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.

  • $\begingroup$ Could you explain why $F_{e}$ is countable? $\endgroup$ – Vishal Gupta Jul 23 '13 at 8:32
  • $\begingroup$ @Vishal Did so. $\endgroup$ – martini Jul 23 '13 at 9:28
  • $\begingroup$ 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... $\endgroup$ – Vishal Gupta Jul 23 '13 at 14:15

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