# Completeness

• Jan 11th 2010, 09:53 PM
Drexel28
Completeness
I am supposed to prove that the set of all bounded real functions on a metric space $X$ with metric $d(x,y)=\left\|f-g\right\|_{\infty}=\sup_{x\in X}\left|f(x)-g(x)\right|$ is complete.

So we have some Cauchy sequence $\left\{f_n\right\}_{n\in\mathbb{N}}$. If we fix $\xi\in X$ we can clearly see that $\left\{f_n(\xi)\right\}_{n\in\mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$ and thus there exists some $\alpha\in\mathbb{R}$ such that $f_n(\xi)\to\alpha$. So, define $\bar{f}:X\mapsto\mathbb{R}$ by $\bar{f}(\xi)=\lim\text{ }f_n(\xi)$. It is an elementary concept from analysis that a Cauchy sequence of functions converges uniformly. Thus, we can make $\left|f_n(x)-\bar{f}(x)\right|<\varepsilon$ by making $n$ sufficiently large regardless of what value $x$ is. Thus, for every $\varepsilon>0$ there exists some $N\in\mathbb{N}$ such that $n\leqslant N\implies \left|f_n(x)-\bar{f}(x)\right|<\varepsilon,\text{ }\forall x\in X$ and so $n\leqslant N\implies \left\|f_n(x)-\bar{f}(x)\right\|_{\infty}<\varepsilon$. So, it remains to show that $\bar{f}(x)$ is bounded. But, this is trivial, since there exists some $n\in\mathbb{N}$ such that $\left|\bar{f}(x)\right|-\left|f_n(x)\right|\leqslant\left|\bar{f}(x)-f_n(x)\right|\leqslant 1$ for every $x\in X$ and thus $\left|\bar{f}(x)\right|\leqslant\left|f_n(x)\right |+1$ but since $f_n(x)$ is bounded for every $n$ we see that there exists some $M\in\mathbb{R}$ such that $\left|\bar{f}(x)\right|\leqslant M+1$ and thus $\bar{f}(x)$ is bounded and thus it is in our metric space.

So, we may conclude that this space is complete.

How's that look?
• Jan 12th 2010, 12:30 AM
Shanks
It is correct except some typo.
• Jan 12th 2010, 07:17 AM
Drexel28
Quote:

Originally Posted by Shanks
It is correct except some typo.

Thanks, Shanks! What specific typos?