2
$\begingroup$

This is question is simple to write but (I think) hard to solve. Does the following equality hold? $$\bigcup_{{\beta}<\beth_{\omega_1}}\beth_{\omega+1}(|\beta+\omega|) =\beth_{\omega_1} $$

Where $\beta$ is an ordinal.

Proof:

Suppose that, $\bigcup_{{\beta}<\beth_{\omega_1}}\beth_{\omega+1}(|\beta+\omega|) > \beth_{\omega_1}$. Since $\beth_{\omega_1}$ is a limit cardinal, we have that there exists some $\beta<\beth_{\omega_1}$ such that $\beth_{\omega+1}(|\beta+\omega|)>\beth_{\omega_1}$. Since $\beta<\beth_{\omega_1}$, we assume WLOG that $\beta=\beth_\alpha$ where $\alpha<\omega_1$. Then, $\beth_{\omega+1}(|\beta+\omega|)=\beth_{\omega+1}(| \beth_\alpha|)= \beth_{\omega+1} (\beth_\alpha)<\beth_{\omega_1}$ which is a contradiction.

Note the other direction is almost trivial, and hence we have equality.

Edit: I originally asked for a proof of this question until Andres Caicedo convinced me that it was easy. Is the above proof correct?

$\endgroup$
  • 2
    $\begingroup$ The equality holds. It is actually not hard to show. $\endgroup$ – Andrés E. Caicedo Jul 2 '14 at 5:19
  • 1
    $\begingroup$ @AndresCaicedo: I took your advice and attempted a proof! $\endgroup$ – Kyle Jul 2 '14 at 5:36
2
$\begingroup$

The proof is fine. Here is a suggestion for improvement.

Don't assume towards contradiction. Argue directly. It amounts to the same proof and just makes things clearer (because you don't have to worry about the contradiction).

If $\beta<\beth_{\omega_1}$, then there is some $\alpha$ such that $\beta\leq\beth_\alpha$, therefore $$\beth_{\omega+1}(|\beta+\omega|)\leq\beth_{\omega+1}(\beth_\alpha)=\beth_{\alpha+\omega+1}<\beth_{\omega_1}.$$ It follows that the supremum of those cardinals cannot be more than $\beth_{\omega_1}$. The other direction is easy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.