0

Define

$L_0=Q$

$L_1=\lbrace x \in C; e^{x} \in L_0 \rbrace$

$L_{-1}=\lbrace x \in C; \ln{x} \in L_0 \rbrace$

$L_{n+1}=\lbrace x \in C; e^{x} \in L_n \rbrace$

$0$ is in $L_1$ and $L_0$. Do any other numbers belong to more than one of these sets? Are all complex numbers in at least one of the sets?

J. M. ain't a mathematician
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Angela Pretorius
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    The sets are all countable, right? So their union is countable? So it can't be $\bf C$? – Gerry Myerson Dec 04 '11 at 12:37
  • I believe it is unknown whether, say, $\log\log2$ is irrational, which means pretty much nothing is known about intersections of your sets. – Gerry Myerson Dec 04 '11 at 12:40
  • See the discussion of http://math.stackexchange.com/questions/13054/how-to-show-eee79-is-not-an-integer for some thoughts on what is and what isn't known about these sets. – Gerry Myerson Dec 04 '11 at 12:52
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    This set of tags is just off. – Asaf Karagila Dec 04 '11 at 13:49
  • It is standard to use $\mathbb{C}$ for the complex numbers. (right click on $\mathbb{C}$ and choose "Show Source"). And $\mathbb{Q}$ for rationals. – robjohn Dec 04 '11 at 19:16
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    It is (in my humble opinion) substandard to use $\Bbb C$ for anything. Real mathematicians (and complex ones, too) use $\bf C$. – Gerry Myerson Dec 04 '11 at 23:05
  • @Gerry: What happened to your wonderful, and this time relevant too, comment about how [descriptive-set-theory] is not about describing sets? :-) – Asaf Karagila Dec 04 '11 at 23:30
  • @AsafKaragila, once burned, twice shy. But let me encourage you to re-tag. I'd suggest number-theory, and diophantine-analysis (if there is such a tag). – Gerry Myerson Dec 04 '11 at 23:33
  • @Gerry: I did the first too, there are diophantine tags, but not analysis. I would suggest you retag that if needed, since I can't tell for myself... – Asaf Karagila Dec 04 '11 at 23:37
  • @AsafKaragila, good. I've added transcendence-theory. – Gerry Myerson Dec 04 '11 at 23:42
  • @Gerry: Excellent. Should we remove this comment discussion, or should we leave it for future generations to see and tell whether or not our actions were justified? – Asaf Karagila Dec 04 '11 at 23:47
  • I see no reason to remove it (and no reason to continue it). – Gerry Myerson Dec 05 '11 at 00:44

1 Answers1

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Collecting my comments into an answer:

The rationals are countable, so all the sets $L_n$ are countable, so their union is countable, but the complex numbers are uncountable, so not every complex number is in at least one of the sets. That being said, I'm not sure there's even a single number $x$ of which one could say that it has been proven that $x$ is not in the union of the $L_n$.

For a number $x$ to be in more than one of the sets, there would have to be a rational number $y$ such that at least one of the numbers $\exp(y),\exp(\exp(y)),\exp(\exp(\exp(y))),\dots$ is rational. It is known that $y$ and $e^y$ are both rational if and only if $y=0$. To the best of my knowledge, no one knows whether there is any rational $y$ such that some iterated exponential of $y$ is rational. For example, I believe it is unknown whether $e^e=\exp(\exp(1))$ is rational.

Gerry Myerson
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