09 - Completeness of the Space of Bounded Continuous Functions

Completeness of the Space of Bounded Continuous Functions

Let $X$ be a metric space then

$$C_b(X)\text{ is complete}$$

where $C_b(X)$ denotes normed vector space of bounded continuous functions $f:X\to \mathbb{R}$ with

$$||f|{|}_{\infty }=\underset{x\in X}{\sup }|f(x)|$$

Proof

By Completeness of Normed Vector Spaces for Bounded Functions, $B(X)$ is complete
So by Relation between completeness and closure then need to show

$$C_b(X)\text{ is c alosed subset of }B(X)$$

By Interior points and convergent sequences then just need to show that
If $(f_n{)}_{n=1}^{\infty }$ is a sequence of elements of $C_b(X)$ then $f\in C_b(X)$ ($f$ is continuous)

Let $a\in X$ and let $ϵ>0$
Since $f_n\to f$ in the $||\cdot |{|}_{\infty }$-norm then there exists $n$ such that

$$||f_n-f||\le \frac{ϵ}{3}$$

As $f_n$ is continuous then there is $\delta >0$ such that

$$|f_n(x)-f_n(a)|<\frac{ϵ}{3}\text{ for all }x\in B(a,\delta )$$

But for $x\in B(a,\delta )$ then

$$\begin{array}{rl}|f(x)-f(a)& |\le |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|\\ & <\frac{ϵ}{3}+\frac{ϵ}{3}+\frac{ϵ}{3}=ϵ\end{array}$$

Hence $f$ is continuous at $a$ and as $a$ was arbitrary then $f$ is a continuous function on $X$