TITLE: field generated by a set
QUESTION [18 upvotes]: Let $S$ be the set of real numbers which can be written in the form $  \sum_{n\geq0}{ \frac{\epsilon_{n}}{n!}}$ ,where ${\epsilon_n}^2=\epsilon_n$ and let $K$ be the field generated by $S$ , help me to prove or disprove that $K=\mathbb{R}$ where $\mathbb{R}$ is the set of real numbers.
Thanks

REPLY [4 votes]: The set $S$ is a compact set of Hausdorff dimension zero.  Even more, all Cartesian powers $S^n$ of $S$ have Hausdorff dimension zero.  The field $K$ it generates still has Hausdorff dimension zero, so it is not $\mathbb R$.  The basic  idea:  $K$ is a countable union of sets $f(E)$, where $E \subseteq S^n$ for some $n$, $f$ is a rational function in $n$ variables, and the gradient of $f$ is bounded on $E$.  Since $f$ satisfies a Lipschitz condition on $E$, the dimension of $f(E)$ is still zero.  
(A more sophisticated version of this argument is in: Edgar & Miller, Real Analysis Exchange 27 (2001) 335--339, Lemma 3.)