A question about fields of real numbers Assume that the continuum hypothesis holds. If $F$ is an uncountable field of real numbers, does $F$ always contain a proper uncountable subfield? Are there many specific uncountable fields of real
numbers whose existence can be proved without assuming the axiom of choice?
 A: I think the following argument ought to answer your first question, but I haven't checked the details. An uncountable subfield F of R will contain an uncountable polynomially independent subset (by Zorn's lemma). And any proper subset of that polynomially independent subset will generate a proper subfield of F.
A: Take a compact Cantor set $K \subseteq \mathbb{R}$ of Hausdorff dimension zero.  Actually we need all cartesian powers $K^n$ of dimension zero as well.  The field $\mathbb{Q}(K)$ generated by it is uncountable, but still of Hausdorff dimension zero, so it is a proper subfield.
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That field consists of the values of rational functions $w(x_1,\dots,x_n)$ of many variables with rational coefficients, where the variables range over $K$.  There are countably many such things, so you just have to show any one of them has dimension zero.  The domain of any such $w$ (that is, the set where the denominator does not vanish) consists of an increasing countable union $\bigcup_k A_k$ of sets where the gradient is bounded, so that $w$ is Lipschitz continuous on each $A_k$.  So the image of $w$ on $K^n$ is again a countable union of sets of dimension zero.
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G. A. Edgar & Chris Miller, Borel subrings of the reals.  Proc. Amer. Math. Soc. 131 (2003) 1121-1129
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Borel sets that are subrings of $\mathbb R$ either have Hausdorff dimension zero as described, or else are all of $\mathbb R$.
