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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?

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    $\begingroup$ What kind of basis? An algebraic basis? It's important at this point. $\endgroup$ Oct 20, 2012 at 17:53

2 Answers 2

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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 $X_n=[v_1,\dots,v_n]$, i.e., the linear span of the first $n$ vectors.

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

Online

Books

  • 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
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    $\begingroup$ I have found a proof that a proper subspace ofa normed space has empty interior here: matthewhr.files.wordpress.com/2012/08/… $\endgroup$ Aug 15, 2014 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, 2015 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. $\endgroup$ Jun 26, 2015 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, 2015 at 12:46
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    $\begingroup$ @HelloWorld Exactly as you wrote, $X_n$ is the linear span of the vectors $v_1,\dots,v_n$. $\endgroup$ Jan 14, 2023 at 15:32
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One easy proof might the that following:

If possible let hamel basis of $X$ is countable and the basis is $\{v_1,\cdots,v_n,\cdots\}.$ WLOG we can assume $\|x_n\|=1.$ Now, take $x=\sum_0^\infty \frac{x_n}{2^n}$ which is not linearly combination of the above bases, but it is in $X$ by completeness of $X$, a contradiction.

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  • $\begingroup$ How do you know that the limit you have is not in the finite linear span of your basis? or instead how do you know you can always find one such that it's not? I thought about this proof but this is where I was stucked. $\endgroup$
    – cct
    Oct 10, 2023 at 23:07

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