Could someone give me an explicit basis of $\mathbf{R}$ as a vector space over $\mathbf{Q}$?

I know some linearly independent subset, namely $1,e,e^2,\dots$ but this seems to be a deep result already (this is just the transcendence of $e$, right?), but no maximal linearly independent subset.

Let me clarify: I know how to prove that every vector space (over a division ring) has a basis, it is by Zorn's lemma and equivalent to AC. I don't think that this directly implies that one cannot describe some basis explicitly; can one prove that the existence of a basis of $\mathbf{R}$ over $\mathbf{Q}$ is equivalent to AC? This would convince me.

  • 1
    $\begingroup$ There are no explicit basis, but the axiom of choice is to do it. $\endgroup$ – Hamou Oct 11 '14 at 20:55
  • 1
    $\begingroup$ The dimension of $\bf R$ over $\bf Q$ is uncountably infinite, so a basis has to be uncountably infinite. That limits how "explicit" a basis can be. $\endgroup$ – Lee Mosher Oct 11 '14 at 20:56
  • $\begingroup$ @LeeMosher: I know, but more there still might be a basis which is more explicit than "it exists by Zorn's lemma". Hamou I know how to prove this in general, but you claim something stronger. $\endgroup$ – Rudi Oct 11 '14 at 21:01
  • 1
    $\begingroup$ @Lee: Consider $V=\bigoplus_{i<\omega_5}\Bbb Q$, then this space has an uncountable dimension, but there is a fairly explicit basis. In fact, $V=\bigoplus_{i\in\Bbb R}\Bbb Q$ has an explicit Hamel basis, and it is provably (without choice!) of the same cardinality as $\Bbb R$. So uncountability is not a sufficient reason for any limitation on explicit bases. (If the last example baffles, note that the axiom of choice is needed in order to show that $\Bbb R$ and $\bigoplus_\Bbb R\Bbb Q$ are isomorphic...) $\endgroup$ – Asaf Karagila Oct 12 '14 at 3:47

The axiom of choice is necessary for producing a Hamel basis for $\Bbb R$ over $\Bbb Q$, meaning we can prove that without the axiom of choice we cannot prove the existence of a Hamel basis; and of course it is not equivalent to the existence of such basis.

Let me start with the easy part. The second part. There is no statement which is limited to a particular set that is equivalent to the axiom of choice in general. The reason is that the axiom of choice may fail so far above the real numbers, that the usual proof goes on without a hitch, but the axiom of choice still fails in horrible ways. So no, this is not equivalent to the axiom of choice.

The reason we cannot produce an explicit definition that $\sf ZF$ proves that is a Hamel basis for $\Bbb R$ over $\Bbb Q$, is that there are models of $\sf ZF+\lnot AC$ where there is no such basis. For example, if there is a Hamel basis then there is a discontinuous function $f\colon\Bbb R\to\Bbb R$ such that $f(x+y)=f(x)+f(y)$. This is really just a linear endomorphism over $\Bbb R$ as a vector space over $\Bbb Q$.

Why is there a discontinuous one? Because a Hamel basis would have $2^{2^{\aleph_0}}$ permutations, each inducing a linear automorphism (not just any endomorphism), but as a separable space there are only $2^{\aleph_0}$ continuous functions from $\Bbb R$ to itself.

And in some models of $\sf ZF+\lnot AC$ we can prove that every linear endomorphism of $\Bbb R$ is continuous. So in such model there are no Hamel bases to $\Bbb R$ over $\Bbb Q$.

Of course this does not mean that you can't write a formula which describe a Hamel basis in some models of $\sf ZF$. But these models will have to satisfy certain assumptions which cannot be proved without the axiom of choice, and usually even more than just that. So the "explicit" part here is not that explicit after all.

  • $\begingroup$ Fascinating what you say here. Thanks, this is exactly what is was looking for. Much more convincing than saying that one uses the axiom of choice for showing it to exist, which doesn't really answer the question $\endgroup$ – Rudi Oct 11 '14 at 22:33
  • $\begingroup$ Yeah, that's not an argument that the axiom of choice is necessary. But it's often a good hint. $\endgroup$ – Asaf Karagila Oct 11 '14 at 22:34

No. To show the existence of such a basis one uses the Axiom of Choice. And anything that uses this axiom explicitly and necessarily defies explicit description. (If one could give an explicit description, one would make choices that can be described in finitely many words). For a deeper insight into this, check out Asaf's answer that will occur here in a few minutes, I'm sure

  • $\begingroup$ I know how to prove the existence of a basis in general, but this does not prove that I cannot find an explicit basis. it just makes it rather unlikely $\endgroup$ – Rudi Oct 11 '14 at 21:00
  • $\begingroup$ I wasn't sure whether or not to post an answer, but then I read your answer, and I didn't want to disappoint a fellow 100k'er. :P $\endgroup$ – Asaf Karagila Oct 11 '14 at 21:33

You are looking for the Hamel Basis. There is no way to write this basis explicitly, since its existence is dependent on the Axiom of Choice.

  • $\begingroup$ Thanks but I know this already. So you claim. $\endgroup$ – Rudi Oct 11 '14 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.