Question

Prove that the Hillbert space is separable

Answer 1

A Hilbert space H is separable (that is, has a countable dense subset) if and only if it has one countable orthonormal basis if and only if every orthonormal basis for H is countable.

Proof.$( \implies )$   Suppose $H$ is separable and consider any orthonormal basis $S \subset H$. Then if $\phi_1,\phi_2\in S$ ∈ and $\phi_1 \mathrlap{\,/}{=}\phi_2$ we conclude $\begin{Vmatrix} \phi_1 - \phi_2 \end{Vmatrix} = \sqrt{\lang \phi_1 - \phi_2 , \phi_1 - \phi_2 \rang} = \sqrt{2}$. Thus $S \subset H$ has the discrete topology. Now $H$ a separable metric space $\implies$ $S$ is a separable metric space $\implies$ $S$ S has a countable dense subset. But $S$ is the only subset of itself that is dense in $S$ and so $S$ must be countable.

$(\impliedby)$  Suppose for the converse that there is one countable orthonormal basis $S = \{\phi_1, \phi_2, . . .\}$

for $H$. We look at the case $\Bbbk = \R$ ($H$ a real Hilbert space) first. It is quite easy to check that the sets 

$D_n =\begin{Bmatrix} \displaystyle\sum_ {i=1} ^n q_i\phi_i : q_i \in Q \ for\ 1 ≤ i ≤ n \end{Bmatrix}$

are each countable and that their union $D = \bigcup_{n=1} ^∞ D_n$ is countable and dense in $H$. The closure of each $D_n$ n is easily seen to be the R-linear span of $\phi_1, \phi_2,...,\phi_n,$ and so the closure of $D$ includes all finite linear combinations $\displaystyle\sum_ {i=1} ^n x_i\phi_i$.  But, each $x \in H$ is a limit of such finite linear combinations. Hence the closure of $D$ is all of $H$.  As $D$ is countable, this shows that $H$ must be separable. In the complex case ($H$ a Hilbert space over $\Bbbk = \Complex$ ) we must take $q_i\in Q+iQ$ instead, so that we can get all finite $\Complex$ -linear combinations of the $\phi_i$ in the closure of $D$.