# Show that $X$ can not be a Banach space

Let $X$ be an in finite dimensional vector space with an algebraic basis $\{ e_1 , ... , e_n, ... \}$. I need to prove that $X$ cannot be a Banach space with respect to any norm.

By the definition of Banach space, I should show that any convergent sequence in $X$ converges in $X$ but I am not getting any starting approach.

• HINT: Can you write the element $$v= \sum_{k=1}^{\infty} 2^{-k} e_k$$ as a finite linear combination of the $e_j$s? – Crostul Sep 24 '16 at 10:23
• This series is absolutely convergent hence convergences to a finite value. So, I suppose it can be written as a finite linear combination of $e_k s$ – user371842 Sep 24 '16 at 10:33

## 1 Answer

Finite dimensional spaces are nowhere dense in infinite dimensional spaces.

X is a countable union of finite dimensional spaces, each of which are nowhere dense in $X$, hence $X$ is of the first category. Therefore, since complete spaces are of the second category, $X$ is not complete, and hence not Banach.