Banach spaces of infinite dimension do 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

 $e_{i}(n)=\begin{cases}1,&\text{if }n=i\\ 0,&\text{otherwise}.\end{cases}$

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.

References

• 1 H. Elton Lacey, The Hamel Dimension of any Infinite Dimensional Separable Banach Space is c, Amer. Math. Mon. 80 (1973), 298.
Title Banach spaces of infinite dimension do not have a countable Hamel basis BanachSpacesOfInfiniteDimensionDoNotHaveACountableHamelBasis 2013-03-22 14:59:12 2013-03-22 14:59:12 yark (2760) yark (2760) 21 yark (2760) Result msc 46B15