Prove $\epsilon_0$ < $\omega_1$

Question

This is a question in ordinal arithmetic. (If anyone has read 'Classic Set Theory' by Derek Goldrei, this question comes from page 252.)

$\epsilon_0$ = sup {$\omega$, $\omega^\omega$, ... } and $\omega$ is the smallest infinite ordinal.

$\omega_1$ is the least uncountable ordinal so that $\alpha < \omega_1$ if and only if $\alpha$ is a countable ordinal.

I will take the definition of an ordinal to be: a set $\alpha$ is an ordinal if (i) $\alpha$ is well-ordered by $\in$ and (ii) if $\beta \in \alpha$ then $\beta$ is a subset of $\alpha$. (or $\alpha$ is $\in$-transitive).

I am also not assuming that $\epsilon_0$ is countable at this point.

Answer 1

As each $\omega^{\omega^\cdots}<\omega_1$, $\epsilon_0\le\omega_1$. Can't be equal because the cofinality of $\epsilon_0$ is obviously $\omega$ and ${\rm cf}(\omega_1)=\cdots$

Answer 2

Each of $\omega, \omega^{\omega}, \dots$ is countable, as you can easily prove by induction; then $\epsilon_0 = \bigcup \{ \omega, \omega^{\omega}, \dots \}$ is a union of countably many countable sets, so it must be countable, hence $<\omega_1$.