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 $x = \lambda_1 \cdot v_1 + \dots + \lambda_n \cdot v_n$ for some $n \in \mathbb{N}$ . Consequently, $$ E = \bigcup \limits_{i=1}^\infty {(\operatorname{span}(v_j)_{j=1}^i)}. $$ 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 space is 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

References

1

H. Elton Lacey, The Hamel Dimension of any Infinite Dimensional Separable Banach Space is c, Amer. Math. Mon. 80 (1973), 298.