Show the set of Bounded continuous maps from a topological space to a normed vector space is closed set in bounded maps Let $X$ be a topological space and $E$ a vector space with norm $|\cdot|$, where $B(X,E)$ denotes the set of bounded maps from $X$ to $E$ (endowed with $\|\cdot\|_\infty$)and $BC(X,E)$ be the set of the bounded continuous map.
(a)Show $BC(X,E)$ is a closed set in $B(X,E)$
(b)A Banach space is a complete normed vector space, show that if $E$ is a Banach space then $b(X, E)$ is complete.
I'm stuck on part (a),  I wanted to show the set is closed by showing it contains all of its limit points: Let $f\in B(X, E)$ be a limit of some sequence $\{f_n\}$ in $BC(X, E)$. Let $\epsilon>0$ then we have $N>0$ such that $\|f_n-f\|<\epsilon$ for all $n>N$. Now, let $x,y \in X$, then it seems that if I pick $\delta>0$ such that it satisfies the continuity of $f_n$ for each $n$, Then we always have $|f(x)-f(y)|=|f(x)-f_n(x)|+|f_n(x)-f(y)|<\|f_n-f\|+|f_n(x)-f_n(y)|<2\epsilon$ for all $n>N$. Hence, the space is closed. But I am unsure whether I can use the argument that if a space contains the limit point, it is closed. According to the chapter of the book, I am reading, closed is defined as a complement of open sets.
Also, it seems that for part (b), the statement follows from the fact that $E$ is complete.
 A: It is not clear as to how you got $|f_n(x)-f(y)| <\epsilon$. Also, you have not chosen $x$ 'close' to $y$ so you cannot expect $|f(x)-f(y)| <\epsilon$.
Since $X$ is not even metric you need a totally different argument.
Fix $x$ and consider continuity of $f$ at $x$. Choose $N$ such that $|f_N(y)-f(y)|<\epsilon$ for all $y$. Since $f_N$ is continuous there exists a neighborhood $U$ of $x$ such that $|f_n(y)-f_n(x)| <\epsilon$ for all $y \in U$. Now $|f(y)-f(x)| \leq |f(y)-f_N(y)|+|f_N(y)-f_N(x)| +|f_N(x) -f(x)|<3\epsilon$ for al $y \in U$. This proves that $f$ is continuous at $x$.
Proof of completeness of $B(X,E)$ when $E$ is complete: Let $(f_n)$ be a Cauchy sequence. Then $(f_n(x)) $ is a  Cauchy sequence in $E$ for each $x$ so $f(x)=\lim f_n(x)$ exists . To see that $f$ is bounded just note that there exists $N$ with $\|f(x)\| \leq 1+ \|f_N(x)\|$ for all $x$ and $f_N$ is bounded.
A: You are close to the right idea in your argument but you don't express it too clearly. You only need to care about one $n$ and fixed $x$. Also, there is no $\delta$ since $X$ is a topological space, possibly non-metric. So, given $\varepsilon>0$ there exists $n$ with $\|f-f_n\|<\varepsilon/3$. For that $n$ and our $x$, there exists a neighbourhood $V$ of $x$ such that $|f_n(x)-f_n(y)|<\varepsilon/3$ for all $y\in V$. Then, for any $y\in V$,
\begin{align}
|f(x)-f (y)|&≤|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f (y)|\\[0.3cm]
&\leq 2\|f -f_n\|+\frac\varepsilon3<2\frac\varepsilon3+\frac\varepsilon3=\varepsilon.
\end{align}
It is true that one defines closed as complement of open. But it is an easy exercise to show that a set is closed if and only if it contains all its limit points. Namely, if $K$ does not contain a certain limit point $x$, then $x\in K^c$ and there exists a net $\{x_j\}\subset K$ with $x_j\to x$. Then every neighbourhood of $x$ contains points of $K$, so $x$ is not interior in $K^c$, which shows that $K^c$ is not open and hence $K$ is not closed. Conversely, if $K$ contains all of its limit points, then for any $x\in K^c$ there exists a neighbourhood $V$ with $V\subset K^c$ ( for otherwise $x$ would be a limit point of $K$ and hence in $K$); that is, $K^c$ is open and therefore $K$ is closed.
As for part (b), you say

the statement follows from the fact that E is complete.

That is like saying "the consequence follows from the hypothesis". Which is true, but it is no proof. You have to show that every Cauchy sequence in $B(X,E)$ has a limit. Which follows rather easily from the fact that $E$ is complete, but you have to construct the limit and show that it is bounded.
