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 whether 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;
$C=\left\{(x_n)\in l^2:\sum_{n=1}^{\infty}x_n^2\leq1\right\}$.
What I know:
1. A subset of a metric space $(X,d)$ is compact if every sequence from $S$ has a subsequence which converges to an element of $S$. This definition is equivalent to the coverings definition in metric space, so I used this one.
2. A closed subspace of a compact space is compact.
I am not sure how to start and which properties to use.
If I'm going to start from the definition, I'm clueless on how to construct the appropriate subsequence.
If I'm going to use closed subspace property that means I have to first show that $l^2$ is compact and figure out whether $A,B,C$ are closed or not. And showing whether or not $l^2$ is compact or not is really a challenge and maybe I need to start from the definition again.
Can anybody share some clue on how to start and the rough steps on the working? Hopefully I can fill in the mathematical writing details myself.
Many many thanks!
I don't understand how your $A$ is related to $k$.
For $B$, it is a subset of a finite-dimensional subspace. You can find a homeomorphism between $B$ and some closed ball in $F^k$, where $F$ is your scalar field.
For $C$, recall how we prove that locally compact vector spaces are finite-dimensional.