Answer to Question #114845 in Real Analysis for Sheela John

Question

Prove that Hilbert space is seperable

Answer 1

Theorem: 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 a separable and consider any orthonormal basis $S\subset H$. Then if $\phi_1,\phi_2\in S$ and $\phi_1{=}\mathllap{/\,}\phi_2$ we conclude $||\phi_1-\phi_2||=\sqrt{(\phi_1-\phi_2,\phi_1-\phi_2)}=\sqrt{2}$. Thus $S\subset H$ has the discrete topology.

Now $H$ a separable metric space $\implies S$ is a separable metric space $\implies S$ has a countable dense subset. But $S$ is the only subset of itself that is dense in $S$ (since $S$ has the discrete topology) 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 $\Kappa=\R$ ($H$ is a real Hilbert space) first. It is quite easy to check that the sets

$D_n=\{\sum$$n\atop j=1$ $q_j \phi_j:q_j\in Q$ for $1\leq j\leq n\}$

are each countable and that their union $D=\bigcup$$\infin\atop n=1$$D_n$ is countable and dense in $H$. The closure of each $D_n$ 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 $\sum$$n\atop j=1$$x_j\phi_j$. 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 ($H$ is a Hilbert space over $K=\Complex$ we must take $q_j\in Q+iQ$ instead, so that we can get all finite $\Complex$-linear combinations of the $\phi_j$ n the closure of $D$ (and there is no other difference in the proof). $\blacksquare$