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}leprod_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 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<kappa}{}^lambdaalpha, and kappa^lambdalesum_alpha|alpha|^lambdalekappacdot(<kappa)^lambda; the other inequality is clear.
  • Suppose that {rm cf}(kappa)lelambda2^lambda<kappa, 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<beta<{rm cf}(kappa). 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=(<kappa)^lambda. The other inequality is clear.
  • Finally, suppose that {rm cf}(kappa)lelambda, 2^lambda<kappa, and rhomapstorho^lambda is not eventually constant as rho approaches kappa. Notice that if rho<kappa then rho^lambda<kappa. Otherwise, tau^lambda=(tau^lambda)^lambdagekappa^lambdagetau^lambda for any rholetau<kappa, and the map rhomapstorho^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^lambdale(prod_alphakappa_alpha)^lambda=prod_alphakappa_alpha^lambdaleprod_alpha (<kappa)^lambda=prod_alphakappa=kappa^{{rm cf}(kappa)}. 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+beta; alpha exists since beta<beta+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<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)lealpha<alpha+alphalealeph_{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=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<beta}aleph_xi=aleph_{|beta|+lambda}^{|beta|}prod_naleph_{|beta|+lambda+n}, 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<beta}aleph_xilealeph_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 rhomapstorho^{|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_nuprod_{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_nuprod_{xi<|beta|+gamma_nu}aleph_xileprod_{xi<beta}aleph_xilealeph_beta^{|beta|}.
  • Finally, if 2^{|beta|}<aleph_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<beta}aleph_xilealeph_beta^{|beta|}, and we are done.
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