TITLE: Unit Ball of $\mathcal{l}_2$
QUESTION [5 upvotes]: Let $B(\mathcal{l}_2) :=\{x \in \mathcal{l}_2 : \|x \| \leq 1 \}$ and $S(\mathcal{l}_2) :=\{x \in \mathcal{l}_2 : \|x \| = 1 \}$ be the unit ball and the unit sphere of $\mathcal{l}_2$, respectively.
I'm trying to show that $B(\mathcal{l}_2)$ contains an infinite set $A$ such that, for every $x,y \in A$ with $x \neq y$, we have $\|x -y \| > \sqrt{2}$.
My intuition tells me that such a set $A$ should lie inside $S(\mathcal{l}_2)$. Next, I've tried to use a point $z$ such that $d(z,S(\mathcal{l}_2)) = \sqrt{2}$ to define $A$. I know that for such a $z$ there exists $w \in S(\mathcal{l}_2)$ such that $\|z - w \| = \sqrt{2}$, and for every $v \in S(\mathcal{l}_2),v \ne w$, we must have $\|z - v \| > \sqrt{2}$.
So, right now I'm confused about how to go ahead with the definition of $A$ in a way that the distance between any two distinct points in $A$ is greater that $\sqrt{2}$. I get the feeling that I should be able to use what I've described in the previous paragraph, but I don't see how.
Could somebody point me in the right way?

REPLY [5 votes]: Let $(e_1,e_2,\ldots)$ be an orthonormal subset of $\ell^2$.  
Let $(a_1,a_2,a_3,\ldots)$ be a sequence of real numbers in $(0,1)$ to be specified later.
Define the sequence $(v_1,v_2,\ldots)$ in the unit ball as follows:
$\begin{align*} 
v_1&=e_1\\
v_2&=-a_1e_1+\sqrt{1-a_1^2}e_2\\
v_3&=-a_2(e_1+e_2)+\sqrt{1-2a_2^2}e_3\\
v_4&=-a_3(e_1+e_2+e_3)+\sqrt{1-3a_3^2}e_4\\
&\vdots\\
v_{k+1}&=-a_k(e_1+\cdots+e_k)+\sqrt{1-ka_k^2}e_{k+1}
\end{align*}$
If you work out the distances and simplify, you find that a sufficient condition for this example to work is that $a_k<\dfrac{1}{\sqrt{k+k^2}}$ for all $k$.  (I used Mathematica to help.)  E.g., if $a_k=\dfrac{1}{2k}$ for all $k$, then $\|v_k-v_j\|>\sqrt{2}$ for all $k\neq j$.