Understand Proof of the space of Bounded function from a topological space $X$ to a normed vector space is Complete Let $B(X,E)$ denote the space of all bounded functions from topological space $X$ to a complete normed vector space $E$. And Suppose $B(X, E)$ is endowed with the sup norm. Show it is Complete.

Start with a Cauchy sequence $\{f_n\}\in B(X,E)$ and pick any arbitrary $x\in X$. Let $\epsilon>0$ then $\exists N>0$ such that $|f_n(x)-f_m(x)|\leq \|f_n-f_m\|< \epsilon$. So, $\{f_n(x)\}$ is Cauchy in $E$, since $E$ is complete, there exists a function $f:X\to E$ such that $\lim_{n\to\infty}f_n(x)=f(x)$.
Now Claim that $f_n\to f$ in sup norm. Let $\epsilon>0$ since $\{f_n\}$ is cauchy there exists $N$ such that $\|f_{N+1}-f_n\|<\frac{\epsilon}{2}$ for $n>N$. Then for any $x\in X$ we have $|f_{N+1}(x)-f(x)|=\lim_{n\to \infty}|f_{N+1}(x)-f_n(x)|$
Hence $|f_{N+1}(x)-f(x)|<\frac{\epsilon}{2}$. Since $x$ was arbitrary and $N$ do not depend on x, we have $\|f_{n}-f\|<\epsilon$ for all $n>N$. Which verifies the proof of convergence.

Now I have some questions, how did the solution conclude that $N$ doesn't depend on $x$? was it Because the $N$ was chosen based on "Cauchyness"? And how does the uniform convergent been concluded from showing the quality hold for some fixed $N+1$? And why the $\frac{\epsilon}{2}$? I dont see how that is useful here.
Thanks for you help!
 A: The condition that $N$ satisfies is that for all $m, n > N, \|f_n - f_m\| < \epsilon$. This has nothing to do with any fixed value $x$. Instead $$\|f_n - f_m\| = \sup \{|f_n(y) - f_m(y)| : y \in X\}$$
There is no $x$. There is not even a $y$, as $y$ is a dummy variable of the notation. This is why $N$ does not depend on $x$.
We know that such an $N$ must exist because it is given that $\{f_n\}$ is Cauchy.
By the proof, we know that $|f_{N+1}(x) - f(x)| \le \frac \epsilon 2$ (the strict inequality in the proof is a minor error). Now because $N$ does not depend on $x$, we can extend that to say, $\forall x \in X, |f_{N+1}(x) - f(x)| \le \frac \epsilon 2$. So $\frac \epsilon 2$ is an upper bound for $\{|f_{N+1}(x) - f(x)| : x \in X\}$. Hence, $f_{N+1} - f$ is a bounded function, and has a least upper bound $\|f_{N+1} - f\|$. As $f_{N+1}$ is also bounded, so is their difference $f = f_{N+1} - (f_{N+1} - f)$. So $f \in B(X,E)$.
But these are statements about a particular $f_{N+1}$. This is still not enough to show $f_n \to f$. For that, we note that for all $n > N, \|f_n - f_{N+1}\| < \frac \epsilon 2$ and $\|f_{N+1} - f\| \le \frac \epsilon 2$. so by the triangle inequality, $\|f_n - f\| < \epsilon$ as required.
