## 580 -Cardinal arithmetic (6)

February 17, 2009

3. The Galvin-Hajnal theorems.

In this section I want to present two theorems of Galvin and Hajnal that greatly generalize Silver’s theorem. I focus on a “pointwise” (or everywhere) result, that gives us information beyond the pointwise theorems from last lecture, like Corollary 23. Then I state a result where the hypotheses, as in Silver’s theorem, are required to hold stationarily rather than everywhere. From this result, the full version of Silver’s result can be recovered.

Both results appear in the paper Fred Galvin, András Hajnal, Inequalities for Cardinal Powers, The Annals of Mathematics, Second Series, 101 (3), (May, 1975), 491–498, available from JSTOR, that I will follow closely. For the notion of $\kappa$-inaccessibility, see Definition II.2.20 from last lecture.

Theorem 1. Let $\kappa,\lambda$ be uncountable regular cardinals, and suppose that $\lambda$ is $\kappa$-inaccessible. Let $(\kappa_\alpha:\alpha<\kappa)$ be a sequence of cardinals such that $\prod_{\alpha<\beta}\kappa_\alpha<\aleph_\lambda$ for all $\beta<\kappa.$ Then also $\prod_{\alpha<\kappa}\kappa_\alpha<\aleph_\lambda.$

The second theorem will be stated next lecture. Theorem 1 is a rather general result; here are some corollaries that illustrate its reach:

Corollary 2. Suppose that $\kappa,\lambda$ are uncountable regular cardinals, and that $\lambda$ is $\kappa$-inaccessible. Let $\tau$ be a cardinal, and suppose that $\tau^\sigma<\aleph_\lambda$ for all cardinals $\sigma<\kappa.$ Then also $\tau^\kappa<\aleph_\lambda.$

Proof. Apply Theorem 1 with $\kappa_\alpha=\tau$ for all $\alpha<\kappa.$ ${\sf QED}$

Corollary 3. Suppose that $\kappa,\lambda$ are uncountable regular cardinals, and that $\lambda$ is $\kappa$-inaccessible. Let $\tau$ be a cardinal of cofinality $\kappa,$ and suppose that $2^\sigma<\aleph_\lambda$ for all cardinals $\sigma<\tau.$ Then also $2^\tau<\aleph_\lambda.$

Proof. Let $(\tau_\alpha:\alpha<\kappa)$ be a sequence of cardinals smaller than $\tau$ such that $\tau=\sum_\alpha\tau_\alpha,$ and set $\kappa_\alpha=2^{\tau_\alpha}$ for all $\alpha<\kappa.$ Then $\prod_{\alpha<\beta}\kappa_\alpha=2^{\sum_{\alpha<\beta}\tau_\alpha}<\aleph_\lambda$ for all $\beta<\kappa,$ by assumption. By Theorem 1, $\prod_{\alpha<\kappa}\kappa_\alpha=2^{\sum_\alpha\tau_\alpha}=2^\tau<\aleph_\lambda$ as well. ${\sf QED}$

Corollary 4. Let $\kappa,\rho,\tau$ be cardinals, with $\rho\ge2$ and $\kappa$ regular and uncountable. Suppose that $\tau^\sigma<\aleph_{(\rho^\kappa)^+}$ for all cardinals $\sigma<\kappa.$ Then also $\tau^\kappa<\aleph_{(\rho^\kappa)^+}.$

Proof. This follows directly from Corollary 2, since $\lambda=(\rho^\kappa)^+$ is regular and $\kappa$-inaccessible. ${\sf QED}$

Corollary 5. Let $\rho,\tau$ be cardinals, with $\rho\ge2$ and $\tau$ of uncountable cofinality $\kappa.$ Suppose that $2^\sigma<\aleph_{(\rho^\kappa)^+}$ for all cardinals $\sigma<\tau.$ Then also $2^\tau<\aleph_{(\rho^\kappa)^+}.$

Proof. This follows directly from Corollary 3 with $\lambda=(\rho^\kappa)^+.$ ${\sf QED}$

Corollary 6. Let $\xi$ be an ordinal of uncountable cofinality, and suppose that $2^{\aleph_\alpha}<\aleph_{(|\xi|^{{\rm cf}(\xi)})^+}$ for all $\alpha<\xi.$ Then also $2^{\aleph_\xi}<\aleph_{(|\xi|^{{\rm cf}(\xi)})^+}.$

Proof. This follows from Corollary 5 with $\rho=|\xi|,$ $\tau=\aleph_\xi,$ and $\kappa={\rm cf}(\xi).$ ${\sf QED}$

Corollary 7. Let $\xi$ be an ordinal of uncountable cofinality, and suppose that $\aleph_\alpha^\sigma<\aleph_{(|\xi|^{{\rm cf}(\xi)})^+}$ for all cardinals $\sigma<{\rm cf}(\xi)$ and all $\alpha<\xi.$ Then also $\aleph_\xi^{{\rm cf}(\xi)}<\aleph_{(|\xi|^{{\rm cf}(\xi)})^+}.$

Proof. This follows from Corollary 4: If $\sigma<{\rm cf}(\xi)$, then $\aleph_\xi^\sigma=\aleph_\xi\sup_{\alpha<\xi}\aleph_\alpha^\sigma,$ by Theorem II.1.10 from lecture II.2. But $\xi<(|\xi|^{{\rm cf}(\xi)})^+,$ so both $\aleph_\xi$ and $\sup_{\alpha<\xi}\aleph_\alpha^\sigma$ are strictly smaller than $\aleph_{(|\xi|^{{\rm cf}(\xi)})^+}.$ ${\sf QED}$

Corollary 8. If $2^{\aleph_\alpha}<\aleph_{(2^{\aleph_1})^+}$ for all $\alpha<\omega_1,$ then also  $2^{\aleph_{\omega_1}}<\aleph_{(2^{\aleph_1})^+}.$

Proof. By Corollary 5. ${\sf QED}$

Corollary 9.  If $\aleph_\alpha^{\aleph_0}<\aleph_{(2^{\aleph_1})^+}$ for all $\alpha<\omega_1,$ then also  $\aleph_{\omega_1}^{\aleph_1}<\aleph_{(2^{\aleph_1})^+}.$

Proof. By Corollary 7. ${\sf QED}$

Notice that, as general as these results are, they do not provide us with a bound for the size of $2^\tau$ for $\tau$ the first cardinal of uncountable cofinality that is a fixed point of the aleph sequence, $\tau=\aleph_\tau,$ not even under the assumption that $\tau$ is a strong limit cardinal.