This is a response to the following exchange at Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?
[Bill constructs a $\aleph_0$ linearly independent set in $\Bbb R$]
Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. – JDH Jun 23 '11 at 2:22
I believe the answer to this question is no, for the following reason. Suppose we have a spanning set for $\Bbb R$ over $\Bbb Q$, of cardinality $\frak n$. Then by definition we have the cardinality equation ${\frak c}\le^*[{\frak n}]^{<\omega}\cdot \aleph_0$ (where $[{\frak n}]^{<\omega}$ is the cardinality of finite subsets, and $\frak m\le^*n$ means there is a surjection from $\frak n$ to $\frak m$).
If $\frak n$ is finite, then we get ${\frak c}\le^*\aleph_0$ which is false, so $\frak n$ is infinite (but not necessarily Dedekind-infinite). If we assume the axiom of choice, then the equation reduces to $\frak c\le n$, hence every Hamel basis is size continuum. But without AC I don't know any way to simplify the original equation to get anything stronger than "$\frak n$ is infinite".
This is at best heuristic evidence for the original question, since it is considering a different problem (spanning sets instead of linearly independent sets). How could one show that there are no ZF-constructible $\Bbb Q$-linearly independent subsets of $\Bbb R$ of cardinality continuum?
