## 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 $(.

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. (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}leprod_alpha2^{ but also $(2^{ and the result follows. If $kappa$ is strong limit, then $2^{, 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) and $2^{kappa_alpha}=2^{kappa_beta}$ for all $alpha. Then $2^kappa=(2^{.

• 2. Write $mu$ as a disjoint union of $mu$ sets $A_alpha$, $alpha, each of size $mu$. This is possible since $mutimesmu=mu$. Since each $A_alpha$ has size $mu$, it is necessarily cofinal in $mu$. We then have $kappa^mugeprod_{iinmu}kappa_i=prod_{alphainmu}prod_{iin A_alpha}kappa_igeprod_alphakappa=kappa^mu$.

• 3. If $kappale 2^lambda$, then $2^lambdalekappa^lambdale(2^lambda)^lambda=2^lambda$.
• If $lambda<{rm cf}(kappa)$ then any function $f:lambdatokappa$ is bounded, so ${}^lambdakappa=bigcup_{alpha, and $kappa^lambdalesum_alpha|alpha|^lambdalekappacdot(; the other inequality is clear.
• Suppose that ${rm cf}(kappa)lelambda$$2^lambda, and $rhomapstorho^lambda$ 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. Then $kappa^lambdale(prod_alphakappa_alpha)^lambda=prod_alphakappa_alpha^lambda=prod_alpha kappa_0^lambda=kappa_0^{lambdacdot{rm cf}(kappa)}=kappa_0^lambda=(. The other inequality is clear.
• Finally, suppose that ${rm cf}(kappa)lelambda$, $2^lambda, and $rhomapstorho^lambda$ is not eventually constant as $rho$ approaches $kappa$. Notice that if $rho then $rho^lambda. Otherwise, $tau^lambda=(tau^lambda)^lambdagekappa^lambdagetau^lambda$ for any $rholetau, and the map $rhomapstorho^lambda$ would be eventually constant below $kappa$ after all. Hence, $(. Choose an increasing sequence of cardinals $(kappa_alpha:alpha<{rm cf}(kappa))$ cofinal in $kappa$. Then $kappa^lambdale(prod_alphakappa_alpha)^lambda=prod_alphakappa_alpha^lambdaleprod_alpha (. The other inequality is clear.

• 5. Suppose that $betageomega$ and for all $alpha$, $2^{aleph_alpha}=aleph_{alpha+beta}$. Let $alpha$ be least such that $beta; $alpha$ exists since $beta. On the other hand, $alpha$ is a limit ordinal, since $1+omega=omega$ and $omegalebeta$, so $(lambda+n)+beta=lambda+(n+beta)=lambda+beta$ for any $lambda$ and any $n, and we would have a contradiction to the minimality of $alpha$. Let $gamma. 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)lealpha). 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 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 $alpha=omegacdotdelta$ 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, by induction. Since $|beta|=|beta|aleph_0$, we have $aleph_{|beta|+lambda}^{|beta|}prod_naleph_{|beta|+lambda+n}=prod_naleph_{|beta|+lambda}^{|beta|}aleph_{|beta|+lambda+n}=prod_naleph_{|beta|+lambda+n}^{|beta|}$, where the last equality holds by Hausdorff’s result that $(kappa^+)^tau=kappa^+cdotkappa^tau$. Finally, $prod_naleph_{|beta|+lambda+n}^{|beta|}=(prod_naleph_{|beta|+lambda+n})^{|beta|}=(aleph_beta)^{aleph_0cdot|beta|}=aleph_beta^{|beta|}$ 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 $aleph_betale 2^{|beta|}$ then $aleph_beta^{|beta|}=2^{|beta|}leprod_{xi.
• 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|} and $rhomapstorho^{|beta|}$ is eventually constant as $rho$ approaches $aleph_beta$. Then $aleph_beta^{|beta|}=(, by induction (it is here that we used the assumption that $alpha$ is a limit of limit ordinals). And $sup_nuprod_{xi<|beta|+gamma_nu}aleph_xileprod_{xi.
• Finally, if $2^{|beta|} and $rhomapstorho^{|beta|}$ 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}leprod_{xi, and we are done.