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

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

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

$0$ 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?

  • 2
    $\begingroup$ The sets are all countable, right? So their union is countable? So it can't be $\bf C$? $\endgroup$ – Gerry Myerson Dec 4 '11 at 12:37
  • $\begingroup$ I believe it is unknown whether, say, $\log\log2$ is irrational, which means pretty much nothing is known about intersections of your sets. $\endgroup$ – Gerry Myerson Dec 4 '11 at 12:40
  • $\begingroup$ See the discussion of math.stackexchange.com/questions/13054/… for some thoughts on what is and what isn't known about these sets. $\endgroup$ – Gerry Myerson Dec 4 '11 at 12:52
  • 1
    $\begingroup$ This set of tags is just off. $\endgroup$ – Asaf Karagila Dec 4 '11 at 13:49
  • 2
    $\begingroup$ It is (in my humble opinion) substandard to use $\Bbb C$ for anything. Real mathematicians (and complex ones, too) use $\bf C$. $\endgroup$ – Gerry Myerson Dec 4 '11 at 23:05

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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