Give an example of an open cover of the closed unit ball in ${l}^2$ that has no finite subcover. I know that many questions similar to that question are answered here on this site:
Give an example of an open cover of the closed unit ball in ${l}^2$ that has no finite subcover. But I do not know the technical details in our case here. Any help will be appreciated.
 A: If we refer to the closed unit ball in $l^{2}$ as $\mathcal{B}$ we have the set $\{e_{i}\}_{i \in \mathbb{N}} \subset \mathcal{B}$ where $$
e_{i} = (x_{1}, x_{2}, x_{3}, \dots, x_{i}, x_{i+1}, x_{i+2}, \dots ) =  (0,0,0,\dots, 1, 0, 0, \dots)
$$
i.e. the sequence that is $1$ at the $i$-th position and $0$ elsewhere. Now using the standard $l^{2}$-norm, notice that $$||e_{i} - e_{j}|| = \sqrt{2}$$
if $i \neq j$. 
Let $r = \frac{\sqrt{2}}{2}$ and consider the open cover 
$$
\bigcup_{x \in \mathcal{B}} B(x, r) 
$$
where $B(x,r)$ is the open ball of radius $r$ about point $x$. It is easy to see that this is indeed an open cover. Now, I claim that this open cover doesn't have a finite subcover. 
Why? Well, let's argue by contradiction. Suppose there was a finite subcover of balls $\{B(x_{i}, r)\}_{i=1}^{k}$ of $\mathcal{B}$. Since $\{e_{i}\}_{i \in \mathbb{N}}$ is an infinite set, this implies that at least two distinct members of this set must be in the same ball of radius $r$. But this would imply the existence of $i, j$ where $i \neq j$ but 
$$
||e_{i} - e_{j}|| < \frac{\sqrt{2}}{2}
$$
which contradicts our observation above. So, we've constructed an open cover of $\mathcal{B}$ that has no finite subcover. 
A: Let $x_1$, $\ldots$, $x_n$ non-zero elements in $l^2$. Take $y \perp \langle x_1, \ldots, x_n\rangle$, $\|y\|=1$. We have $\| y-x_i\|^2=\|y\|+ \|x_i\|^2>1$ for $1\le i \le n$. Therefore: the ball $B(0,1)$ cannot be covered by finitely many balls $B(x,1)$, $x\ne 0$.
