Infinitely Many Disjoint Balls Contained In Unit Ball I am trying to solve a portion Problem 16 of Chapter 2 from Methods of Modern Mathematical Physics by Simon and Reed. The task is to show that the unit ball in an infinite dimensional Hilbert space contains infinitely many disjoint balls of radius $\sqrt{2}/4$. My approach is shown below.
By Theorem 2.5, I know that every Hilbert space contains an orthonormal basis $\{a_n\}_{n = 1}^{\infty}$. I define the sequence of balls $\{B_n\}_{n=1}^{\infty}$ such that $$B_n = \{x\ :\ \|x - a_n/2\| < \sqrt{2}/4\}.$$ If $a \in B_n$, we observe that $$\|a\| \leq \|a - a_n/2\| + \|a_n/2\| < \sqrt{2}/4 + 1/2 < 1.$$ Thus, $B_n$ is contained in the unit ball. For $m \neq n$, we can observe that $$\|a_m/2 - a_n/2\| \leq \|x - a_m/2\| + \|x - a_n/2\| < \sqrt{2}/4 + \sqrt{2}/4 = \sqrt{2}/2.$$ However, this last inequality does not guarantee that $B_m \cap B_n = \emptyset$. Are there any suggestions for how to ensure that $B_m \cap B_n = \emptyset$?
 A: Here is how one works in general:
Indeed, take your ONB $\{a_n\}$. Set $B_n=B(\lambda_n a_n,\frac{\sqrt{2}}{4})$ where $\lambda_n$ are numbers that we will specify soon.
If $a\in B_n$, then $\|a\|\leq\|a-\lambda_n a_n\|+\|\lambda_n a_n\|<\frac{\sqrt{2}}{4}+|\lambda_n|<1$, where we now require that $|\lambda_n|<1-\frac{\sqrt{2}}{4}$.
Now assume that $B_n\cap B_m\neq\emptyset$, say there exists an element $z$ in the intersection. Then we have that $\|\lambda_n a_n-\lambda_m a_m\|\leq\|\lambda_n a_n-z\|+\|z-\lambda_ma_m\|\leq\frac{\sqrt{2}}{2}$, so $\|\lambda_n a_n-\lambda_ma_m\|^2<\frac{1}{2}$. But we have that
$$\|\lambda_na_n-\lambda_ma_m\|^2=|\lambda_n|^2+|\lambda_m|^2.$$
So if you simply take $\lambda_n=\frac{1}{2}$ for all $n$, then we get a contradiction from this last condition, because then $1/4+1/4<1/2$, which is false. Since the first condition we require is satisfied (as $\frac{1}{2}<1-\frac{\sqrt{2}}{4}$) the balls $\{B(\frac{1}{2}a_n,\frac{\sqrt{2}}{4})\}_{n=1}^\infty$ do the trick.
A: The last inequality does guarantee they are disjoint, since it implies $||a_n - a_m|| < \sqrt{2}.$ But $$||a_n - a_m||^2 = ||a_n||^2 + ||a_m||^2 + 0 = 2$$ since the $a_n, a_m$ are orthogonal.
