Compactness of (sub)set of sequences with finite norm

Question

I've got the space $$l^2 = \Big\{(x_n):\sum_{n=1}^{\infty}x_n^2<\infty \Big\}$$ with the norm $$||(x_n)|| = \Big(\sum_{n=1}^{\infty} x_n^2 \Big)^{\frac{1}{2}}$$ and I want to determine whether the subset $$A = \Big\{(x_n)\in l^2:\sum_{n=1}^k x_n^2 \leq 1 \Big\} \quad k \in \mathbb{N}$$ is compact ($k$ is fixed). I'm not sure how to go about solving this problem (haven't covered open cover definition of compactness yet by the way).

My intuition is that this is compact, but what's confusing me is how I can manipulate a finite sum of terms of a sequence to show anything about its infinite sum. Alternatively, I know a continuous map preserves compactness, but I'm not even sure if taking $A$ to a set of corresponding finite sequences is continuous or even how that would help.

Edit: Okay I think I figured it out for a more general subset,

$$C = \Big\{(x_n)\in l^2 : \sum_{n=1}^{\infty}x_n^2 \leq 1\Big\}$$

So this isn't compact, if you consider the sequence $(s_n)$, where $s_1 = (1,0,0...), s_2 =(0,1,0,...)$ and so on. Then each term is a fixed distance $\sqrt{2}$ away from any other term. Hence it cannot be Cauchy or converge, and neither can any subsequence. $(s_n)$ also satisfies the criterion for $A$, so it also isn't compact.

Answer 1

Let $k=1$, for example. Define $A$ as in your question.

What we will do is the following : We know there exists $(a_n) \in l^2$ such that $(a_n)$ has no convergent subsequence. This can be constructed by having $||a_n||$ to be increasing to infinity.

For example: $$ (a_n)_k = \left(\frac{n}{k}\right) $$

(That is , the $k$th term of the sequence $a_n$ is given by $\frac nk$).

Then you can see that $||a_n|| = \sqrt{\sum_{k=1}^\infty ((a_{n})_k)^2} = n\sqrt{\sum_{k=1}^\infty \frac 1{k^2}} = \frac{n\pi}{\sqrt 6}$. This will have no convergent subsequence, as it has no Cauchy subsequence!

Now, pick your favourite number $0 <\epsilon < 1$,and define $b_n$ as follows: $$ (b_n)_k = \begin{cases} \epsilon & k=1 \\ (a_n)_{k-1} & k > 1 \end{cases} $$

What we are doing, is basically appending $\epsilon$ on the left of every sequence $a_n$, this is giving the sequence $b_n$.

What is special about $b_n$, we ask?

$b_n \in A$! This is because $((b_n)_1)^2 = \epsilon^2 < 1$ for all $n$.

Furthermore, what is $||b_n||$? You can calculate this, but one thing will be clear: the number goes to $\infty$ as $n$ goes to $\infty$.

Hence, it follows that $b_n$ does not have a convergent subsequence, completing the counterexample of $A$ being sequentially compact (and hence compact).

For one, you can easily adjust this proof for general $k$, by appending $k$ terms instead of one at the beginning, each term being $\frac \epsilon k$ now. Hence $A$ won't be sequentially compact for any $k$.

The essential reason is this:

Cauchyness, or convergence of a sequence is a property that happens in the limit. Hence, criteria that attempt to constrain only finitely many terms of the sequence (in our case, the first $k$ terms) will usually not have good limit properties such as closure, compactness etc. because in the limit, we can skip all these constrained terms as I did. $b_n$ shared the same limit properties as $a_n$, while vastly differing from it initially. Keep this in mind, it will be helpful.

Another thing : the given set is actually closed in $l^2$, I think. So all you have to show is that it is not totally bounded (which is the equivalent condition for compactness). To check this, you can take vectors in the unit ball, and apply some kind of construction you have made, to construct a counterexample. Or just use the fact that the unit ball isn't totally bounded, by the proof you have given.