Hamel bases without (too much) axiomatic set theory Hamel bases have cropped up on the periphery of my mathematical interests a few times over my mathematical career, but I have never found the time or had a real need to look into them at any depth. Most of what I know comes from the 1966 book "A First Course in Functional Analysis" by M. Davies, in which he uses them to prove the existence of discontinuous solutions of the functional equation $f(x+y) = f(x) + f(y)$.
My questions/queries:


*

*Is it possible to talk (meaningfully) about Hamel bases without invoking the axiom of choice?

*am I correct in my primitive, intuition-led understanding: "we can't explicitly exhibit a Hamel basis because that would be "equivalent" (in some obscure way that I cannot define precisely) to explicitly exhibiting a "choice function"?

*can anyone give me a nice reference where an old-fashioned analyst (well, old, at least) could read up on such matters without getting too heavily involved in axiomatic set theory or foundations of maths texts?
 A: In general, you can't even posit the existence of a Hamel basis without the axiom of choice; and even with the axiom of choice, you can't get your hands on them. If you were to say "pick a basis of $\mathbb{R}$ over $\mathbb{Q}$" then you'd already have invoked the axiom of choice; and then you wouldn't be able to write down what a general vector in the basis looks like. This is characteristic of any construction in mathematics that depends on the axiom of choice, since it's the only axiom which says "this thing exists" without also saying what it looks like.
The sense in which the existence of Hamel bases is equivalent to the axiom of choice is as follows: if every vector space has a (Hamel) basis, then the axiom of choice holds; and if the axiom of choice holds, then every vector space has a basis. This equivalence means that, if the axiom of choice fails, then there is a vector space without a basis; likewise, if there exists a vector space without a basis, then the axiom of choice fails. But the question of whether the axiom of choice holds or fails (and hence of whether Hamel bases exist or don't exist) is unprovable, in a very concrete sense... the response of most mathematicians to this fact is "we'll just admit the axiom of choice".
It's not true that the existence of a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ implies the full axiom of choice. However, it does imply a weaker form of the axiom of choice which is [still] unprovable.
Beyond this, I don't know what to say: it's tough to answer an axiomatic set theoretic question without talking about axiomatic set theory.
A: Hamel basis implies discontinuous real-valued solutions to the functional equation $f(x+y)=f(x)+f(y)$ and such a function is known to  be non-measurable.  Solovay proved that ZF + countable Dependent Choice is consistent with "every set of reals is Lebesgue measurable" so that

a Hamel basis cannot be proved to exist in ZF + DC.  Informally, it cannot be done in classical analysis without uncountable, set-theoretic forms of choice

A: It is shown here that the existence of a Hamel basis for every vector space implies the Axiom of Choice in ZF.
A: To anyone finding this question now, Beriashvili-Schindler-Wu-Yu have recently found a model in which choice fails but which has a Hamel basis: http://wwwmath.uni-muenster.de/u/rds/hamel_basis.pdf.
