Orthonormal Sets in Separable Inner Product Spaces

# Orthonormal Sets in Separable Inner Product Spaces

Theorem 1: Let $H$ be an inner product space. If $H$ is separable and $E \subset H$ is an orthonormal subset of $H$ then $E$ is countable. |

*Recall that a space is said to be separable if it contains a countable and dense subset.*

**Proof:**Suppose instead that $E$ is uncountable.

- Since $H$ is separable, $H$ has a countable and dense subset $D$. Now if $e, e' \in E$ and $e \neq e'$ then since $e$ and $e'$ are orthogonal we have that:

\begin{align} \quad \| e - e' \|^2 = \langle e - e', e - e' \rangle = \langle e, e \rangle -2 \langle e, e' \rangle + \langle e', e' \rangle = \| e \|^2 + \| e' \|^2 = 2 \end{align}

- Therefore, for every $e, e' \in E$ with $e \neq e'$ we see that:

\begin{align} \quad \| e - e' \| = \sqrt{2} \end{align}

- That is, the distance between $e$ and $e'$ is $\sqrt{2}$. Consider the collection of open balls:

\begin{align} \quad \mathcal F = \left \{ B \left (e, \frac{\sqrt{2}}{2} \right ) = \left \{ h \in H : \| e - h \| < \frac{\sqrt{2}}{2} \right \} : e \in E \right \} \end{align}

- Since $E$ is assumed to be uncountable, so is $\mathcal F$. Furthermore, the open balls in $\mathcal F$ are disjoint. Since $D$ is dense in $\mathcal H$, every ball in $\mathcal F$ must contain a point of $D$. But $D$ is countable and so this is impossible since $E$ is uncountable. So the assumption that $E$ is uncountable is false. Thus $E$ must be countable.