 Solution

For a countable fixed point of a normal ordinal function, first show that $\Phi$ maps every countable ordinal to a countable ordinal. Use transfinite induction below ${\omega }_1$. The zero case is assumed, and the successor case is the stated preservation property. If $\lambda$ is a countable limit, normality gives $\Phi (\lambda )={\bigcup }_{\alpha <\lambda }\Phi (\alpha )$, a countable union of countable sets, hence countable. This countable-union step is where axiom of countable choice is used; the assumed axiom of choice supplies it.

Now set ${\alpha }_0=0$, ${\alpha }_{n+1}=\Phi ({\alpha }_n)$. The preceding result makes every term countable, and the domination proved earlier makes the sequence nondecreasing. If it ever stabilizes, it already gives the required fixed point. Otherwise $\alpha ={\sup }_{n<\omega }{\alpha }_n$ is a countable nonzero limit, using axiom of countable choice again for countability of the union. Its defining sequence is cofinal, so monotonicity and continuity giveApply this to $\Phi (\alpha )={\omega }^{\alpha }$. This is normal, begins at $1$, and ${\omega }^{\alpha +1}={\omega }^{\alpha }\cdot \omega$ is countable whenever ${\omega }^{\alpha }$ is countable. Thus a countable $ϵ$ satisfies $\boxed{ϵ={\omega }^{ϵ}}$. Starting at zero gives the familiar tower sequence $0,1,\omega ,{\omega }^{\omega },\dots$, whose supremum is ${ϵ}_0$. The finite recursion, ordinal identities and fixed-point continuity do not require choice; the uses of choice above concern countable unions of countable ordinals.