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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.

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1  
The reason why you can't apply Arzelà-Ascoli here is that $c_0 = C_0(\mathbb{N})$ and $\mathbb{N}$ is not compact –  kahen Oct 10 '11 at 16:04
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Hint: a real sequence that tends to $0$ is the same thing as a function from the one-point compactification of $\mathbb{N}$ to $\mathbb{R}$ that sends $\infty$ to $0$. –  Chris Eagle Oct 10 '11 at 16:24
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Following @kahen's idea: You can apply Arzelà-Ascoli. Let $\mathbb N_\infty = \mathbb N \cup \{\infty\}$ denote $\mathbb N$'s one-point compactification. Then $c_0 \cong \{x \in C(\mathbb N_\infty) \mid x(\infty) = 0\}$. Now $A \subseteq c_0$ is compact, iff it is compact seen as subset of $C(\mathbb N_\infty)$. –  martini Oct 10 '11 at 16:24
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In your edit you probably mean $C(\mathbb{N}_\infty)$ instead of $\mathbb{N}_{\infty}$ several times. I also suggest not to use Arzelà-Ascoli (which is somewhat overkill) in the final solution, but to redo the argument by hand. You will understand Arzelà-Ascoli much better if you spell out the translation in the explicit case at hand here. –  t.b. Oct 11 '11 at 9:21
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(now it is fixed and ok). Here's another thing you should think about. The unit ball is not compact. The sequence $(1,0,0,\ldots), (0,1,0,\ldots), (0,0,1,\ldots),\ldots$ is a closed and bounded set (and contained in the unit ball) but it does not have a convergent subsequence (any two distinct elements have distance $1$, so there is no Cauchy-subsequence). So your "I claim that they have the basis the closed balls in the metric induced by $\|\cdot\|_\infty$." must be incorrect. –  t.b. Oct 11 '11 at 17:40

1 Answer 1

up vote 2 down vote accepted

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.

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