About a subset of $\mathbb{R}^n$: The other direction of Arzela-Ascoli Theorem Arzela-Ascoli Theorem. Let $(E,d)$ be a compact metric space and denote by $C(E)$ the space of all continuous real valued functions defined on $E$, with respect to the supremum norm $\|\cdot\|_\infty$ on $C(E).$ Let $\varnothing \neq A\subseteq C(E)$. Assume that $A$ is closed and bounded. Then $A$ is compact iff A is equicontinuous.
I'm reading a proof of the above theorem. The $(\Rightarrow)$ direction is fairly easy to follow. The $(\Leftarrow)$ direction is also fined except for some minor question. It goes this way. 
Assume that $A$ is equicontinuous. Let $\epsilon >0.$ Then we can find a $\delta >0$ such that for any $x,y\in E$, the inequality
$$d(x,y)<\delta \quad \Rightarrow \quad |f(x)-f(y)|<\epsilon$$
holds $\forall f\in A.$ Since $E$ is compact, $E$ is totally bounded. Thus, (corresponding to $\delta$), we can find $x_1, \cdots, x_n \in E$ such that
$$E\subseteq \bigcup_{j=1}^nB(x_j,\delta).$$
For each $f\in A$, we write
$$(f(x_2), \cdots, f(x_n))=\hat{f}\in \mathbb{R}^n.$$
Let
$$\hat{A}=\{\hat{f}: f\in A\}.$$
I got no problem of showing that $\hat{A}$ is a bounded subset of $\mathbb{R}^n$, with respect to supremum metric. My question is, how do we show that $\hat{A}$ is also a closed subset of $\mathbb{R}^n$?
NOTE. The continuation of the above discussion goes this way. Because $\hat{A}$ is a closed and bounded subset of $\mathbb{R}^n$, $\hat{A}$ is compact and hence totally bounded. This leads into a conclusion that $A$ is totally bounded. But the hypotheses also imply that $A$ is complete. Therefore, $A$ is compact.
 A: Assume $f_k$ is a sequence of functions in $A$ with the property that $\hat{f}_k$ has a limit in $\mathbb{R}^n$.  By definition this means that $f_k(x_i)$ converges for each $i$.  I claim that in fact $f_k(x)$ converges for every $x \in E$.  Given $x$, choose an $i$ such that $d(x,x_i) < \delta$.  For all sufficiently large $k, l$ we have 
$$|f_k(x) - f_l(x)| \leq |f_k(x) - f_k(x_i)| + |f_k(x_i) - f_l(x_i)| + |f_l(x_i) - f_l(x)| < 3\epsilon$$
This shows that the sequence $f_k(x)$ is Cauchy for each $x$.  But we can do better: according to the estimates above, we will have that $|f_k(x) - f_l(x)| < 3\epsilon$ as long as $k$ and $l$ are chosen large enough to make $|f_k(x_i) - f_l(x_i)| < \epsilon$ for each $i$.  Since there are only finitely many $x_i$'s to worry about, $f_k$ is in fact uniformly Cauchy.
It follows that $f_k$ has a uniform limit $f$ in $A$ since $A$ was assumed to be closed, and clearly $\hat{f}_k \to \hat{f}$ in $\mathbb{R}^n$.
A: Here is an argument of a different flavor. 
We know that a subset $S$ of $\mathbb{R}^n$ is closed and bounded iff $S$ is compact. This is known as the Bolzano-Weirstress theorem. This theorem actually immediately follows from Arzela-Ascoli. To see this, consider the following:
recall that a set $S$ is  sequentially  compact iff compact. So we don't need to worry about covering and subcovering, we can work directly with sequences.
Every sequence in $S$ has a convergent subsequence (definition of sequentially compact).
Also note that a sequence converges iff the sequence is Cauchy. We can tie this directly with equicontinuity. So in other words, the Bolzano-Weirstrass theorem about compact subsets of $\mathbb{R}^n$ is just a special case of Arzela-Ascoli. Here is a semi-sketchy proof. 
Let $K$ be a compact subset of $C([a,b])$. In particular, it is bounded and equicontinuous by Arzela-Ascoli. Now suppose $K$ is defined by the space of bounded functions $f_k:\mathbb{N}\rightarrow\mathbb{R}$ (bounded because $K$ was assumed compact). Note that every function is continuous, since functions from $\mathbb{N}\rightarrow\mathbb{R}$ are continuous. But note that an arbitrary sequence ${(x_k)}$ in $\mathbb{R}$ is defined in just this way. For every position in the sequenced indexed by $n \in \mathbb{N}$, we have a real number in the position $n_k$. 
By Arzela-Ascoli, we have equicontinuity, so $|f_k-f_l|< \epsilon$. But this implies that the sequence is convergent, since it is Cauchy. So we have 
$|f_k-L|<\epsilon$ for sufficiently large $k$. In other words, we have a convergent sequence ${(x_k)}$ from $\mathbb{R}$. In particular, we have a convergent subsequence $\forall f_k\in K$ (just the entire sequence), which shows that Arzela-Ascoli implies Bolzano-Weirstrass when continuous functions are restricted in a compact subset $K$ from $f:\mathbb{N}\rightarrow\mathbb{R}$.
in $\mathbb{R}^n$, a similar argument works, just replace absolute values with norms. And conclude that since the set $K$ is compact, it is closed and bounded, which holds in general for any subset. 
