# Which cardinal tower is larger?

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

so

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

and

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

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?

• It's correct.${}$ – Simply Beautiful Art Apr 13 at 12:35
• The question is not incorrect. The answer to “which is larger” simply is ”neither”. – celtschk Apr 13 at 14:03
• @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. – Adam Apr 13 at 15:21
• Beware the trick question. Which is greater: One dozen or two six-packs? – DanielWainfleet Apr 19 at 19:52

## 1 Answer

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 $$\eta>\xi$$, $$\xi^\eta=2^\eta$$).