Show that $B(X)=\{f:X \rightarrow \mathbb{R}: f \text{ is bounded}\}$ is complete. 
Given a metric space $(X,d)$ consider the metric space $B(X)=\{f:X \rightarrow \mathbb{R}: f \text{ is bounded}\}$ with the distance $d_{\infty}(f,g)=sup_{x\in X}|f(x)-g(x)|$. Show that $(B(X),d_{\infty})$ is complete.

Ok so I am having some problems in completing this proof. I was thinking about the following: Given $((f_{n}))_{n \in \mathbb{N}}\subset B(X)$ a Cauchy sequence, I want to see there exists a function $f$ bounded, such that $f_{n}\rightarrow f$. However, I know nothing about my Cauchy sequence in $B(X)$ so there is no way I can say something about my hypotethical limit $f$. So I thought, ok, $d_{\infty}:B(X)\times B(X)\rightarrow \mathbb{R}$ is uniformly continuous. So there should be some way in which I should be able to relate my Cauchy sequence $((f_{n}))_{n \in \mathbb{N}}$ with $d_{\infty}$ to produce a Cauchy sequence in $\mathbb{R}$ wich of course is complete. 
However, I don't know if this strategy is even in the correct path. I couldn't think about anything else. Any hint or comment would be greatly appreciated!
 A: Hint: given $x \in X$, the sequence of real numbers $a_n = f_n(x)$ is a Cauchy sequence in $\mathbb{R}$. That should tell you all you need to know about the value of the desired limit function $f$ at $x$.
A: Ok, here I am with the answer to my question.
Let $(f_{n})\subset B(X)$ be a Cauchy sequence. Given $x \in X$ we have that $(f_{n}(x)) \subset \mathbb{R}$ is Cauchy (in $\mathbb{R}$). Lets see this. Since $(f_n) \subset B(X)$ is Cauchy, given $\epsilon > 0$ there exist $n_{0}\in \mathbb{N}$ such that $n,m \geq n_0$ implies $d_{\infty}(f_n,f_m)=sup_{x \in X}|f_{n}(x)-f_{m}(x)|<\epsilon$. Therefore, for any $x \in X$ (in particular the one given before) we have that $|f_{n}(x)-f_{m}(x)|\leq sup_{x \in X}|f_{n}(x)-f_{m}(x)|<\epsilon$ which implies $(f_{n}(x))\subset \mathbb{R}$ is Cauchy as we said before and since $\mathbb{R}$ is complete, $f_{n}(x) \rightarrow f(x)$. 
Now we have to check two things to complete the demonstration, first that for any $x\in X, f(x)$ is bounded and that $f_n \rightarrow f$ uniformly.
EDIT:
Ok, we want to see that $ f $ is bounded. We have that given $\epsilon =\frac {1}{2}$ for every $ x \in X $, $|f_n (x) -f_m (x)| < \frac{1}{2}$ if $ m, n > N $ with $N$ sufficiently big. Therefore, $|f(x)-f_N (x)| \leq \frac {1}{2}$ for every $ x \in X $ and therefore $ f (x)$ is bounded by $b_N + \frac {1}{2}$ were $ b_N $ is a bound for $|f_N|$
Finally we want to see that $f_n \rightarrow f$ uniformly. Since $(f_{n}) $is Cauchy in $X$, given $\epsilon>0$ there exists $n_0 \in \mathbb{N}$ such that $m,n \geq n_0$ implies $d_{\infty}(f_n,f_m)=sup_{x \in X}|f_{n}(x)-f_{m}(x)|<\epsilon$. Making $m\rightarrow \infty$, we obtain that $sup_{x \in X}|f_{n}(x)-f(x)|<\epsilon$ concluding that $f_n \rightarrow f$ uniformly. Therefore $B(X)$ is complete. 
