## 116c- Homework 4

Due Wednesday, May 7 at 2:30 pm.

Update. In exercise 2, $\lambda=\mu$, of course. In exercise 3, $(\kappa)^\lambda$ in the right hand side must be $(<\kappa)^\lambda$.

Hint for exercise 5: Consider the smallest $\alpha$ such that $\alpha+\beta>\beta$ (ordinal addition).

Update. Here is a quick sketch of the solutions:

• 1.(a). Write $\kappa=\sum_{\alpha<{\rm cf}(\kappa)}\kappa_\alpha$ for some (not necessarily strictly) increasing ${\rm cf}(\kappa)$-sequence of cardinals $\kappa_\alpha<\kappa$. (If $\kappa$ is regular, take $\kappa_\alpha=1$ for all $\alpha$. If $\kappa$ is singular, let the $\kappa_\alpha$ be increasing and cofinal in $\kappa$.) Then $2^\kappa=2^{\sum_{\alpha<{\rm cf}(\kappa)}\kappa_\alpha}=\prod_\alpha2^{\kappa_\alpha}\le\prod_\alpha2^{<\kappa}=(2^{<\kappa})^{{\rm cf}(\kappa)}$ but also $(2^{<\kappa})^{{\rm cf}(\kappa)}\le (2^\kappa)^\kappa=2^\kappa$ and the result follows. If $\kappa$ is strong limit, then $2^{<\kappa}=\kappa$, and we get $2^\kappa=\kappa^{{\rm cf}(\kappa)}$.
• 1.(b). Under the assumption, we may choose the $\kappa_\alpha$ as in 1.(a) such that ${\rm cf}(\kappa)<\kappa_0$ and $2^{\kappa_\alpha}=2^{\kappa_\beta}$ for all $\alpha<\beta<{\rm cf}(\kappa)$. Then $2^\kappa=(2^{<\kappa})^{{\rm cf}(\kappa)}=(2^{\kappa_0})^{{\rm cf}(\kappa)}=2^{\kappa_0}=2^{<\kappa}$.

• 2. Write $\mu$ as a disjoint union of $\mu$ sets $A_\alpha$, $\alpha<\mu$, each of size $\mu$. This is possible since $\mu\times\mu=\mu$. Since each $A_\alpha$ has size $\mu$, it is necessarily cofinal in $\mu$. We then have $\kappa^\mu\ge\prod_{i\in\mu}\kappa_i=\prod_{\alpha\in\mu}\prod_{i\in A_\alpha}\kappa_i\ge\prod_\alpha\kappa=\kappa^\mu$.

• 3. If $\kappa\le 2^\lambda$, then $2^\lambda\le\kappa^\lambda\le(2^\lambda)^\lambda=2^\lambda$.
• If $\lambda<{\rm cf}(\kappa)$ then any function $f:\lambda\to\kappa$ is bounded, so ${}^\lambda\kappa=\bigcup_{\alpha<\kappa}{}^\lambda\alpha$, and $\kappa^\lambda\le\sum_\alpha|\alpha|^\lambda\le\kappa\cdot(<\kappa)^\lambda$; the other inequality is clear.
• Suppose that ${\rm cf}(\kappa)\le\lambda$$2^\lambda<\kappa$, and $\rho\mapsto\rho^\lambda$ is eventually constant as $\rho$ approaches $\kappa$. Choose a strictly increasing sequence $(\kappa_\alpha:\alpha<{\rm cf}(\kappa))$ cofinal in $\kappa$ such that $\kappa_\alpha^\lambda=\kappa_\beta^\lambda$ for $\alpha<\beta<{\rm cf}(\kappa)$. Then $\kappa^\lambda\le(\prod_\alpha\kappa_\alpha)^\lambda=\prod_\alpha\kappa_\alpha^\lambda=\prod_\alpha \kappa_0^\lambda=\kappa_0^{\lambda\cdot{\rm cf}(\kappa)}=\kappa_0^\lambda=(<\kappa)^\lambda$. The other inequality is clear.
• Finally, suppose that ${\rm cf}(\kappa)\le\lambda$, $2^\lambda<\kappa$, and $\rho\mapsto\rho^\lambda$ is not eventually constant as $\rho$ approaches $\kappa$. Notice that if $\rho<\kappa$ then $\rho^\lambda<\kappa$. Otherwise, $\tau^\lambda=(\tau^\lambda)^\lambda\ge\kappa^\lambda\ge\tau^\lambda$ for any $\rho\le\tau<\kappa$, and the map $\rho\mapsto\rho^\lambda$ would be eventually constant below $\kappa$ after all. Hence, $(<\kappa)^\lambda=\kappa$. Choose an increasing sequence of cardinals $(\kappa_\alpha:\alpha<{\rm cf}(\kappa))$ cofinal in $\kappa$. Then $\kappa^\lambda\le(\prod_\alpha\kappa_\alpha)^\lambda=\prod_\alpha\kappa_\alpha^\lambda\le\prod_\alpha (<\kappa)^\lambda=\prod_\alpha\kappa=\kappa^{{\rm cf}(\kappa)}$. The other inequality is clear.

