Showing subsets of $L^2=\{(x_n) : \sum_{n=1}^\infty x_n^2 < \infty \}$ are compact Let $L^2=\{(x_n) : \sum_{n=1}^\infty x_n^2 < \infty \}$, with $||(x_n)||=(\sum_{n=1}^\infty x_n^2)^{\frac{1}{2}}$. Show whether the following subsets of $l^2$ are compact.
Let $$l^2=\left\{(x_n):\sum_{n=1}^{\infty}x_n^2<\infty\right\}$$
equipped with the norm
$$\|(x_n)\|=\left(\sum_{n=1}^{\infty}x_n^2\right)^{1/2}.$$
State and explain if the following subsets of $l^2$ are compact:
$A=\left\{(x_n)\in l^2:\sum_{n=1}^{k}x_n^2\leq1  \right\}$ where $k\in\mathbb{N}$ is fixed;
$B=\left\{(x_n)\in l^2:\sum_{n=1}^{\infty}x_n^2\leq1, x_n=0\text{ for all } n>k  \right\}$ where $k\in\mathbb{N}$ is fixed;
$C=\left\{(x_n)\in l^2:\sum_{n=1}^{\infty}x_n^2\leq1\right\}$.
I only know the sequentially compact definition of compactness. 
I also have the theorem which states that all compact metric spaces are closed and bounded.
I think that A is not compact, as it is not bounded (this was easy to show using a sequence which became a constant after k)
I think C might not be compact, but only because I have seen that the unit ball in an infinite dimensional vector space is not compact, but I don't know how to prove it.
I think B might be compact, but I'm unsure. Any help would be appreciated!
 A: For metric spaces compactness and sequential compactness are identical.
I assume that you are working over the reals, but it's really not very important.
First, let me state that your conjectures are correct. $A$ is non-compact, $B$ is compact, and $C$ is non-compact.
You argument for $A$ being non-compact is correct.
Let us prove that $B$ is compact. Let $k$ be fixed. Now consider the finite dimensional vector space $\mathbb{R}^{k}$, equipped with the usual norm. It is not hard to see that the inclusion map
\begin{align}
\mathbb{R}^{k} & \rightarrow l^{2}, \\
(x_{1},...,x_{k}) & \mapsto (x_{1},...,x_{k},0,0,...),
\end{align}
is continuous. Furthermore, the subspace $B \subset l^{2}$ is the image of the unit sphere in $\mathbb{R}^{k}$ under this map. The unit sphere in $\mathbb{R}^{k}$ is compact, hence $B$ is compact.
For the set $C$, do you recall why the unit ball in infinite dimensions is non-compact? The space $C$ is exactly the unit sphere in $l^{2}$, so the proof for that fact applies here. Hint: Find a sequence of vectors in $C$ without a convergent subsequence.
A: For $A$, if you set terms to be constant after first $k$ terms, the resulting sequence is not in $\ell^2$ because sum is not square-convergent. Instead, you can use things like $me_m$ where $m>k$, $e_m$ is an elementary sequence having $1$ at $m$'th entry and other terms are $0$.
Set $B$ is in fact homoeomorphic to the unit ball in $\mathbb{R}^k$ via obvious map;
$C$ can be deduced to be non-compact as a consequence of general fact that you quoted. This is not quite trivial. I'd like to quote the following exercise introducing this fact for you:

Let $X$ be a normed space, and let $Y$ be a proper closed subspace. Choose $x\in X\setminus Y$.
(i) show that $\text{dist}(x,Y)=\inf\{\|x-y\|:y\in Y\}>0$
(ii) Conclude that there is $y\in Y$ with $\|x-y\|<2\text{dist}(x,y)$ and $z=x-y$ satisfies $\|z\|\le 2\text{dist}(z,Y)$
(iii) Deduce that if $Y$ is a proper closed subspace of $X$ then there exists $x$ in the closed unit ball such that $\text{dist}(x,Y)\ge \frac{1}{2}$
(iv) using (iii), Deduce that a closed unit ball of a infinite dimensional normed space $X$ is non-compact (Start by considering an element $x_1$ in the unit ball. The subspace spanned by $x_1$ is closed proper)

A: For metric spaces compactness and sequential compactness are identical.
I assume that you are working over the reals, but it's really not very important.
First, let me state that your conjectures are correct.  is non-compact,  is compact, and  is non-compact.
You argument for  being non-compact is correct.
Let us prove that  is compact. Let  be fixed. Now consider the finite dimensional vector space ℝ, equipped with the usual norm. It is not hard to see that the inclusion map
ℝ(1,...,)→2,↦(1,...,,0,0,...),
is continuous. Furthermore, the subspace ⊂2 is the image of the unit sphere in ℝ under this map. The unit sphere in ℝ is compact, hence  is compact.
For the set , do you recall why the unit ball in infinite dimensions is non-compact? The space  is exactly the unit sphere in 2, so the proof for that fact applies here. Hint: Find a sequence of vectors in  without a convergent subsequence.
