda h\|_\infty = |\lambda|\,\|h\|_\infty$. For $g,h \in C(K;\mathbb{F})$, the scalar triangle inequality gives $|g(x)+h(x)| \le |g(x)|+|h(x)| \le \|g\|_\infty+\|h\|_\infty$ for every $x \in K$, hence $\|g+h\|_\infty \le \|g\|_\infty+\|h\|_\infty$. Thus $(C(K;\mathbb{F}),\|\cdot\|_\infty)$ is a [normed vector space](/page/Normed%20Vector%20Space). By the first step, the [continuous function](/page/Continuous%20Function) $f: K \to \mathbb{F}$ has finite uniform norm. Hence $f \in C(K;\mathbb{F})$ as an element of the normed space under consideration. The previous steps show that the arbitrary Cauchy sequence $(f_n)_{n=1}^{\infty}$ converges to $f$ in the norm $\|\cdot\|_\infty$. Therefore every Cauchy sequence in $(C(K;\mathbb{F}),\|\cdot\|_\infty)$ converges in that space. Hence $(C(K;\mathbb{F}),\|\cdot\|_\infty)$ is complete, so it is a [Banach space](/page/Banach%20Space). [/step]