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?

  • $\begingroup$ 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"? $\endgroup$ Commented Aug 26, 2015 at 6:12
  • $\begingroup$ interior of a proper subspace of a nls is empty: which is false in R @HagenvonEitzen $\endgroup$
    – Learnmore
    Commented Aug 26, 2015 at 6:13
  • 1
    $\begingroup$ The only proper subspace of $\mathbb R$ is $0$ and has empty interior ... $\endgroup$ Commented Aug 26, 2015 at 6:15
  • $\begingroup$ yes sorry ;you are right @HagenvonEitzen $\endgroup$
    – Learnmore
    Commented Aug 26, 2015 at 6:17
  • 3
    $\begingroup$ Ok, denumerable = countably infinite. Then it is true. $\endgroup$
    – A.Γ.
    Commented Aug 26, 2015 at 7:06

2 Answers 2


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 $M$.


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


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