• 5. Suppose that $\beta\ge\omega$ and for all $\alpha$, $2^{\aleph_\alpha}=\aleph_{\alpha+\beta}$. Let $\alpha$ be least such that $\beta<\alpha+\beta$; $\alpha$ exists since $\beta<\beta+\beta$. On the other hand, $\alpha$ is a limit ordinal, since $1+\omega=\omega$ and $\omega\le\beta$, so $(\lambda+n)+\beta=\lambda+(n+\beta)=\lambda+\beta$ for any $\lambda$ and any $n<\omega$, and we would have a contradiction to the minimality of $\alpha$. Let $\gamma<\alpha$. Then $\gamma+\beta=\beta$ and $2^{\aleph_{\alpha+\gamma}}=\aleph_{\alpha+\beta}$. By exercise 1.(b), $2^{\aleph_{\alpha+\alpha}}=\aleph_{\alpha+\beta}$ since $\aleph_{\alpha+\alpha}$ is singular (it has cofinality ${\rm cf}(\alpha)\le\alpha<\alpha+\alpha\le\aleph_{\alpha+\alpha}$). This is a contradiction, since by assumption, $2^{\aleph_{\alpha+\alpha}}=\aleph_{\alpha+\alpha+\beta}>\aleph_{\alpha+\beta}$.

• 4. We want to show that $\prod_{\xi<\beta}\aleph_\xi=\aleph_\beta^{|\beta|}$ for all limit ordinals $\beta$. Write $\beta=|\beta|+\alpha$ where $\alpha=0$ or a limit ordinal, and proceed by induction on $\alpha$, noticing that the case $\alpha=0$ follows from exercise 2.
• Notice that any limit ordinal $\alpha$ can be written as $\alpha=\omega\cdot\delta$ for some nonzero ordinal $\delta$. This is easily established by induction. If $\delta$ is a limit ordinal, then any sequence converging to $\delta$ gives rise in a natural way to a sequence of limit ordinals converging to $\alpha$. On the other hand, if $\delta$ is a successor, then $\alpha=\lambda+\omega$ for some $\lambda=0$ or limit. In summary: any limit ordinal is either a limit of limit ordinals, or else it has the form $\lambda+\omega$ for $\lambda=0$ or limit.
• We divide our induction on $\alpha$ in two cases. Suppose first $\alpha=\lambda+\omega$, as above. Then $\prod_{\xi<\beta}\aleph_\xi=\aleph_{|\beta|+\lambda}^{|\beta|}\prod_n\aleph_{|\beta|+\lambda+n}$, by induction. Since $|\beta|=|\beta|\aleph_0$, we have $\aleph_{|\beta|+\lambda}^{|\beta|}\prod_n\aleph_{|\beta|+\lambda+n}=\prod_n\aleph_{|\beta|+\lambda}^{|\beta|}\aleph_{|\beta|+\lambda+n}=\prod_n\aleph_{|\beta|+\lambda+n}^{|\beta|}$, where the last equality holds by Hausdorff’s result that $(\kappa^+)^\tau=\kappa^+\cdot\kappa^\tau$. Finally, $\prod_n\aleph_{|\beta|+\lambda+n}^{|\beta|}=(\prod_n\aleph_{|\beta|+\lambda+n})^{|\beta|}=(\aleph_\beta)^{\aleph_0\cdot|\beta|}=\aleph_\beta^{|\beta|}$ as wanted, where the previous to last equality is by exercise 2.
• Now suppose that $\alpha$ is a limit of limit ordinals. Choose a strictly increasing sequence $(\gamma_\nu:\nu<{\rm cf}(\alpha))$ of limit ordinals cofinal in $\alpha$. We argue according to which of the four possibilities described in exercise 3 holds. If $\aleph_\beta\le 2^{|\beta|}$ then $\aleph_\beta^{|\beta|}=2^{|\beta|}\le\prod_{\xi<\beta}\aleph_\xi\le\aleph_\beta^{|\beta|}$.
• Notice that ${\rm cf}(\alpha)={\rm cf}(\beta)={\rm cf}(\aleph_\beta)$ and ${\rm cf}(\beta)\le|\beta|$, so the second possibility does not occur.
• Suppose now that we are in the third possibility, i.e., $2^{|\beta|}<\aleph_\beta$ and $\rho\mapsto\rho^{|\beta|}$ is eventually constant as $\rho$ approaches $\aleph_\beta$. Then $\aleph_\beta^{|\beta|}=(<\aleph_\beta)^{|\beta|}=\sup_{\nu<{\rm cf}(\alpha)}\aleph_{|\beta|+\gamma_\nu}^{|\beta|}=\sup_\nu\prod_{\xi<|\beta|+\gamma_\nu}\aleph_\xi$, by induction (it is here that we used the assumption that $\alpha$ is a limit of limit ordinals). And $\sup_\nu\prod_{\xi<|\beta|+\gamma_\nu}\aleph_\xi\le\prod_{\xi<\beta}\aleph_\xi\le\aleph_\beta^{|\beta|}$.
• Finally, if $2^{|\beta|}<\aleph_\beta$ and $\rho\mapsto\rho^{|\beta|}$ is not eventually constant as $\rho$ approaches $\aleph_\beta$, then $\aleph_\beta^{|\beta|}=\aleph_\beta^{{\rm cf}(\alpha)}$ and $\aleph_\beta^{{\rm cf}(\alpha)}=\prod_{\nu<{\rm cf}(\alpha)}\aleph_{|\beta|+\gamma_\nu}$, by exercise 2, but $\prod_{\nu<{\rm cf}(\alpha)}\aleph_{|\beta|+\gamma_\nu}\le\prod_{\xi<\beta}\aleph_\xi\le\aleph_\beta^{|\beta|}$, and we are done.