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I came across the following theorem:

A Banach Space cannot have a denumerable basis which has been proven in my book.

I can't understand why is it true since $\mathbb R$ is a banach space over $\mathbb R$ and it has a countable basis i.e $\{1\}$

Where am I missing the link?

Learnmore
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  • If your book proves this, which step of the proof fails in the case of $\mathbb R$? Could the statement really say "cannot have a countably infinite base"? – Hagen von Eitzen Aug 26 '15 at 06:12
  • interior of a proper subspace of a nls is empty: which is false in R @HagenvonEitzen – Learnmore Aug 26 '15 at 06:13
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    The only proper subspace of $\mathbb R$ is $0$ and has empty interior ... – Hagen von Eitzen Aug 26 '15 at 06:15
  • yes sorry ;you are right @HagenvonEitzen – Learnmore Aug 26 '15 at 06:17
  • this is the only fact used and Baire Category theirem ;so where am i missing @HagenvonEitzen – Learnmore Aug 26 '15 at 06:18
  • See $,$ B.V.Limaye Functional Analysis (1996),$,$Th. 8.4, p. 132 . – Tony Piccolo Aug 26 '15 at 06:57
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    Ok, denumerable = countably infinite. Then it is true. – A.Γ. Aug 26 '15 at 07:06
  • @DavidMitra I also wrote this comment, but then google that denumerable means precisely cardinality of $\mathbb{N}$. A finite basis is then not denumerable, so the claim is true for any Banach space. – A.Γ. Aug 26 '15 at 11:50
  • Suppose that $ E $ is a Banach space of infinite dimension and has a countable basis $ (e_n)_{n\in \mathbb N} $ for $ n \in \mathbb {N} $.

    Then define $F_n $ , vector space generated by $e_i $ for $ i \leq n $.

    As $ F_n$ is of finite dimensional then it is a closed space with empty interior. But the union of $F_n $ is equals to $ E $. Thanks to Baire's theorem, $ E $ should be empty interior which is absurd.

     – Zbigniew Aug 26 '15 at 17:03

2 Answers2

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Suppose that $\{e_1,e_2,...\}$ is a basis of the Banach spaces $M$. Let $M_n=\text{span}\{e_1,e_2,...,e_n\}$. So $M_n$ is closed, and is a proper subspace of $M$. So $\text{int}(M_n)=\emptyset$. Given $x\in M$, since $\{e_1,e_2,...\}$ is a Hamel basis of $M$ there exists $n$ such that $x=\sum_{j=1}^n\alpha_je_j$, so $x\in M_n$. This prove that $M=\bigcup_nM_n$. Then $M$ is a countable union of sets with empty interior, by Baire's theorem $M$ needs to satisfy $\text{int}(M) \neq \emptyset$, contradiction.

Note: $\text{int}(M)$ is the interior of $M$.

kimchi lover
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Euler88 ...
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Of course every finite dimensional Banach space has a finite basis. As a consequence of Baires category theorem every infintedimensional Banach space cannot have a countable Hamel Basis, but very often a Schauder basis.There exist separable Banach spaces that don´t even have a Schauder basis so as was shown by Per Enflo. So You are right, the word "infinitedimensional" is missing in the statement

Peter Melech
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