502 – The Banach-Tarski paradox

December 17, 2009

1. Non-measurable sets

In these notes I want to present a proof of the Banach-Tarski paradox, a consequence of the axiom of choice that shows us that a naive understanding of the concept of volume can lead to contradictions. A good reference for this topic is the very nice book The Banach-Tarski paradox by Stan Wagon.

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580 -Cardinal arithmetic (11)

March 12, 2009

4. Strongly compact cardinals and {{\sf SCH}}

 

Definition 1 A cardinal {\kappa} is strongly compact iff it is uncountable, and any {\kappa}-complete filter (over any set {I}) can be extended to a {\kappa}-complete ultrafilter over {I.}

 

The notion of strong compactness has its origin in infinitary logic, and was formulated by Tarski as a natural generalization of the compactness of first order logic. Many distinct characterizations have been found.

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580 -Cardinal arithmetic (3)

February 9, 2009

It is easy to solve negatively the question immediately following Homework problem 5 on lecture II.1. I asked whether if X is Dedekind-finite but {\mathcal P}(X) is Dedekind-infinite, then it followed that there is an infinite Dedekind-finite set Y such that {\mathcal P}(Y)\preceq X.

To exhibit a counterexample, it is enough to know that it is consistent to have an infinite Dedekind finite set X that is the countable union of finite sets (in fact, sets of size 2). Notice that \omega is a surjective image of X, so {\mathcal P}(X) is Dedekind-infinite. Suppose that {\mathcal P}(Y)\preceq X. Then certainly Y\preceq X, so Y is a countable union of finite sets Y_n. If Y is infinite then Y_n\ne\emptyset for infinitely many values of n. But then \omega is also a surjective image of Y, so \omega (and in fact P(\omega)) injects into {\mathcal P}(Y) and therefore into X, contradiction.

At the end of last lecture we showed Theorem 10, a general result that allows us to compute products \kappa^\lambda for infinite cardinals \kappa,\lambda, namely:

Let \kappa and \lambda be infinite cardinals. Let \tau=\sup_{\rho<\kappa}|\rho|^\lambda. Then 

\displaystyle \kappa^\lambda=\left\{\begin{array}{cl} 2^\lambda & \mbox{if }\kappa\le 2^\lambda,\\ \kappa\cdot\tau & \mbox{if }\lambda<{\rm cf}(\kappa),\\ \tau & \begin{array}{l}\mbox{if }{\rm cf}(\kappa)\le\lambda,2^\lambda<\kappa,\mbox{ and }\\ \rho\mapsto|\rho|^\lambda\mbox{ is eventually constant below }\kappa,\end{array}\\ \kappa^{{\rm cf}(\kappa)} & \mbox{otherwise.}\end{array}\right.

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