Completeness of C(K) under the sup norm

Let $(K,d)$ be a compact metric space and let $C(K,\mathbb{R})$ denote the set of continuous functions $f:K\to\mathbb{R}$.

Define the sup norm by $\|f\|_\infty=\sup_{x\in K}|f(x)|,$ and the induced metric $d_\infty(f,g)=\|f-g\|_\infty=\sup_{x\in K}|f(x)-g(x)|.$

Theorem: The metric space $\bigl(C(K,\mathbb{R}),d_\infty\bigr)$ is complete .

This is the basic Banach-space fact behind many compactness and approximation arguments: uniform Cauchy sequences of continuous functions converge uniformly to a continuous function.

Proof sketch: Let $(f_n)$ be Cauchy in $d_\infty$. Then for each fixed $x\in K$, the real sequence $(f_n(x))$ is Cauchy in $\mathbb{R}$, hence converges ; define $f(x)=\lim_{n\to\infty} f_n(x).$ The uniform Cauchy property implies uniform convergence $f_n\to f$, i.e. $\|f_n-f\|_\infty\to 0$. Since each $f_n$ is continuous and the convergence is uniform, $f$ is continuous (uniform limit theorem ). Thus $f\in C(K,\mathbb{R})$ and $f_n\to f$ in $d_\infty$.