Hilbert space has countable basis

Question

Let $H$ be Hilbert space. I want to show that if $H$ has a countable orthonormal basis, then every orthonormal basis for $H$ must be countable. I spent almost a day, but could not solve this problem. I am trying to solve it with some kind of contradiction. I started by assuming that there exist an uncountable basis. Then I want to derive something crazy but cannot find it. Could you give me some hints or suggestions?

Answer 1

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.