Some ordinal arithmetic exercises

Question

I reckon it's easier if I don't open an extra thread for each of these small exercises. I'm trying to get a better grasp of ordinal arithmetic. If someone could please give me feedback on my solutions. :)

1. Every ordinal is of the form $\alpha+n$, where $\alpha$ is a limit ordinal and $n\in\omega$.
2. Every ordinal is of the form $\omega^\alpha\cdot m+n$, where $\alpha$ is an ordinal and $m,n\in\omega$.
3. If $\alpha=\omega\cdot\alpha$ (edited, previously: $\omega=\omega\cdot\alpha$), then $\alpha=\omega^\omega\cdot\beta$ for some $\beta$.
4. If $ \alpha=\omega^\alpha $, then $\alpha$ is uncountable.
5. If $\alpha>1$ and $\alpha=\alpha^\omega$, then $\alpha$ is uncountable.

1. True:
   • $\lambda$ a limit $\implies\lambda=\lambda +0$
   • $\beta=\alpha + n\implies \beta+1=\alpha+(n+1)$
2. False. Suppose $\omega^2+\omega = \omega^\alpha\cdot m + n$. Clearly need $\alpha=2$ and $m=1$ (since $\omega^2 < \omega^2 + \omega^2 = \omega^2 \cdot 2$. But $n\in\omega$, so doesn't work.
3. I need a hint for this one. I'm trying to prove it using induction but I struggle with the "if then" format. Given $\beta+1=\omega(\beta+1)$, I cannot assume $\beta=\omega\cdot\beta$ so induction seems to be pointless?
4. and 5. Both false. I think $\epsilon_0=\sup\{\omega,\omega^\omega,\omega^{\omega^\omega},\ldots\}$ is a counterexample, or do we not have $\epsilon_0^\omega = \epsilon_0$? My idea is a tower of infinitely many $\omega$ so that taking the power of one more $\omega$ cannot make a difference. I'm not sure if this is a valid ordinal, though.

Answer 1

Hint for 3. Ordinal multiplication is associative, so if $\omega = \omega \cdot \alpha$ and $\alpha = \omega^\omega \cdot \beta$, then $$\begin{align} \omega &= \omega \cdot \alpha \\ &= \omega \cdot (\omega^\omega \cdot \beta) \\ &= (\omega \cdot \omega^\omega) \cdot \beta \\ &= \omega^\omega \cdot \beta. \\ \end{align}$$ What could $\beta$ possibly be?

PS after you revised 3. — You've figured it out from @bof's hints in comments.

Re 4. Yes, $\epsilon_0$ is a counterexample. It's the least fixed point of $\alpha\mapsto \omega^\alpha$, so it's the limit (union) of the $\omega$-sequence $(0, 1, \omega, \omega^\omega, \omega^{\omega^\omega}, \cdots)$. By induction, every term of that sequence is countable; so $\epsilon_0$ is countable, as it's a countable union of countable sets.

Re 5. As per @bof's comments, $\alpha^\omega > \alpha$ for $\alpha > 1$, so 5. is false. (It's not true that $\epsilon_0^\omega = \epsilon_0$.)