4
$\begingroup$

How to prove that every infinite cardinal is equal to $\omega_\alpha$ for some $\alpha$ in Kunen's book, I 10.19?

I will appreciate any help on this question. Thanks ahead.

  • 0
    @ineff: The definitions can be seen in the chapter I of Kunen's book:)2012-01-14

1 Answers 1

9

I took the trouble to read through Kunen in order to understand the problem, as well the definitions which you can use for this.

  1. Cardinal is defined to be an ordinal $\kappa$ that there is no $\beta<\kappa$ and a bijection between $\kappa$ and $\beta$.

  2. The successor cardinal $\kappa^+$ is the least cardinal which is strictly larger than $\kappa$.

  3. $\aleph_\alpha=\omega_\alpha$ defined recursively, as the usual definitions go: $\aleph_0=\omega$; $\aleph_{\alpha+1}=\omega_{\alpha+1}=\omega_\alpha^+$; at limit points $\aleph_\beta=\omega_\beta=\sup\{\omega_\alpha\mid\alpha<\beta\}$.

Now we want to show that:

Every cardinal is an $\omega_\alpha$ for some $\alpha$.

Your question concentrates on the second part of the lemma.

Suppose $\kappa$ is an infinite cardinal. If $\kappa=\omega$ we are done. Otherwise let $\beta=\sup\{\alpha+1\mid\omega_\alpha<\kappa\}$. I claim that $\kappa=\omega_\beta$.

Now suppose that $\omega_\beta<\kappa$ then we reach a contradiction since this means that $\beta<\sup\{\alpha+1\mid\omega_\alpha<\kappa\}=\beta$ (since $\beta$ is in this set, then $\beta<\beta+1\le\sup{\cdots}=\beta$).

If so, $\kappa\le\omega_\beta$. If $\beta=\alpha+1$ then $\omega_\alpha<\kappa\le\omega_\beta$ and by the definition of a successor cardinal we have equality. Otherwise $\beta$ is a limit cardinal and we have that $\omega_\alpha<\kappa$ for every $\alpha<\beta$, then by the definition of a supremum we have that $\omega_\beta\le\kappa$ and again we have equality.

  • 0
    @AsafKaragila I should've read the editing history! Thanks for the comment, Asaf.2012-01-16