1. ## 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?

2. It is correct except some typo.

3. Originally Posted by Shanks
It is correct except some typo.
Thanks, Shanks! What specific typos?