I'm revising for my set theory final, and I've been asked to find an ordinal $\mu > \omega$ with $2^\mu = \mu$, then to answer whether $3^\mu = \mu$.
The ordinal I picked as $\mu$ was the union of all the ordinals $\mu_i$ where $\mu_0 = \omega + 1$ and $\mu_{n^+} = 2^{\mu_n}$.
I could show that then $2^\mu = \mu$, but I don't know whether $3^\mu = \mu$ or not, and I was wondering if anyone could help, either with the $\mu$ I picked, or with a different $\mu$ which is easier to work with. Another idea I had for $\mu$ was the limit of $\mu+0 = \omega$ and $\mu_{n^+} = \omega^{\mu_n}$, but I couldn't prove $2^\mu = \mu$ or not.
Thanks!
If $2^\mu=\mu$, then $\mu$ is a limit; so $2^\mu\le 3^\mu\le 4^\mu=2^{(2\mu)}=2^\mu$. Note that a similar argument works with 2 and 3 replaced by any finite $m, n>1$.
Exercise: what about $m, n$ possibly infinite?