field generated by a set 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
 A: 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.)
A: My proof is flawed. I will update it if I find a correct one.
$S$ indeed generates $\mathbb{R}$. 
First we establish, with relative ease, that  $\epsilon^2=\epsilon \implies\epsilon=0$ or $1$. Clearly, as $1 \in S$, it additively generates $\mathbb{Z}$ and before you know it, $\mathbb{Q} \in S$, since it is the smallest field containing $\mathbb{Z}$.
Consider then $a= \displaystyle\sum_{n=1}^\infty \frac{1}{n!}.$ So, $a\in S$. We now show that any real number in $(1,a)$ is in $S$. 
Pick $r \in (1,a)$. Now, Let $\epsilon_1=1$. Now inductively define $\epsilon_k$ as follows 
If $\displaystyle \sum_{n=1}^k\frac{1}{n!}>a$, let $\epsilon_k=0$. 
If not, let $\epsilon_k=1$. If $a=\displaystyle \sum_{n=1}^k\frac{1}{n!}$, we're done, let all other $\epsilon_n=0$, and $a \in S$ as desired. 
We constructed the $\epsilon_n$ sequence to ensure that $\displaystyle \sum_{n=1}^\infty\frac{\epsilon_n}{n!}=r$. So,$r \in S$. (This is where the mistake is: I only know that the sum of our subseries is  less than $r$, as Jason shows in the comment.)
Then, since we have an interval, by suitably rescaling and translating it using the rational numbers already in the set, it follows that $\mathbb{R} \subset S$.
Acknowledgements:
My sincere thanks to Brian.M.Scott who showed me this approach when we were discussing another problem.
