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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?

Martin Sleziak
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mintu
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1 Answers1

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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 $X_n=[v_1,\dots,v_n]$.

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
Martin Sleziak
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    I have found a proof that a proper subspace ofa normed space has empty interior here: http://matthewhr.files.wordpress.com/2012/08/uncountability-of-banach-hamel-basis.pdf – Self-teaching worker Aug 15 '14 at 17:56
  • Can't we prove it without Baire Category Theory in other words without axiom of dependent choice – Sushil Jun 26 '15 at 12:18
  • @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](http://chat.stackexchange.com/transcript/message/22401261#22401261) for some comments. – Martin Sleziak Jun 26 '15 at 12:39
  • 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. – Sushil Jun 26 '15 at 12:46
  • 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) – Sushil Jun 26 '15 at 12:50
  • @Sushil: Without the axiom of choice it is consistent for a vector space to have two bases which have different cardinalities. I don't know if it consistent that one of them is well-ordered, though. – Asaf Karagila Jun 26 '15 at 12:52
  • @AsafKaragila Can we also prove cardinality of Hamel bases(if exist) are equal then it implies AC. – Sushil Jun 26 '15 at 12:56
  • @Sushil: [Not even remotely](http://mathoverflow.net/questions/93242/sizes-of-bases-of-vector-spaces-without-the-axiom-of-choice). – Asaf Karagila Jun 26 '15 at 12:57
  • @MartinSleziak Ok but as I noticed from link mentioned by Asaf "vector space dimension is well-defined" is much weaker than DC. Hence now my question makes sense to ask: 'Prove that every Hamel basis in an infinite dimensional Banach Space of X is uncountable without Baire Category Theory assuming vector space dimension is well-defined". I think I can post it now. – Sushil Jun 26 '15 at 13:12
  • @MartinSleziak http://math.stackexchange.com/questions/1340333 – Sushil Jun 26 '15 at 17:00