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.


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?

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

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.


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