# Show that $\aleph_{\alpha}^{\aleph_{1}} = \aleph_{\alpha}^{\aleph_{0}}\cdot 2^{\aleph_1}$

So basically I want to show:

$\aleph_{\alpha}^{\aleph_{1}} = \aleph_{\alpha}^{\aleph_{0}}\cdot2^{\aleph_1}$ , Where $\omega\leq\alpha<\omega_1$

There is a proof that I found;

$\aleph_\alpha^{\aleph_1} = \aleph_\alpha^{\aleph_0}\cdot 2^{\aleph_1}$ for all $\omega \le \alpha < \omega_1$

But I don't understand his proof for the limit case of $\alpha$.

To be more specific, I dont understand why $\left( \sup_{\omega \le \beta < \alpha} \aleph_{\beta}\right)^{\aleph_1} \le \left( \prod_{\omega \le \beta < \alpha} \aleph_{\beta}\right)^{\aleph_1}$

Any insights or help is deeply appreciated.

Cheers

• I guess I could have provided a little more detail in my old question... ;-) – Stefan Mesken Jan 29 '18 at 14:54
• Haha, thx, but since you replied, I left one last inquiry on that post, perhaps you could take a look at it ? – some1fromhell Jan 30 '18 at 3:06

For each $\beta$ satisfying $\omega \leq \beta < \alpha$,

$$\aleph_\beta \leq \prod_{\omega \leq \beta < \alpha} \aleph_\beta$$

therefore

$$\sup_{\omega \leq \beta < \alpha} \aleph_\beta \leq \prod_{\omega \leq \beta < \alpha} \aleph_\beta$$

Remember that $\kappa\le\lambda$ implies $\kappa^\mu\le\lambda^\mu$. So this boils down to showing that $$\sup_{\omega\le\beta<\alpha}\aleph_\beta\le\prod_{\omega\le\beta<\alpha}\aleph_\beta.$$

More generally, can you show that if $X$ is a set of nonzero cardinals then $\sup X\le\prod_{\kappa\in X}\kappa$?

• But what if $0\in X$? :) – Asaf Karagila Jan 29 '18 at 14:47
• (God, I hate mathematicians sometimes...) – Asaf Karagila Jan 29 '18 at 14:47
• @AsafKaragila ... dangit. – Noah Schweber Jan 29 '18 at 15:11