Let $X$ be an infinite dimensional Banach space. Prove that every basis of $X$ is uncountable.

Can anyone help how can I solve the above problem?

  • 7
    $\begingroup$ What kind of basis? An algebraic basis? It's important at this point. $\endgroup$ – Patrick Da Silva Oct 20 '12 at 17:53

It seems that the proof using the Baire category theorem can be found in several places on this site, but none of those questions is an exact duplicate of this one. Therefore I'm posting a CW-answer, so that this question is not left unanswered.

We assume that a Banach space $X$ has a countable basis $\{v_n; n\in\mathbb N\}$. Let us denote $X_n=[v_1,\dots,v_n]$.

Then we have:

So we see that $\operatorname{Int} \overline{X_n} = \operatorname{Int} X_n=\emptyset$, which means that $X_n$ is nowhere dense. So $X$ is a countable union of nowhere dense subsets, which contradicts the Baire category theorem.

Some further references:

Other questions and answers on MSE



  • Corollary 5.23 in Infinite Dimensional Analysis: A Hitchhiker's Guide by Charalambos D. Aliprantis, Kim C. Border.
  • A Short Course on Banach Space Theory By N. L. Carothers, p.25
  • Exercise 1.81 in Banach Space Theory: The Basis for Linear and Nonlinear Analysis by Marián Fabian, Petr Habala, Petr Hájek, Vicente Montesinos, Václav Zizler
  • 1
    $\begingroup$ I have found a proof that a proper subspace ofa normed space has empty interior here: matthewhr.files.wordpress.com/2012/08/… $\endgroup$ – Self-teaching worker Aug 15 '14 at 17:56
  • $\begingroup$ Can't we prove it without Baire Category Theory in other words without axiom of dependent choice $\endgroup$ – Sushil Jun 26 '15 at 12:18
  • $\begingroup$ @Sushil You have a much better chance of getting some answer if you post your question as a question, not just as a comment. However, before posting such question, some clarifications are needed in my opinion. See here for some comments. $\endgroup$ – Martin Sleziak Jun 26 '15 at 12:39
  • $\begingroup$ Oh I see. But I want some clarity. Cardinality of Hamel basis(if exist) are equal does it imply AC(or ADC). If this implication is wrong I may ask Let X be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable without Baire Category Theory. $\endgroup$ – Sushil Jun 26 '15 at 12:46
  • $\begingroup$ Infact A vector space can't have countable basis and uncountable basis simultaneously. Do we really need AC here or can we prove it with Axiom of countable choice(if needed) $\endgroup$ – Sushil Jun 26 '15 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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