$\left\{x\in H: 2\leq \|x\|\leq 5\right\}$ is compact?

Question

In a Hilbert space $H$ of dimention infinite, $A=\left\{x\in H:2\leq \|x\|\leq 5\right\}$ is compact? (totally bounded and complete) Thanks in advance.

Answer 1

This is not compact. Consider the points $$ A_1=(2,\cdots), A_2=(1,1,\cdots ), A_3=(\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}}\cdots) $$ where for $A_{n}$, each coordinate in $1\cdots n$ dimension has value $\sqrt{\frac{2}{n}}$ and all other coordinates' value is $0$. Then we have: $$ |A_{n}|=2, \forall n $$ as well as $$ A_{n}\rightarrow 0 $$ assume $A_{n}$ is convergent.

If the set you described is compact, then $A_{n}$ must have a convergent subsequence. But this does not hold since $|0|=0$. Here I am assuming $H\cong l^{2}$.

Answer 2

The answer is no. The general proof in a normed space can be done using Riesz's lemma.

In Hilberst spaces, you can use an easier proof as follows: Take $\{v_1,v_2,\cdots v_n, \cdots \}$ to be an o.n.b. Then considen $u_n=2v_n$. $||u_n||=2$ but $||u_n-u_m||^2=4+4=8$ (Check).

So, $u_n$ can't have a convergent subsequence.