Banach spaces of infinite dimension do not have a countable Hamel basis
ABanachSpaceOfInfiniteDimensionDoesntHaveACountableAlgebraicBasis
2005-01-31 04:13:06
2006-12-27 07:24:09
Result
Banach space
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A Banach space of infinite dimension does not have a countable Hamel basis.
\textbf{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.
\textbf{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\cite{hel}.
\textbf{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.
\begin{thebibliography}{9}
\bibitem{hel}
H.\ Elton Lacey,
{\it The Hamel Dimension of any Infinite Dimensional Separable Banach Space is c},
Amer.\ Math.\ Mon.\ 80 (1973), 298.
\end{thebibliography}