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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 $S$)2012-05-03
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    basicly if you know the factorial base , every real $x$ such that $0$d_n$ of integeres such that$ 0 \leq d_n < n$ and $ x= \sum{ \frac{d_{n}}{n!}}$2012-05-03
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    yes, but is it in $K$ if $d_n→∞$?2012-05-03
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    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 $K$ , regards2012-05-03
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    This is why I provided $a$ with an unbounded yet small representation in factorial base: it is more likely to have a unique canoncical way of getting it with elements of $S$. Now have you an answer to the second part of my comment? ($R≠ℝ$? .. or even $G≠ℝ$?).2012-05-03
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    no I don't have an answer2012-05-03
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    Not sure if "$\sum_{n \ge 0}$" is referring to a finite or infinite sum, but I'm going to assume infinite. Obviously $\epsilon_{n} \in \{0,1\}$, so $\displaystyle \sum_{n=0}^{\infty} \frac{\epsilon_{n}}{n!} \in \{0, \sum_{n=0}^{\infty} \frac{1}{n!} \} = \{0,e \}$. $e$ is nonzero, so $\{0,\frac{1}{e},1,e\} \subseteq \langle S \rangle$. Since 1 is in there, $\mathbb{Q} \subseteq \langle S \rangle$. So now we know that $\mathbb{Q} \subset \langle S \rangle \subseteq \mathbb{R}$. Not sure where to go from here.2012-06-10
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    @JacksonWalters: I think you've misinterpeted the question. By your interpretation, $K$ would be $\mathbb{Q}(e)$, which is countable and thus not $\mathbb{R}$. Notice that it's $\varepsilon_n$, not $\varepsilon$; each one is zero or one, but they don't all have to be the same.2012-09-16
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    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...2012-09-16

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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: http://www.math.osu.edu/~edgar.2/preprints/miller/?2012-09-26
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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)2012-05-03
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    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.2012-05-03
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    The last occurance of "this example", should be "your answer" - sorry to be confusing!2012-05-03
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    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 ?2012-05-03
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    Well, unfortunately I really have no idea how to approach the problem. I'll keep thinking about it, though.2012-05-03
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    @FortuonPaendrag You say that your proof is flawed. I think it would be good if you can probably indicate why do you think it is flawed in the answer itself.2012-09-26
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    @FortuonPaendrag Also, I am offering a bounty on this question, so if you have an idea on how to prove or disprove the result, I encourage you to update your answer.2012-09-26
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    @JasonDeVito I guess your objection is invalid, since we are talking about $S$ generating $R$, rather than $S$ being $R$ itself. Since, if $\frac{1}{10} \in S$ then $\frac{k}{10} \in G_{S}$ where $k \in Z$.2012-09-26
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    @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.2012-09-26
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    Where I mean $K$ by $G_{S}$2012-09-26