# 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

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Can you prove that $a=\sum\frac{[\log n]}{n!}$ is in $K$? I have trouble decomposing $a$ into elements of $S$. Also, is the question $R=ℝ$ trivial? (where $R$ is the ring generated by $S$) – jmad May 3 '12 at 14:08
basicly if you know the factorial base , every real $x$ such that $0<x<1$ , there is a sequence $d_n$ of integeres such that$0 \leq d_n < n$ and $x= \sum{ \frac{d_{n}}{n!}}$ – mathfan May 3 '12 at 14:12
yes, but is it in $K$ if $d_n→∞$? – jmad May 3 '12 at 14:24
dear jmad that is why I'm asking help , actually if $d_n < M$ for some $M$ than we can easily prove that x is in $K$ , regards – mathfan May 3 '12 at 14:29
Is there a reason to write '$\epsilon_n^2=\epsilon_n$' rather than '$\epsilon_n\in\{0,1\}$'? The latter seems substantially clearer and it's not taking up a whole lot more space... – Steven Stadnicki Sep 16 '12 at 20:07

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.

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I don't quite follow something. If, say, the denominator was $10^n$ instead of $n!$, and you try the same argument, it's NOT true that you get everything in between $\frac{1}{10}$ and $a' = \sum_{n=1}^\infty \frac{1}{10^n} = .1111111...$. In fact, you get precisely the decimals that has only 1s or 0s in their decimal expansion. So, for example, you'll never get $.102$ which is between $.1$ and $.11111....$. (continued) – Jason DeVito May 3 '12 at 18:36
In your argument applied to this example, what goes wrong is that it may be the case that with your choice of $\epsilon_n$ so far, $\sum_{n=1}^k \frac{\epsilon_n}{10^n} < r$, but $\sum_{n=k+1}^\infty \frac{1}{10^n}$ is too small so $\sum_{n=1}^k \frac{\epsilon_n}{10^n} + \sum_{n=k+1}^\infty \frac{1}{10^n} < r$ as well. Since $n!$ eventually grows faster than $10^n$, I'm afraid the same thing will happen in this example, but I haven't actually seen an explicit counterexample yet. – Jason DeVito May 3 '12 at 18:38
The last occurance of "this example", should be "your answer" - sorry to be confusing! – Jason DeVito May 3 '12 at 18:44
No worries @Jason, I got what you meant the second I saw 1.02. But, I think that by using a similar method there is a way to show that a hamel basis is in S ? – Galois Group May 3 '12 at 18:50
@JayeshBadwaik: Dear Jayesh, I have updated my answer as per your request. Also, the case to show that the subsequence sums of $\displaystyle \sum_{n=1}^\infty \frac{1}{10^n}$ is not so bad. But we cannot do anything similar here. I think Jason was only objecting to my claim that there is an entire interval $(1,a)$, for which he does give a good counterexample. – Galois Group Sep 26 '12 at 16:51

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.)

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You mean this: math.osu.edu/~edgar.2/preprints/miller? – M Turgeon Sep 26 '12 at 18:18