Hilbert space has countable basis

Solution:

Let $$(e_i)_{i\in\Bbb N}$$ and $$(v_\alpha)_{\alpha\in X}$$ be orthonormal bases of $$H$$ with an arbitrary index set $$X$$.

First observe that $$\{\alpha\in X:\,\forall i\in\Bbb N\,(e_i\perp v_\alpha)\}=\emptyset$$ since $$e_i$$ is an orthonormal basis.

Then, for each $$i,n\in\Bbb N$$, consider $$A_{i,n}:=\{\alpha\in X: |\langle e_i,v_\alpha\rangle|\ge \frac1n\}$$.
Since for any $$\alpha_1,\dots,\alpha_k\in X$$, we have $$1=\|e_i\|^2\ \ge\ \left\|\sum_{j=1}^k\langle e_i,v_{\alpha_j}\rangle\cdot v_{\alpha_j}\right\|^2 \ =\ \sum_{j=1}^k|\langle e_i,v_{\alpha_j}\rangle|^2\,,$$ we obtain that each $$A_{i,n}$$ must be finite.

Consequently, $$X\ =\ X\,\setminus\,\{\alpha:\,\forall i\in\Bbb N\,(e_i\perp v_\alpha)\}\ =\ \displaystyle\bigcup_{i,n\in\Bbb N}A_{i,n}$$ is countable.

Hint:

I found this statement in wikipedia:
(the 'if' part isn't too difficult but is a good first step before tackling the 'only if' part)

A Hilbert space is separable if and only if it admits a countable orthonormal basis.

and the statement can be proven without using the axiom of choice.


Let $$(X,d)$$ be a metric space, and let $$A\subseteq X$$ be a dense subset. Then, every dense subset $$B$$ of $$X$$ has a dense subset of cardinality at most $$\lvert A\rvert$$.

Proof: If $$X$$ is finite, then $$X$$ is discrete and $$A=B=X$$. If $$X$$ is infinite, then $$A$$ and $$B$$ must be infinite. Moreover, the set $$S=\{B(a,q)\,:\, a\in A\land q\in\Bbb Q_+\}$$ is a basis of $$X$$. Since $$A$$ is infinite, $$\lvert S\rvert\le \lvert A\times \Bbb Q\rvert=\lvert A\rvert$$. For all $$D\in S$$, select some $$b_D\in B$$, given by density of $$B$$, such that $$b_D\in D$$. The set $$\{b_D\,:\, D\in S\}$$ intersects all the elements of a basis, and therefore it is dense. Moreover its cardinality is at most $$\lvert S\rvert\le\lvert A\rvert$$.

Now, let $$\mathcal B_1,\mathcal B_2\subseteq H$$ be two subsets of orthonormal vectors each of which generates a dense subspace. Also, let $$\kappa_i=\lvert\mathcal B_i\rvert$$ and $$\kappa_1\le\kappa_2$$. If $$\kappa_1<\aleph_0$$, then proving $$\kappa_2=\kappa_1$$ is elementary linear algebra. If $$\kappa_1\ge\aleph_0$$, then call $$\span_{\Bbb Q}(\mathcal B_1)$$ the $$\Bbb Q$$-linear subspace generated by $$\mathcal B_1$$. It is clear that $$\overline{\span_{\Bbb Q}(\mathcal B_1)}=\overline{\span_{\Bbb R}(\mathcal B_1)}= H$$ and that $$\kappa_1=\lvert \span_{\Bbb Q}(\mathcal B_1)\rvert$$. By the above lemma, $$\span_{\Bbb R}(\mathcal B_2)$$ must have a dense subset $$G$$ of cardinality at most $$\kappa_1$$. However, since every vector in $$\span_{\Bbb R}(\mathcal B_2)$$ is linear combination of finitely many elements of $$\mathcal B_2$$, there is a subset $$G'\subseteq\mathcal B_2$$ of cardinality at most $$\lvert G\rvert$$ such that $$\span_{\Bbb R}(G')\supseteq G$$. However, $$\mathcal B_2\setminus G'\subseteq (\span_{\Bbb R}(G'))^\perp=\left(\overline{\span_{\Bbb R}(G')}\right)^\perp\subseteq\left(\overline G\right)^\perp=H^\perp=\{0\}$$

which means that $$\mathcal B_2=G'$$ and therefore $$\kappa_2=\lvert G'\rvert\le\lvert G\rvert\le \kappa_1$$.