# 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?

• Perhaps you would get a more interesting question by asking whether there is an uncountable subfield $F$ of $\mathbb{R}$ for which the existence of a proper uncountable subfield cannot be proved without using some form of the the Axiom of Choice. – Simon Thomas Jun 7 '10 at 15:13
• Correct me if I am wrong. I believe we can define a field $K$ by adjoining to $\mathbb{Q}$ all elements in $\mathbb{R}$ transcendental over $\mathbb{Q}$, without axiom of choice. The extension $\mathbb{R}/K$ must then be algebraic and hence $K$ or any extension of $K$ must be uncountable. I think this answer the second part of the question. It is not obvious to me how to define a uncountable subfield of $K$ without $AC$. Suppose, I want to prove that it is impossible to prove the existence of a uncountable proper subfield of $K$ is there any obvious path to proceed? – abcdxyz Jun 7 '10 at 15:40
• If $a$ is transcendental and $b$ is algebraic and nonzero over $\mathbb{Q}$ then $ab$ is transcendental. If $K$ is obtained by adjoining all real transcendentals to $\mathbb{Q}$ then $K$ contains $a$ and $ab$ and so also $b$. Thus $K=\mathbb{R}$. – Robin Chapman Jun 7 '10 at 16:10

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
Borel sets that are subrings of $\mathbb R$ either have Hausdorff dimension zero as described, or else are all of $\mathbb R$.
• Interesting! This answer mixes field theory and metric geometry in a way I have not seen. Could you give some more details and/or references as to why $\mathbb{Q}(K)$ has Hausdorff dimension zero? – Pete L. Clark Jun 7 '10 at 15:10
• Note that the field $\mathbb{Q}(K)$ is actually Borel. – François G. Dorais Jun 7 '10 at 16:17