TITLE: A Banach Space cannot have a denumerable basis:Why is it true?
QUESTION [10 upvotes]: 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?

REPLY [7 votes]: 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 $M$.