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. 🙂
Best Answer
Here is a linearly independent subset of $\mathbb{R}$ with size $2^{\aleph_0}$.
Let $q_0, q_1, \ldots$ be an enumeration of $\mathbb{Q}$. For every real number $r$, let $$T_r = \sum_{q_n < r} \frac{1}{n!}$$ The proof that these numbers are linearly independent is similar to the usual proof that $e$ is irrational. (It's a cute problem; there's spoiler below.)
I think a similar trick might work for algebraic independence, but I don't recall having seen such a construction. Actually, John von Neumann showed that the numbers $$A_r = \sum_{n=0}^\infty \frac{2^{2^{[nr]}}}{2^{2^{n^2}}}$$ are algebraically independent for $r > 0$. [Ein System algebraisch unabhängiger zahlen, Math. Ann. 99 (1928), no. 1, 134–141.] A more general result due to Jan Mycielski seems to go through in ZF + DC perhaps just ZF in some cases. [Independent sets in topological algebras, Fund. Math. 55 (1964), 139–147.]
As for subspaces and subfields isomorphic to $\mathbb{R}$, the answer is no. (Since I'm not allowed to post any logic here, I'll refer you to this answer and let you figure it out.)
Well, I'll bend the rules a little... Consider a $\mathbb{Q}$-linear isomorphism $h:\mathbb{R}\to H$, where $H$ is a $\mathbb{Q}$-linear subspace of $\mathbb{R}$ (i.e. $h$ is an additive group isomorphism onto the divisible subgroup $H$ of $\mathbb{R}$). If $h$ Baire measurable then it must be continuous by an ancient theorem of Banach and Pettis. It follows that $h(x) = xh(1)$ for all $x \in \mathbb{R}$ and therefore $H = \mathbb{R}$. Shelah has produced a model of ZF + DC where all sets of reals have the Baire property, so any such $h$ in this model must be Baire measurable. A similar argument works if Baire measurable is replaced by Lebesgue measurable, but Solovay's model of ZF + DC where all sets of reals are Lebesgue measurable uses the existence of an inaccessible cardinal, and this hypothesis was shown necessary by Shelah.
Spoiler
Suppose for the sake of contradiction that $r_1 > r_2 > \cdots > r_k$ and $a_1,a_2,\ldots,a_k \in \mathbb{Z}$ are such that $a_1T_{r_1} + a_2T_{r_2} + \cdots + a_kT_{r_k} = 0$. Choose a very large $n$ such that $r_1 > q_n > r_2$. If $n$ is large enough that $$(|a_1| + |a_2| + \cdots + |a_k|) \sum_{m=n+1}^\infty \frac{n!}{m!} < 1$$ then the tail terms of $n!(a_1T_{r_1}+\cdots+a_kT_{r_k}) = 0$ must cancel out, and we're left with $$a_1 = -\sum_{m=0}^{n-1} \sum_{q_m < r_i} a_i \frac{n!}{m!} \equiv 0 \pmod{n}$$ If moreover $n > |a_1|$, this means that $a_1 = 0$. Repeat to conclude that $a_1 = a_2 = \cdots a_k = 0$.