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[parent] Banach spaces of infinite dimension do not have a countable Hamel basis (Result)

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,

$\displaystyle 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

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



"Banach spaces of infinite dimension do not have a countable Hamel basis" is owned by yark. [ full author list (2) | owner history (1) ]
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Cross-references: metric, induced, metric space, complete, norm, operations, vector space, sequences, continuum hypothesis, cardinality of the continuum, infinite-dimensional, Baire category theorem, first category, nowhere dense, normed space, finite dimensional, finite, subspaces, union, linear combination, proof, Hamel basis, countable, infinite, Banach space

This is version 18 of Banach spaces of infinite dimension do not have a countable Hamel basis, born on 2005-01-31, modified 2006-12-27.
Object id is 6691, canonical name is ABanachSpaceOfInfiniteDimensionDoesntHaveACountableAlgebraicBasis.
Accessed 3526 times total.

Classification:
AMS MSC: 46B15 (Functional analysis :: Normed linear spaces and Banach spaces; Banach lattices :: Summability and bases)
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question for recent banach space of infinite dimension entry by mathforever on 2005-01-31 14:22:41
I don't get the result: take any separable infinite dimensional Hilbert space (which is defenitely Banach space) and you get countable algebraic basis. Am I not right?
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