Consider the cardinals $\kappa:=2^{2^{\aleph_0}}$, $\lambda:=2^\kappa$, $\mu:=2^\lambda$. Which cardinal is larger, $\kappa^{\mu^\lambda}$ or $\lambda^{\mu^\kappa}$?

The only rules I believe I need are $2^c>c$ and $a.b=\max \{a,b\}$ for infinite cardinals $a,b,c$. I see that

$$\mu^\lambda = (2^{2^{2^{2^{\aleph_0}}}})^{2^{2^{2^{\aleph_0}}}} = 2^{(2^{2^{2^{\aleph_0}}}.2^{2^{2^{\aleph_0}}})} = 2^{2^{2^{2^{\aleph_0}}}}$$ and $$ \mu^\kappa = (2^{2^{2^{2^{\aleph_0}}}})^{2^{2^{\aleph_0}}} = 2^{(2^{2^{2^{\aleph_0}}}.2^{2^{\aleph_0}})} = 2^{2^{2^{2^{\aleph_0}}}} = \mu $$


$$ \kappa^{\mu^\lambda} = (2^{2^{\aleph_0}})^{2^{2^{2^{2^{\aleph_0}}}}} = 2^{2^{2^{2^{2^{\aleph_0}}}}} $$


$$ \lambda^{\mu^\kappa} = {(2^{2^{2^{\aleph_0}}})}^{2^{2^{2^{2^{\aleph_0}}}}} = 2^{2^{2^{2^{2^{\aleph_0}}}}}$$

but then I get that $\kappa^{\mu^\lambda} = \lambda^{\mu^\kappa}$, which means that either I or the question I'm doing is incorrect. Which one is it?

  • 1
    $\begingroup$ It's correct.${}$ $\endgroup$ – Simply Beautiful Art Apr 13 at 12:35
  • $\begingroup$ The question is not incorrect. The answer to “which is larger” simply is ”neither”. $\endgroup$ – celtschk Apr 13 at 14:03
  • $\begingroup$ @celtschk: There's more context to the question than I've given, which says that particular cardinals (of which I've provided two) can be ordered strictly in ascending order. $\endgroup$ – Adam Apr 13 at 15:21
  • $\begingroup$ Beware the trick question. Which is greater: One dozen or two six-packs? $\endgroup$ – DanielWainfleet Apr 19 at 19:52

Your solution is correct, here is a more readable way of looking at it, remembering that $\mu>\lambda>\kappa$ by Cantor's theorem to simplify the products:

$$\mu^\lambda=(2^\lambda)^\lambda=2^{\lambda\cdot\lambda}=2^\lambda=\mu$$ $$\mu^\kappa=(2^\lambda)^\kappa=2^{\lambda\cdot\kappa}=2^\lambda=\mu$$

Now we're interested in comparing $\kappa^\mu$ and $\lambda^\mu$, but those are both $2^\mu$, since $\mu>\kappa$ and $\mu>\lambda$ (whenever you have two cardinals $\eta>\xi$, $\xi^\eta=2^\eta$).

| cite | improve this answer | |

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.