0
$\begingroup$

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?

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    The sets are all countable, right? So their union is countable? So it can't be $\bf C$?2011-12-04
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    I believe it is unknown whether, say, $\log\log2$ is irrational, which means pretty much nothing is known about intersections of your sets.2011-12-04
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    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.2011-12-04
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    This set of tags is just off.2011-12-04
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    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.2011-12-04
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    It is (in my humble opinion) substandard to use $\Bbb C$ for anything. Real mathematicians (and complex ones, too) use $\bf C$.2011-12-04
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    @Gerry: What happened to your wonderful, and this time relevant too, comment about how [descriptive-set-theory] is not about describing sets? :-)2011-12-04
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    @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).2011-12-04
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    @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...2011-12-04
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    @AsafKaragila, good. I've added transcendence-theory.2011-12-04
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    @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?2011-12-04
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    I see no reason to remove it (and no reason to continue it).2011-12-05

1 Answers 1

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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.