A question about fields of real numbers. - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T14:09:23Zhttp://mathoverflow.net/feeds/question/27352http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/27352/a-question-about-fields-of-real-numbersA question about fields of real numbers.Garabed Gulbenkian2010-06-07T14:08:40Z2010-06-07T15:32:59Z
<p>Assume that the Continuum Hypothesis holds. If F is an uncountable field of real numbers, does F always
necessesarily contain a proper uncountable sub-field? Are there many specific uncountable fields of real
numbers whose existence can be proved without assuming the Axiom of Choice?</p>
http://mathoverflow.net/questions/27352/a-question-about-fields-of-real-numbers/27355#27355Answer by gowers for A question about fields of real numbers.gowers2010-06-07T14:45:49Z2010-06-07T14:45:49Z<p>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.</p>
http://mathoverflow.net/questions/27352/a-question-about-fields-of-real-numbers/27358#27358Answer by Gerald Edgar for A question about fields of real numbers.Gerald Edgar2010-06-07T14:58:05Z2010-06-07T15:22:03Z<p>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.</p>
<p><strong>edit</strong><br>
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.</p>