On the complexity of Hamel bases of infinite-dimensional Banach spaces

Abstract

We call a subset S of a topological vector space V linearly Borel, if for every finite number n, the set of all linear combinations of S of length n is a Borel subset of V . It will be shown that a Hamel base of an infinite dimensional Banach space can never be linearly Borel. This answers a question of Anatolij Plichko. In the sequel, let X be any infinite dimensional Banach space. A subset S of X is called linearly Borel (w.r.t. X), if for every positive integer n, the set of all linear combinations with n vectors of S is a Borel subset of X. Since X is a complete metric space, X is a Baire space, i.e., a space in which non-empty open sets are not meager (cf. [1, Section 3.9]). Moreover, all Borel subsets of X have the Baire property, i.e., for each Borel set S, there is an open set O such that O∆S is meager, where O∆S = (O S) ∪ (S O). This is already enough to prove the following. Theorem. If X is an infinite dimension Banach space and H is a Hamel base of X, then H is not linearly Borel (w.r.t. X). I would like to thank the Swiss National Science Foundation for its support during the period in which the research for this paper has been done. On The Complexity Of Hamel Bases 2 Proof: Let X be any infinite dimensional Banach space over the field F and let H be any Hamel base of X. For a positive integer n, let [H] be the set of all n-element subsets of H and let