Fields of intermediate cardinality Assuming the existence of a cardinal $\aleph_0 <\mathfrak{m} < 2^{\aleph_0}$, does it follow that there is a subfield of $\mathbb{R}$ of cardinality $\mathfrak{m}$?
 A: Certainly. You can adjoint the subset of cardinality $\mathfrak{m}$ to $\mathbb{Q}$.
First by axiom of choice, you can enumerate the subset of cardinality $\mathfrak{m}$, say $x_0, x_1, ..., x_\mathfrak{m}$. Then construct the tower of extension $\mathbb{Q} \rightarrow \mathbb{Q}(x_0) \rightarrow \cdots$. Let me write $F_\alpha = \mathbb(x_\gamma | \gamma < \alpha)$. It can be shown that for any ordinal $\alpha$, $|F_\alpha| \leq \aleph_0 |\alpha|$ by transfinite induction. Hence, the eventual field is of cardinality $\leq \aleph_0 \mathfrak{m} = \mathfrak{m}$.
A: A standard theorem on finitary algebraic structures says that if $A$ is an infinite subset of size $\kappa$ of such a structure, the minimal substructure that contains $A$ also has size $\kappa$ (one can deduce this from Löwenheim-Skolem, or use standard set theory closure techniques). 
So take any subset $A$ of size $\mathfrak{m}$, and take the minimal field containing $A$. This fits the bill (we automatically get $\mathbb{Q}$ as a subfield of course).
