116c- Lecture 7

April 22, 2008

We presented the proof that “k-trichotomy” implies choice. The following is still open:

Question. ({\sf ZF}) Assume that X is non-well-orderable. Is there a countably infinite family of pairwise size-incomparable sets?

We mentioned a few (familiar) statements that fail in the absence of choice, like the existence of bases for any vector space, Tychonoff’s theorem, or the “surjective” version of the Schröder-Bernstein theorem.

We defined addition, multiplication and exponentiation of cardinals, and verified that addition and multiplication are trivial. We stated the continuum hypothesis {\sf CH}, and the generalized continuum hypothesis {\sf GCH}.

We want to prove (in subsequent lectures) a few non-trivial results about the behavior of exponentiation. In order to do this, we need the key notion of cofinality. We proved a few basic facts about cofinality and defined regular cardinals.


116c- Homework 3

April 22, 2008

Homework 3.

Update. Now due Wednesday, April 30 at 2:30 pm.

Corrections. (Thanks to Fedor Manin for noticing these.)

  • On Exercise 2.(c), assume in addition that (\lambda,\alpha) satisfies the conditions of (\alpha,\beta) in item 2.(a); this should really be all that is needed of 2.(c) for later parts of the exercise.
  • On Exercise 2.(f), we also need n>0.

Update. Here is a quick sketch of the proof of the Milner-Rado paradox.

First notice that the result is clear if \kappa=\omega, since we can write any \alpha<\omega_1 as a countable union of singletons. So we may assume that \kappa is uncountable.

Notice that |\kappa^{\cdot\rho}|=\kappa. This can be checked either by induction on \rho<\kappa^+, or by using the characterization of ordinal exponentiation in terms of functions of finite support.

Notice that the function \alpha\mapsto\omega^{\cdot \alpha} is normal. By the above, it follows that \omega^{\cdot\lambda}=\lambda for all uncountable cardinals \lambda. In particular, it suffices to prove the result for ordinals \alpha that are an ordinal power of \kappa, since these ordinals are cofinal in \kappa^+, and a representation as desired for an ordinal gives (by restriction) such a representation for any smaller ordinal.

By the above, \kappa^{\cdot\rho}=\omega^{\cdot\kappa\cdot\rho} for any \rho. It is easy to see that for any \delta, if \alpha<\omega^{\cdot\delta}, then the interval \null[\alpha,\omega^{\cdot\delta}) is order isomorphic to \omega^{\cdot\delta}; this can be proved by a straightforward induction on \delta.

For any \alpha<\kappa^+, we can write \kappa^{\cdot\alpha+1}=\bigcup_{\nu<\kappa}A_\nu, where A_\nu=[\kappa^{\cdot \alpha}\cdot\nu,\kappa^{\cdot\alpha}\cdot(\nu+1)) so it has order type \kappa^{\cdot\alpha}.

Also, if \alpha is a limit ordinal below \kappa^+, then we can write \alpha=\sup_{\nu<\mbox{\small cf}(\alpha)}\beta_\nu for some strictly increasing continuous sequence (\beta_\nu:\nu<\mbox{cf}(\alpha)) cofinal in \alpha. Let A_0=\kappa^{\cdot\beta_1} and A_\nu=[\kappa^{\cdot\beta_\nu},\kappa^{\cdot\beta_{\nu+1}}) for 0<\nu<\mbox{cf}(\alpha). Then \kappa^{\cdot\alpha}=\bigcup_\nu A_nu and each A_\nu has order type \kappa^{\cdot\beta_{\nu+1}}.

[That the sequence \beta_\nu is continuous (at limits) ensures that the A_\nu cover \kappa^{\cdot \alpha}. That they have the claimed order type follows from the “straightforward inductive argument” three paragraphs above.]

So we have written each \kappa^{\cdot\rho} as an increasing union of \sigma many intervals whose order types are ordinal powers of \kappa, and \sigma is either \kappa or \mbox{cf}(\rho). Now proceed by induction. We may assume that each ordinal below \kappa^{\cdot\rho} can be written as claimed in the paradox. In particular, each A_\nu, having order type an ordinal smaller than \kappa^{\cdot\rho}, can be written that way, say A_\nu=\bigcup_n A_{\nu,n} where \mbox{ot}(A_{\nu,n})<\kappa^{\cdot n}. If \sigma=\omega, this immediately gives the result for \kappa^{\cdot\rho}: Take X_0=\emptyset and X_{2^m(2n+1)}=A_{m,n}. Clearly their union is \kappa^{\cdot\rho} and they have small order type as required. If \sigma>\omega, take X_0=X_1=\emptyset and X_{n+2}=\bigcup_\nu A_{\nu,n}. Again, their union is \kappa^{\cdot\rho}, and \mbox{ot}(X_{n+2}) is at most the order type of concatenating \kappa many copies of \kappa^{\cdot n} [it is here that we use that A_\nu<A_\mu for \nu<\mu], so \mbox{ot}(X_{n+2})\le\kappa^{\cdot n+1}.