# A Banach Space cannot have a denumerable basis:Why is it true?

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?

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

Suppose that $$\{e_1,e_2,...\}$$ is a basis of the Banach spaces $$M$$. Let $$M_n=\{e_1,e_2,...,e_n\}$$. So $$M_n$$ is closed and $$int(M_n)=\emptyset$$. 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 $$M=\cup_nM_n$$. Then $$M$$ is a countable union of sets with empty interior, by Baire'ss theorem $$M$$ needs to satisfy $$int(M) \neq \emptyset$$, contradiction.