I've found a problem in Set theory that I can't really get my head around. The problem is: Under ordinal exponentiation find an ordinal $\mu$ such that $\omega < \mu$ and $2^{\mu } = \mu$ (where $\omega$ is the inductive set of natural numbers and < is $ \in$)
I think the correct approach would be to use transfinite recursion in some way, but I don't really have any practice in actually applying transfinite recursion for ordinals.
Thanks
The following is indicative of a general technique for finding (arbitrarily large) fixed points of some normal function $F:\mathrm{Ord}\rightarrow \mathrm{Ord}$ (i.e. one that continuous, in the sense that if $\beta=\sup B$ then $F(\beta)=\sup\{F(\alpha)\mid \alpha\in B\}$, is inflationary, in the sense that $\alpha<F(\alpha)$, and is increasing, though you can actually weaken these conditions to just requiring that $F$ be order-preserving, continuous, and satisfy $\alpha\leq F(\alpha)$ to ensure arbitrarily large fixed points, though the naming convention for such a function on ordinals doesn't seem to be standard).
Take $\mu_0=\omega+1$, then let $\mu_1=2^{\omega+1}$, and $\mu_2=2^{2^{\omega+1}}$, and so on, with $\mu_{i+1}=2^{\mu_i}$ for each $i\in \omega$. Then take the supremum of the set $\{\mu_i \mid i\in \omega\}$ to get an order $\mu$. Now, $\mu_0=\omega+1$ is already larger than $\omega$, so we know that $\omega<\mu$. Additionally, $$2^\mu=\sup\{2^{\mu_i} \mid i\in \omega\}=\sup\{\mu_{i+1} \mid i\in \omega\}=\sup\{\mu_i \mid i\in \omega\}=\mu$$
The crux of the argument is that $2^\mu=\sup\{2^{\mu_i} \mid i\in \omega\}$. In general, given any ordinal $\alpha$ and $\beta=\sup B$, then $\alpha^\beta=\sup\{\alpha^\gamma \mid \gamma\in B\}$.
But of course, recall that in the inductive definition of ordinal exponentiation, if $\beta$ is a (non-zero) limit ordinal, then our definition of $\alpha^\beta$ is $\sup\{\alpha^\gamma \mid \gamma<\beta\}$. Since $\beta=\sup B$, we know that $B$ is cofinal in $\{\gamma \mid \gamma<\beta\}$ and so because the function $\gamma\mapsto \alpha^\gamma$ is order-preserving for each $\gamma<\beta$ there is $\delta\in B$ with $\gamma\leq \delta$, and so $\alpha^\gamma\leq \alpha^\delta$, showing that $\{\alpha^\gamma \mid \gamma\in B\}$ is cofinal in $\{\alpha^\gamma \mid \gamma<\beta\}$, and as such $\sup\{\alpha^\gamma \mid \gamma<\beta\}\leq \sup\{\alpha^\gamma \mid \gamma\in B\}$. Since $B\subset \{\gamma \mid \gamma<\beta\}$, the fact that $\sup\{\alpha^\gamma \mid \gamma\in B\}\leq \sup\{\alpha^\gamma \mid \gamma<\beta\}$ is immediate, so we find that $\alpha^\beta=\sup\{\alpha^\gamma \mid \gamma\in B\}$.
On the other hand, if $\beta$ is not a limit ordinal, then either $\beta=0$ and the result is obvious (as then $B$ is either $\emptyset$ or $\{0\}$ and the result follows from the base case for the definition of ordinal exponentiation) or $\beta$ is a successor ordinal. In this latter case, we must have that $\beta\in B$, so because ordinal exponentiation is order-preserving, we find that $\sup\{\alpha^\gamma \mid \gamma\in B\}=\alpha^\beta$.