Order type of the set of all the limit elements in $\omega^{\omega}$

217 Views Asked by At

What do you think is the order type of $\{\alpha\in\omega^{\omega}:\alpha\ \text{is a limit ordinal number}\}$?

My first thought was $\omega$, but when I visualize the above mentioned set, $\omega^2$ seems to have countably many limit elements, let alone $\omega^{\omega}$. So the other guess is $\omega^{\omega}$, but it's not possible. I'm clueless now...

Any hints?

1

There are 1 best solutions below

2
On BEST ANSWER

The answer is actually $\omega^\omega$.

You can find a bijection form $\{\alpha\in\omega^{\omega}:\alpha\ \text{is a limit ordinal number}\}$ to $\omega^\omega$ by mapping the ordinal $\omega^{n+1}\cdot a_n+\omega^{n}\cdot a_{n-1}+\omega^{n-1}\cdot a_{n-2}\cdots \omega^2\cdot a_1+\omega \cdot a_0$ to $\omega^n\cdot a_n+\omega^{n-1}\cdot a_{n-1}+\omega^{n-2}\cdot a_{n-2}\cdots \omega\cdot a_1+a_0$