116c- Homework 4

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\lambda2^\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.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: