# Prove $\omega^{\omega_1}=\omega_1$ and $2^{\omega_1}=\omega_1$

Prove that $\omega^{\omega_1}=\omega_1$ and $2^{\omega_1}=\omega_1$.

I also found an exercise asking to compute the ordinal number $\omega_1^{\omega}$, but I do not even understand what I am supposed to do, any help?

-
Is there any chance, you are supposed to calculate the cantor normal form of $\omega_1^\omega$? – user79202 May 25 '13 at 19:59

Ordinal exponentiation is not cardinal exponentiation.

Recall the definition of $\alpha^\beta$:

• $\alpha^0=1$.
• $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$.
• $\alpha^\beta=\sup\{\alpha^\gamma\mid\gamma<\beta\}$ for a limit ordinal $\beta$.

So now we note the following:

1. If $\alpha,\beta$ are countable ordinals then $\alpha^\beta$ is countable. We can prove this by induction.
2. If $\beta>0$ then $\alpha<\alpha^\beta$.

From this it is somewhat easy to calculate $2^{\omega_1}=\omega^{\omega_1}$, since for every $\beta<\omega_1$ we have $2^\beta$ and $\omega^\beta$ to be countable ordinals, and this is a strictly increasing sequence of length $\omega_1$.

As for the additional exercise for $\omega_1^\omega$, I'm not clear about what it means "to compute the ordinal number", because $\omega_1^\omega$ is an ordinal number.

If one interprets "the ordinal number" as the Cantor normal form of $\omega_1^\omega$, then one can note that:

$$\omega_1^\omega=\left(\omega^{\omega_1}\right)^\omega=\omega^{\omega_1\cdot\omega}$$

I don't know if that's simpler, though.

-
Is it possible to write $\omega_1^\omega$ in a simpler form? – Martin May 25 '13 at 19:59
It depends on your understanding of "simpler". This might be of interest: en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form – user79202 May 25 '13 at 20:05
@Martin: I also interpreted this question as a possible reference to the Cantor normal form of $\omega_1^\omega$, and I have updated my answer about that. It's not simpler in any way, though. – Asaf Karagila May 25 '13 at 20:09