## 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.

## 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.

## 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

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