Banach spaces of infinite dimension do not have a countable Hamel basis

A Banach space of infinite dimension does not have a countable Hamel basis.

Proof

Let $E$ be such space, and suppose it does have a countable Hamel basis, say $B={(v_k)}_{k\in \mathbb{N}}$.

Then, by definition of Hamel basis and linear combination, we have that $x\in E$ if and only if $x={\lambda }_1\cdot v_1+\mathrm{\dots }+{\lambda }_n\cdot v_n$ for some $n\in \mathbb{N}$. Consequently,

$$E=\bigcup _{i=1}^{\mathrm{\infty }}(\mathrm{span}{(v_j)}_{j=1}^i).$$

This would mean that $E$ is a countable union of proper subspaces of finite dimension (they are proper because $E$ has infinite dimension), but every finite dimensional proper subspace of a normed space is nowhere dense, and then $E$ would be first category. This is absurd, by the Baire Category Theorem.

Note

In fact, the Hamel dimension of an infinite-dimensional Banach spaceis always at least the cardinality of the continuum(even if the Continuum Hypothesis fails).A one-page proof of this has been given by H. Elton Lacey[1].

Examples

Consider the set of all real-valued infinite sequences $(x_n)$such that $x_n=0$ for all but finitely many $n$.

This is a vector space, with the known operations. Morover, it has infinite dimension: a possible basis is ${(e_k)}_{k\in \mathbb{N}}$, where

So, it has infinite dimension and a countable Hamel basis.Using our result, it follows directly that there is no way to define a norm in this vector space such that it is a complete metric space under the induced metric.