Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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

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.)

share|cite|improve this answer
You mean this: – M Turgeon Sep 26 '12 at 18:18

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.

share|cite|improve this answer
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 ? – Ravi 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. – Ravi Sep 26 '12 at 16:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.