Characterisation of compact subsets of Banach spaces I have the following homework question:
Characterize the compact subsets of the following Banach spaces:
(1) The space $c_0$ of null sequences (that is, sequences $(x_n)$ of scalars with $ | x_n | \rightarrow 0$ as $n \rightarrow \infty$) with the norm $\|(x_n)\| = \sup_{n \geq 1} |x_n| = \max_{n \geq 1} |x_n|$.
What I think could be an answer:
If $(c_0, \| \cdot\|_\infty)$ is compact then by Ascoli-Arzelà a set is compact if and only if it is closed, bounded and equicontinuous. So I need to show that $c_0$ is compact and once I have that I claim that the closed bounded sets have the basis the closed balls in the metric induced by $\|\cdot\|_\infty$. Then I need to show that the sequences in these balls are equicontinuous.
Edit
Using your comments:
Let $\mathbb{N}_\infty$ denote $\mathbb{N} \cup \{ \infty \}$. Then this space is compact (it is homeomorphic to $\{ 0 \} \cup \{ \frac{1}{n} | n \in \mathbb{N} \}$). Then $c_0$ is homeomorphic to $\{ x | x( \infty ) = 0 \} \subset C(\mathbb{N}_\infty)$. Therefore $K \subset c_0$ is compact iff $K$ is compact in $C(\mathbb{N}_\infty)$. Now Ascoli-Arzelà applies and so $K$ is compact iff $K$ is closed, bounded and equicontinuous.
So I need to write down what closed, bounded, equicontinuous sets look like. I claim that they have the basis the closed balls in the metric induced by $\|\cdot\|_\infty$.
Many thanks for your help.
 A: I'd like to redo the argument used in the proof of Arzelà-Ascoli by hand for this homework:
For $S \subset c_0(\mathbb{N})$ we want to show that $S$ is compact if it is closed, bounded and such that for $\varepsilon > 0$ there exists an $N$ such that for $n > N$ we have $|s(n)| < \varepsilon$ for all $s$ in $S$.
Let $S \subset c_0(\mathbb{N})$ be such a set. We'll show that $S$ is compact by showing that every sequence $s_n$ in $S$ has a convergent subsequence $s^\prime_n$. ($c_0(\mathbb{N})$ is a metric space so compactness and sequential compactness coincide.) For $s^\prime_n$ to be convergent in $S$ it is enough to show that it is Cauchy because $c_0(\mathbb{N}) $ is complete so every Cauchy sequence converges in it. $S$ is closed by assumption so it contains all its limit points, in particular the limit of $s^\prime_n$.
$S$ is also bounded by assumption, i.e. there exists an $M \in \mathbb{R}$ such that $\|s \|_\infty \leq M$ for all $s$ in $S$. So in particular, $s_n(1)$ is a bounded sequence in $\mathbb{R}$ hence by Bolzano-Weierstrass contains a convergent subsequence $s_{n1}(1)$. $s_{n1}(2)$ is again a bounded sequence in $\mathbb{R}$ hence again contains a convergent subsequence $s_{n2}(2)$ which converges at $1$ and $2$. Repeating this argument we get $s_{nk}$, a sequence of functions converging at $1, \dots , k$.
Now we define $g_k (i) := s_{kk}(i)$ then $g_k$ is a function that converges pointwise on all of $\mathbb{N}$. To finish the proof we show that $g_k$ is a Cauchy sequence with respect to $\|\cdot \|_\infty$.
We can use the triangle inequality to get the following:
$$ |g_k(n) - g_j(n)| \leq |g_k(n)| + |g_j(n)|$$ 
By assumption on $S$ there exists an $N_\varepsilon$ such that for $n > N_\varepsilon$ we have $|g_k(n)| < \frac{\varepsilon}{2}$ and $|g_j(n)| < \frac{\varepsilon}{2}$ for all $j,k$.
For $n \leq N_\varepsilon$ we know that by construction $g_k(n)$ is a Cauchy sequence in $\mathbb{R}$ so there exists an $N_n$ such that for $k,j>N_n$ we have $|g_k(n) - g_j(n)| < \varepsilon$.
Hence we have that for all $j,k > \max \{ N_1, \dots, N_{N_\varepsilon}, N_\varepsilon \}$ and all $n$:
$$ |g_k(n) - g_j(n)| < \varepsilon $$
and hence $\sup_{n \in \mathbb{N}} |g_k(n) - g_j(n)| = \| g_k - g_j\|_\infty \leq \varepsilon$ which means that $g_k$ is Cauchy with respect to $\|\cdot\|_\infty$.
So $S$ is compact.
