Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have seen a simple proof that no Banach space over $\mathbb{R}$ can be of countably infinite dimension. However since the space of all square integrable functions on the unit interval forms a Hilbert space, and all Hilbert Spaces are Banach, this space must not be of countable dimension. However we know that each point has a unique decomposition as a sum complex exponentials $e^{2n\pi}$ were $n\in\mathbb{Z}$. Thus these complex exponentials form a basis. But since there is a countable number of such exponentials there must be a contradiction. Where is that contradiction?

share|cite|improve this question
Basis $\ne$ Schauder basis. You’re talking about a countable Schauder basis for the space; it’s not a basis, since you allow infinite sums. – Brian M. Scott Sep 27 '12 at 23:10
Better to say "Hamel basis $\ne$ Schauder basis". A mathematician will use "basis" by itself for his favorite one. – GEdgar Jan 5 at 12:45
up vote 5 down vote accepted

The unique decomposition is as an infinite combination (series) of the complex characters, while a linear basis is a set of vectors such that any element of the space can be expressed as a finite combination.

The set $B$ such that any element of the space $V\supseteq B$ can be uniquely expressed as an infinite combination of elements of $B$ is called Schauder basis, as indicated by Brian M. Scott in the comment, as opposed to linear basis, usually called Hamel basis in functional analytic contexts.

For any cardinal number $\kappa$, there is a Banach space with a Schauder basis of cardinality $\kappa$ -- the Hilbert space of Hilbert dimension $\kappa$, for instance, but none of them have linear dimension $\aleph_0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.