[Math] If $X$ is compact and $C(X)$ is the space of all continuous real valued functions. Prove $C(X)$ is a complete metric space.

complete-spacesgeneral-topologymetric-spacesreal-analysis

Let $X$ be a compact metric space and define $C(X)$ to be the space of all continuous real valued functions on $X$ with a metric defined by
$$d(f,g)=\sup_{x \in X} |f(x) -g(x)|.$$ Show that $C(X)$ is a complete metric space.

How can I start this problem?

Best Answer

You have to show that every Cauchy sequence $(f_n)$ of $C(X)$ converges. Let $(f_n)$ be a Cauchy sequence of $C(X)$, for every $c>0$, there exists $N>0$ such that $n,m>N$ implies that $\sup_{x\in X}\mid f_n(x)-f_m(x)\mid<c,$ this implies that $(f_n(x))$ is a Cauchy sequence. Since $R$ is complete, $f_n(x)$ converges towards $f(x)$.

It remains to show that the function defined on $X$ by $x\rightarrow f(x)$ is continuous. Firstly, remark that there exists $N>0$, such that for every $x\in X$, $n,m>N$, $\sup_{x\in X}\mid f_n(x)-f_m(x)\mid <c/4$. This implies that $$\lim_{m\to\infty}\mid f_n(x)-f_m(x)\mid =\mid f_n(x)-f(x)\mid \leq c/4$$ for $n>N$.

Let $n>N$, since $f_n$ is continuous, there exists $e>0$ such that $d(x,y)<e$ implies that $\mid f_n(x)-f_n(y)\mid< c/4$. This implies that for $d(x,y)<e$ we have: $$\mid f(x)-f(y)\mid \leq \mid f(x)-f_n(x)\mid +\mid f_n(x)-f_n(y)\mid+\mid f_n(y)-f(y)\mid < c/4+c/4+c/4=3c/4<c.$$ Henceforth, $f$ is continuous

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