Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question contains 2 stages:
1. Prove that ${\aleph_n} ^ {\aleph_1} = {2}^{\aleph_1}\cdot\aleph_n$
This one is pretty clear by induction and by applying Hausdorf's formula.

2. Prove ${\aleph_\omega} ^ {\aleph_1} = {2}^{\aleph_1}\cdot{\aleph_\omega}^{\aleph_0}$

There exist an easy proof for the second one?

Thanks in advance.

share|improve this question

1 Answer 1

up vote 7 down vote accepted

As you mention, the first equation is a consequence of Hausdorff's formula and induction.

For the second: Clearly the right hand side is at most the left hand side. Now: Either $2^{\aleph_1}\ge\aleph_\omega$, in which case in fact $2^{\aleph_1}\ge{\aleph_\omega}^{\aleph_1}$, and we are done, or $2^{\aleph_1}<\aleph_\omega$.

I claim that in this case we have ${\aleph_\omega}^{\aleph_1}={\aleph_\omega}^{\aleph_0}$. Once we prove this, we are done.

Note that $\aleph_\omega={\rm sup}_n\aleph_n\le\prod_n\aleph_n$, so $$ {\aleph_\omega}^{\aleph_1}\le\left(\prod_n\aleph_n\right)^{\aleph_1}=\prod_n{\aleph_n}^{\aleph_1}. $$

Now use part 1, to conclude that ${\aleph_n}^{\aleph_1}<\aleph_\omega$ for all $n$, since we are assuming that $2^{\aleph_1}<\aleph_\omega$. This shows that the last product is at most $\prod_n \aleph_\omega={\aleph_\omega}^{\aleph_0}$.

This shows that ${\aleph_\omega}^{\aleph_1}\le {\aleph_\omega}^{\aleph_0}$. But the other inequality is obvious, and we are done.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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