How to prove $B(X)$ is complete? 
In the Folland textbook P121 claims:
  The function $\Vert f-g\Vert_u$(uniform norm) is easily seen to be a metric on $B(X)$(all bounded functions) and $B(X)$ is complete. But how to prove it?

Suppose $\{f_n\}$ is uniformly Cauchy, that is, $d(f_m, f_n)\to 0$, if $m, n$ is enough large. But why $\Vert f_n-f\Vert_u\to 0$?
 A: @Math1000 and @TYG. In the Folland's book $B(X) : = B(X; \mathbb{C})$ the space of all bounded complex (or real) valued functions on $X$ (Here $X$ can be any set). In general, one can consider $X$ a topological space. By definition, $\| f \|_{u} = \sup_{x \in X} |f(x)|$. @Math1000 ansewed that one can consider the pointwise limit function $f(x) = \lim_{n} f_n(x)$ as a candidate. But he no mentioned that in the Folland's book $(f_n)_n$
is uniformly Cauchy. So, 
for any $\epsilon > 0$ there exists $n_0 > 0$ such that for all $x \in X$, we have
$|f_n(x) - f_m(x)| < \epsilon$ for all $m,n > n_0 \in \mathbb{N}$. Thus, for all $x \in X$, 
$$ |f(x) - f_n(x)| = \lim_m |f_m(x) - f_n(x)| \leq  \lim_m \| f_n - f_m \|_u  < \epsilon, $$  if $n > n_0$. Thus, 
$\|f - f_n \|_{u} < \epsilon$ if $n > n_0$, where $n_0$ no depends on $x$. Thus, the convergence is uniform.  
Furthermore, (for completude of the this statement) we can prove  that if  $B_{\alpha}(X; M)$ denotes the set of the maps $f:X \rightarrow M$ such that $ \sup_{x \in X} d(f(x), \alpha(x)) < \infty$ where $M$ is a (complete) metric space, then $B_{\alpha}(X; M)$  is also complete for any set $X$ and any $\alpha: X \rightarrow M $.  
