1. ## Basis

What is the basis for $\mathbb{R}$ viewed as a vector space over the rational field? It certainly does exist, but has anyone dared to find it? Has anyone dared to find the dimension? (If not then, this is one of the things I love about math, nobody found it but shown that it exists).

2. Originally Posted by ThePerfectHacker
What is the basis for $\mathbb{R}$ viewed as a vector space over the rational field? It certainly does exist, but has anyone dared to find it? Has anyone dared to find the dimension? (If not then, this is one of the things I love about math, nobody found it but shown that it exists).
The cardinality of the Hamel Basis of R over Q is C. I can't find a good
online reference at the moment, but I'm hope rgep will be able to point
to a suitable reference when he's next online.

RonL

3. The proposition that every vector space has a basis is equivalent to the Axiom of Choice (the implication from AC is easy: apply Zorn's Lemma to the collection of subspaces-with-a-basis ordered by inclusion -- this is due to Hamel). There is no 'natural' basis for R over Q and the existence of one must surely require some weak form of Choice: it is an open question whether it is equivalent to AC. See Peter Johnstone, "Notes on Set Theory and Logic", Cambridge University Press.

Incidentally there's some confusion in the use of the term Hamel basis: some authors use it precisely to mean a basis for R over Q whereas others use it to mean a basis in the algebraic sense as opposed to a Schauder basis in a topological vector space which has the property that every element is uniquely a possibly infinite, but convergent, sum.

4. Originally Posted by CaptainBlack
The cardinality of the Hamel Basis of R over Q is C.
How about the following argument to show this? The Hamel Basis of R cannot be countable because if it were, R would be a set of finite linear combinations of a countable set using a countable set of coefficients, so R itself would be countable. Then since the Hamel Basis of R is not countable, its cardinality is C by the Continuum Hypothesis .

5. Originally Posted by JakeD
How about the following argument to show this? The Hamel Basis of R cannot be countable because if it were, R would be a set of finite linear combinations of a countable set using a countable set of coefficients, so R itself would be countable. Then since the Hamel Basis of R is not countable, its cardinality is C by the Continuum Hypothesis .
Looks okay to me.