Compactness of a Ball

Question

Let $l^2:\lbrace x=\left( x_i \right)_{i\in \mathbb{N}} \mid x_i\in \mathbb{R}$ for $i\in \mathbb{N}$ and $\sum_{i=1}^\infty x_i^2<\infty\rbrace $

and the usual distance $||x||_2=d_2$

Now, let $\theta\in l^2$ denote the zero sequence $(x_i=0$ $\forall$ $i\in \mathbb{N})$

Is $\overline{B(\theta,1)}=\lbrace x\in l^2 \mid d_2(\theta,x)\leq 1 \rbrace$ compact?

Now, I am trying to approach this by the Heine-Borel theorem, so I want to prove $\overline{B(\theta,1)}$ is closed and bounded but don't quite get the zero sequence part and how I can start from this to prove it is closed and bounded...

Any tips?

Answer 1

The Heine-Borel theorem is valid for $\mathbb{R}^n$ for any natural number $n$, but not for $\ell^2$. In fact, the set $\overline{B(\theta,1)}$ which you are considering is the standard example of a subset of $\ell^2$ which is closed and bounded but not compact. To show that it is not compact, try to find a sequence in $\overline{B(\theta,1)}$ with no convergent subsequence.

Answer 2

Ball in $l^2$ is indeed closed and bounded (almost by definition) but it is not compact. For that consider the sequence

$$(A_n)=(a^1_n, a^2_n,\ldots)$$ $$a^i_n=\begin{cases}1 & n=i \\0 &\text{otherwise}\end{cases}$$

This sequence obviously belongs to $\overline{B(0,1)}$ but it has no convergent subsequence. That's because it is not a Cauchy sequence. Indeed $\lVert A_n-A_m\rVert=\sqrt{2}$ for $n\neq m$.