# Prove that $\omega + \omega_1 = \omega \cdot \omega_1 = \omega^{\omega_1} = \omega_1$

I am assuming already that a) the union of countably many countable sets is countable and b) $\omega_1$ is the least uncountable ordinal, so $x < \omega_1$ if and only if $x$ is a countable ordinal.

I'm not sure if it is relevant but this question also allows The Axiom of Choice.

Thanks

• @Antoine: These are ordinal operations. – Asaf Karagila Jan 25 '14 at 11:38
• @C.K: Do you mean $\omega^{\omega_1}$? Because $\omega_1^\omega$ is strictly larger than $\omega_1$. – Asaf Karagila Jan 25 '14 at 11:39
• Also, what have you tried? – Asaf Karagila Jan 25 '14 at 11:39
• @Antoine: I have read about this briefly in a Classic Set Theory textbook but I thought this was Ordinal Arithmetic, unless I am mistaken. – C.K Jan 25 '14 at 11:40
• @AsafKaragila: Yes sorry, the title has now been edited. – C.K Jan 25 '14 at 11:47

## 1 Answer

HINT: Recall the induction definition of all ordinal arithmetic. If $\delta$ is a limit ordinal, then $\alpha\square\delta=\sup\{\alpha\square\gamma\mid\gamma<\delta\}$ (where $\square$ can be addition, multiplication or exponentiation). Now prove the following lemma, and you're about done:

Lemma. If $\square$ is any ordinal arithmetic operation, and at least one of $\alpha,\beta$ is infinite, then $|\alpha\square\beta|=\max\{|\alpha|,|\beta|\}$.

• I have a kind of different but related question. Is it true that $\sum_{i\in\omega_1}\alpha_i=\sup\{\sum_{i\in\gamma}\alpha_i : \gamma\lt\omega_1\}$? – Bumblebee Jul 7 '17 at 4:05
• Are you talking about ordinals or cardinals? – Asaf Karagila Jul 7 '17 at 6:43
• Ordinals. sorry I forgot to mention it. – Bumblebee Jul 7 '17 at 14:31