# powers of singular cardinals

I am trying to solve the following two problems:

1) if $\beta <\omega_1$, $2^{\aleph_1}<\aleph_{\omega_1}$, and $\aleph_\alpha^{\aleph_0} \leq \aleph_{\alpha +\beta}$ for a stationary set of $\alpha$'s, then $\aleph_{\omega_1}^{\aleph_1} \leq \aleph_{\omega_1+\beta}$

2) if $2^{\aleph_{\omega_1+\alpha}} < \aleph_{\omega_1+\alpha+\alpha}$ for all $\alpha <\omega_1$, then $2^{\aleph_{\omega_1 + \omega_1}} < \aleph_{\omega_1 +\omega_1 +\omega_1}$.

I was trying to prove 2) by modifying Galvin-Hajnal theorem, but I get stuck at a certain point. As far as 1) is concerned, I suppose one has to go by contradiction, but I don't see how to get one.

## 1 Answer

The first proof appear in Introduction to Cardinal Arithmetic by Holz, Steffens, Weitz. The first question appears as a corollary to the Galvin-Hajnal theorem.

The first question is Corollary 2.3.7, appearing on page 124 in the edition open in front of me. While the second does not appear in this form I deduced it from similar lemmas/exercises appearing in the same chapter as well common cardinal arithmetic (appearing in the first chapter of the book)

I have given the proof in a relatively self-contained form here:

First question: Suppose $2^{\aleph_1}<\aleph_{\omega_1}$, and $\beta<\omega_1$ such that the set $\{\alpha<\omega_1\mid\aleph_\alpha^{\aleph_0}\le\aleph_{\alpha+\beta}\}$ is stationary in $\omega_1$ then $\aleph_{\omega_1}^{\aleph_1}\le\aleph_{\omega_1+\beta}$.

By the assumption $2^{\aleph_1}=\aleph_\xi<\aleph_{\omega_1}$ for some $\xi<\omega_1$, therefore for $\alpha>\xi\cdot\aleph_0=\xi'$ we have $\aleph_\alpha^{\aleph_1}=2^{\aleph_1}\cdot\aleph_\alpha^{\aleph_0} = \aleph_\alpha^{\aleph_0}\le\aleph_{\alpha+\beta}$.

This means that both $\aleph_{\omega_1}$ is closed under $\aleph_1$ powers, as well the following equality: $$\{\xi'<\alpha<\omega_1\mid\aleph_\alpha^{\aleph_0}\le\aleph_{\alpha+\beta}\} = \{\xi'<\alpha<\omega_1\mid\aleph_\alpha^{\aleph_1}\le\aleph_{\alpha+\beta}\}$$

We can, if so, prove the same fact about the RHS of this equality. We do so by defining the function $\Phi(\alpha)$ to be such that $\aleph_\alpha^{\aleph_1}=\aleph_{\alpha+\Phi(\alpha)}$.

Since we have that $\aleph_{\omega_1}$ is closed under powers of $\aleph_1$ this is a function from $\omega_1$ to $\omega_1$, and by the assumption we have that $\{\alpha<\omega_1\mid\Phi(\alpha)\le\beta\}$ is stationary, therefore the rank of $\Phi$ is at most $\beta$.

We have the needed conditions for the Galvin-Hajnal lemma, taking the normal sequence simply as $\langle \alpha\mid\alpha<\omega_1\rangle$, we have that $\aleph_\alpha^{\aleph_1} = \aleph_{\alpha+\Phi(\alpha)}$ and therefore $\aleph_{\omega_1}^{\aleph_1} \le \aleph_{\omega_1+\beta}$.

Second question: If $2^{\aleph_{\omega_1+\alpha}}<\aleph_{\omega_1+\alpha+\alpha}$ for all $\alpha<\omega_1$, then $2^{\aleph_{\omega_1+\omega_1}}<\aleph_{\omega_1+\omega_1+\omega_1}$. (The assumption cannot be true if $\alpha<2$. I will assume that as well, and will conveniently ignore these cases when I say "for all $\alpha$")

By the assumption that if $\alpha<\omega_1$ then $2^{\aleph_{\omega_1+\alpha}}<\aleph_{\omega_1+\alpha+\alpha}<\aleph_{\omega_1+\omega_1}$ we have that latter is a strong limit with cofinality $\aleph_1$, and therefore $\aleph_{\omega_1+\omega_1}^{\aleph_1} = 2^{\aleph_{\omega_1+\omega_1}}$.

We can, if so, reduce back to the first question, apply the same argument with $\Phi(\alpha)$ defined to be such that $\aleph_{\omega_1+\alpha}^{\aleph_1}=\aleph_{\omega_1+\alpha+\Phi(\alpha)}$.

Since we have $\aleph_{\omega_1+\alpha}^{\aleph_1}\le 2^{\aleph_{\omega_1+\alpha}} < \aleph_{\omega_1+\alpha+\alpha}$ we have that $\Phi(\alpha)<\alpha$. for all $\alpha<\omega_1$. Therefore the rank of $\Phi$ over the non-stationary ideal of $\omega_1$ is some $\rho\le\omega_1$.

As before we have by the Galvin-Hajnal lemma that $\aleph_{\omega_1+\omega_1}^{\aleph_1} \le \aleph_{\omega_1+\omega_1+\rho}$.

If $\rho<\omega_1$ we are done.

Otherwise, we have $2^{\aleph_{\omega_1+\omega_1}}= \aleph_{\omega_1+\omega_1}^{\aleph_1} \le \aleph_{\omega_1+\omega_1+\omega_1}$.

This cannot be an equality since by Konig's lemma we have that $\operatorname{cf}(2^\kappa)>\operatorname{cf}\kappa$, and both $\aleph_{\omega_1+\omega_1}$ and $\aleph_{\omega_1+\omega_1+\omega_1}$ are of the same cofinality.

We have if so that $2^{\aleph_{\omega_1+\omega_1}}<\aleph_{\omega_1+\omega_1+\omega_1}$ as needed.

• I apologize in advance for any repeated editing, after two and a half hours of trying to solve this question directly into the answer box there are bound to be typos and extraneous statements. – Asaf Karagila May 23 '11 at 21:38