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


Let E be such space, and suppose it does have a countable Hamel basis, say B=(vk)k.

Then, by definition of Hamel basis and linear combinationMathworldPlanetmath, we have that xE if and only if x=λ1v1++λnvn for some n. Consequently,


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


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


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

This is a vector spaceMathworldPlanetmath, with the known operationsMathworldPlanetmath. Morover, it has infinite dimension: a possible basis is (ek)k, where

ei(n)={1,if n=i0,otherwise.

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.


  • 1 H. Elton Lacey, The Hamel Dimension of any Infinite Dimensional SeparablePlanetmathPlanetmath Banach Space is c, Amer. Math. Mon. 80 (1973), 298.
Title Banach spaces of infinite dimension do not have a countable Hamel basis
Canonical name BanachSpacesOfInfiniteDimensionDoNotHaveACountableHamelBasis
Date of creation 2013-03-22 14:59:12
Last modified on 2013-03-22 14:59:12
Owner yark (2760)
Last modified by yark (2760)
Numerical id 21
Author yark (2760)
Entry type Result
Classification msc 46B15