Separability Criterion for Hilbert Spaces

# Separability Criterion for Hilbert Spaces

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

When dealing with Hilbert spaces, there is an easy criterion to determine if the space is separable or not.

Theorem 1: Let $H$ be a Hilbert space. Then $H$ is separable if and only if $H$ has a countable Hilbert basis. |

**Proof:**$\Rightarrow$ Suppose that $H$ is separable. Let $\mathcal F$ be the collection of all orthonormal subsets of $H$ ordered by inclusion and let $\mathcal C$ be a chain in $\mathcal F$. Consider the set:

\begin{align} \quad C = \bigcup_{S \in \mathcal C} S \end{align}

- We first show that $C$ is contained in $\mathcal F$, that is, $C$ is orthonormal. Let $x, y \in C$. Then there exists some $S \in \mathcal C$ such that $x, y \in S$ (since the elements of $\mathcal C$ are ordered by inclusion). But $S$ is an orthonormal subset of $H$, so $\langle x, y \rangle = 0$ and $\| x \|, \| y \| = 1$. Therefore $C \in \mathcal F$.

- Clearly $C$ is an upper bound for $\mathcal C$. By Zorn's lemma there must exist a maximal element $B \in \mathcal F$.

- We first establish that $B$ is an Hilbert basis of $H$. Let $y \in B^{\perp}$ be such that $\| y \| = 1$. Then $B \cup \{ y \} \in \mathcal F$. But then $B \subset B \cup \{ y \}$ which contradicts the maximality of $B$. Therefore we must have that:

\begin{align} \quad B^{\perp} = \{ 0 \} \end{align}

- By the theorem on the Hilbert Bases (Orthonormal Bases) for Hilbert Spaces page we have that $B$ is an Hilbert basis for $H$.

- We now show that $B$ is countable. Let $x, y \in B$ with $x \neq y$. Then:

\begin{align} \quad \| x - y \|^2 = \langle x - y, x - y \rangle = \| x \|^2 - 2 \mathrm{Re} \langle x, y \rangle + \| y \|^2 = 1 - 0 + 1 = 2 \end{align}

- Suppose that $B$ is uncountable. Then there can be no countable and dense subset of $H$ which contradicts the assumption that $H$ is separable. So $B$ is countable Hilbert basis for $H$.

- $\Leftarrow$ Let $(x_n)_{n=1}^{\infty}$ be an Hilbert basis for $H$. Then we must have that $\mathrm{span} (x_1, x_2, ... )$ is dense in $H$ and in particular the set:

\begin{align} \quad \left \{ \sum_{j=1}^{N} (a_j + ib_j)x_j : n \in \mathbb{N}, \: a_j, b_j \in \mathbb{Q} \right \} \end{align}

- is a countable and dense subset of $H$. So $H$ is separable. $\blacksquare$