MathOverflow Asked by Alexander Pruss on September 6, 2020
Consider these two axioms:
Question: Does 2 imply 1 in ZF?
Notes:
A. In ZF, Boolean Prime Ideal implies 1 (but not conversely if ZF is consistent), and 1 implies 2 (take the quotient of the preordered set under $sim$, where $xsim y$ iff $xlesssim y$ and $ylesssim x$, and apply 1).
B. Also, 2+(every set has a linear order) implies 1. (Use 2 to extend the partial order $le$ to a total preorder $lesssim$ preserving strict orderings. We now need to turn $lesssim$ into a linear order. To do that, we just need to linearly order within each equivalence class under $sim$. Do that by taking a linear order on the whole set and using that to induce the order in each equivalence class—though not between them.)
C. Claim 2 implies Banach-Tarski and thus the existence of nonmeasurable sets (Pawlikowski’s proof of Banach-Tarski from Hahn-Banach can be adapted), and hence 2 is not provable in ZF (if ZF is consistent).
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP