# Uncountable basis and separability

We know that a Hilbert space is separable if and only if it has a countable orthonormal basis.

What I want to ask is

If a Hilbert space has an uncountable orthonormal basis, does it mean that it is not separable? Or equivalently, does it imply that the Hilbert space does not have a countable basis?

I know that if a vector space has infinite number of linearly independent vectors then it cannot have a finite (Hamel) basis. But here we do not deal with Hamel basis but with a complete orthonormal set, do I cannot apply the usual techniques.

Any ideas?

Here's a brute-force approach that doesn't mention other bases:

The open balls of radius $\frac{1}{2\sqrt2}$ around the orthonormal basis vectors are disjoint. A countable set can't intersect them all if there are uncountably many, so it isn't dense, and the space isn't separable.

Any two orthonormal bases of a Hilbert space $H$ are equipotent as sets.

This is obvious for finite-dimensional Hilbert spaces, so let $H$ be infinite dimensional, and let $E$ and $F$ be orthonomal bases of $H$. Then for each $e \in E$, we have $$e = \sum_{f \in F} \def\s#1{\left<#1\right>}\s{e,f}f$$ Hence, the set $F_e := \{f \in F\mid \s{e,f}\ne 0\}$ is countable. We have $F = \bigcup_{e \in E} F_e$, giving $\def\abs#1{\left|#1\right|}$$\abs F \le \abs E \sup_{e \in E}\abs{F_e} \le \abs E\abs{\mathbb N} = \abs E$$ So$\abs F \le \abs E$. Exchanging the roles of$E$and$F$in the above argument, we have$\abs E \le \abs F$, so$E$and$F$are equipotent. So, if any orthonormal basis is uncountable, all are. Addendum: To see that$F_e$is countable, one can argue as follows: As$\sum_f \s{e,f}f$converges, and$F$is orthogonal, we have$\def\norm#1{\left\|#1\right\|}$$$1 = \norm e^2 = \sum_f \abs{\s{e,f}}^2$$ So for$n \in \mathbb N$, for only finitely many$e \in E$we can have$\abs{\s{e,f}} \ge \frac 1n$. Let$F_{e,n} = \{f \in F \mid \abs{\s{e,f}} \ge \frac 1n\}$. Then$F_{e,n}$is finite and$F_e = \bigcup_n F_{e,n}$is a countable union of finite sets, hence countable. • Could you explain why$F_{e}$is countable? – Vishal Gupta Jul 23 '13 at 8:32 • @Vishal Did so. – martini Jul 23 '13 at 9:28 • Thanks. That part is clear now. However, there is one more point I would like some clarification. I do not know cardinal arithmetic, so I do not understand how you write$|F| \leq |E|sup|F_{e}|\$ etc... – Vishal Gupta Jul 23 '13 at 14:15