the space of continuous functions is complete

Solution 1:

I will use the notation $C$ for $(C(X\to Y), d_{\infty})$. Let $\{f_n\}$ be a Cauchy sequence in $C$.

Then for each $\epsilon \gt 0$, there is $N\in \mathbb Z^+$ s.t. for all $m,n\gt N$, $||f_n-f_m||_{\infty} \lt \epsilon$.

Thus for all $x\in X$, $\{f_n(x)\}$ is a Cauchy sequence in $Y$. Since $Y$ is complete define $f:X\to Y$ as

$f(x)=\lim\limits_{n\to\infty}f_n(x)$. Then clearly $f_n$ converges to $f$.

The only thing we have to show is the continuity of $f$.

Let $t\in X$ and let $\epsilon\gt0$.

Since $f(t)=\lim\limits_{n\to\infty}f_n(t)$ there is $N\in \mathbb Z^+$ s.t for all $n\gt N$, $||f_n-f||_{\infty}\lt\frac{\epsilon}{3} \qquad (*)$

Choose $n_0\gt N$. Then since $f_{n_0}$ is continious there exist $\delta\gt0$ s.t for all $x\in X$ $|x-t|\lt \delta\implies |f_{n_0}(x)-f_{n_0}(t)|\lt \frac{\epsilon}{3} \qquad (**)$

then by $(*),(**)$

for all $x\in X$, if $|x-t|\lt \delta$ then

$\begin{split}|f(x)-f(t)|& =|f(x)-f_{n_0}(x)+f_{n_0}(x)-f_{n_0}(t)+f_{n_0}(t)-f(t)|\\ &\lt |f(x)-f_{n_0}(x)|+|f_{n_0}(x)-f_{n_0}(t)|+|f_{n_0}(t)-f(t)|\\ &\lt \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} =\epsilon\\ \end{split}$.

Thus $f$ is continuous.

Solution 2:

Since convergence in the sup norm metric is equivalent to uniform convergence, and uniform convergence preserves continuity and boundedness, it remains to show that every Cauchy sequence in $C(X\to Y)$ converges in the sup norm metric.

Let $\{f_n\}$ be a Cauchy sequence in $C(X\to Y)$, then for all $\epsilon >0$ there is an $M$ such that
$m,n\geq M$ implies $d_Y(f_m(x),f_n(x))\leq d_{\infty}(f_m,f_n)\leq\epsilon$ for all $x\in X$, which shows $\{f_n(x)\}_{n\in\mathbb{N}}$ is a Cauchy sequence in $Y$. By the completeness of $Y$, $\{f_n(x)\}_{n\in\mathbb{N}}$ converges to some $f(x)\in Y$ for all $x\in X$.

Since $\{f_n\}$ is a Cauchy sequence, for every $\epsilon >0$ there is an $N\in\mathbb{N}$ such that $m,n\geq N$ implies $d_{\infty}(f_m,f_n)\leq\epsilon$. Let $x\in X$ be arbitrary. Since $\{f_n\}$ converges pointwise to $f$, there is a $K\in\mathbb{N}$ such that $n\geq K$ implies $d_Y(f_n(x),f(x))\leq\epsilon$. Take $m=\max \{N,K\}$, then for all $n\geq N$ we have\begin{align}d_Y(f_n(x),f(x))\leq d_Y(f_n(x),f_m(x))+d_Y(f_m(x),f(x))\leq d_{\infty}(f_m,f_n)+\epsilon\leq2\epsilon, \end{align} which means $d_{\infty}(f_n,f)\leq 2\epsilon$ for all $n\geq N$ since $x\in X$ is arbitrary. This yields the claim.