TITLE: explicit big linearly independent sets
QUESTION [65 upvotes]: In the following, I use the word "explicit" in the following sense: No choices of bases (of vector spaces or field extensions), non-principal ultrafilters or alike which exist only by Zorn's Lemma (or AC) are needed. Feel free to use similar (perhaps more precise) notions of "explicit", but reasonable ones! To be honest, I'm not so interested in a discussion about mathematical logic. If no example is there, well, then there is no example. ;-)
Can you give explicit large linearly independent subsets of $ \mathbb{R}$ over $\mathbb{Q}$? For example, $\{\ln(p) : p \text{ prime}\}$ is such a set, but it's only countable and surely is no basis. You can find more numbers which are linearly independent, but I cannot find uncountably many. AC implies $\dim_\mathbb{Q} \mathbb{R} = |\mathbb{R}|$. Perhaps $ZF$ has a model in which every linearly independant subset of $ \mathbb{R}$ is countable?
The same question for algebraically independent subsets of $ \mathbb{R}$ over $\mathbb{Q}$? Perhaps the set above is such a subset? But anyway, it is too small.
Closely related problems: Can you give an explicit proper subspace of $ \mathbb{R}$ over $\mathbb{Q}$, which is isomorphic to $ \mathbb{R}$? If so, is the isomorphism explicit? Same question for subfields.
That would be great if there were explicit examples. :-)

REPLY [24 votes]: Here's an answer that's similar in spirit to Pietro Majer's, but not quite the same.
As a first step, choose an uncountable family of infinite subsets of positive integers such that any two distinct sets in the family have finite intersection. This can be done explicitly in many ways. One I like is as follows. Since one can take an explicit bijection between $\mathbb{N}$ and $\mathbb{Z}^2$, it's good enough to create a family of subsets of $\mathbb{Z}^2$ instead. And to do that, for each real number $\alpha\in[0,\pi)$ take the set of all points in $\mathbb{Z}^2$ that are within a distance 2 (say) of the line that makes an angle $\alpha$ with the x-axis.
Once we have such a family F, we define a real number $r_X$ for each X in F as follows. It is a number between 0 and 1 that has only 0s and 1s in its decimal expansion. And it has a 1 at the nth place if and only if $n=m^2$ for some $m\in X$. (The reason for restricting to the squares is simply that we want the gaps between successive places where there might be a 1 to get larger and larger, so that we can ignore carrying problems.)
The numbers $r_X$ are linearly dependent over $\mathbb{Q}$ only if we can find a non-zero integer combination of finitely many of them that gives zero. But we can't: if we've got a non-zero coefficient t, then after a while the gaps will be longer than the number of digits of t (or even of the sum of the absolute values of the coefficients, say), and we'll be able to find an element of the corresponding set $X\in F$ that belongs to none of the other sets, thereby proving that that integer combination is not zero.