March 22
Arthur Apter,
CUNY
A choiceless answer to a question of Woodin
In a lecture presented in July 2023, Moti Gitik discussed the following question from the 1980s due to Woodin, as well as approaches to its solution and why it is so difficult to solve:
Question: Assuming there is no inner model of ZFC with a strong cardinal, is it possible to have a model $M$ of ZFC such that $M \vDash$'$2^{\aleph_\omega} > \aleph_{\omega + 2}$ and $2^{\aleph_n} = \aleph_{n + 1}$ for every $n \lt\omega$', together with the existence of an inner model $N^* \subseteq M$ of ZFC such that for the $\gamma, \delta$ so that $\gamma = (\aleph_\omega)^M$ and $\delta = (\aleph_{\omega + 3})^M,$ $N^* \vDash$'$\gamma$ is measurable and $2^\gamma \ge \delta$'?I will discuss how to find answers to this question, if we drop the requirement that $M$ satisfies the Axiom of Choice. I will also briefly discuss the phenomenon that on occasion, when the Axiom of Choice is removed from consideration, a technically challenging question or problem becomes more tractable. One may, however, end up with models satisfying conclusions that are impossible in ZFC.
Reference: A. Apter, 'A Note on a Question of Woodin', Bulletin of the Polish Academy of Sciences (Mathematics), volume 71(2), 2023, 115--121.