Was Sierpinski right? IV
逻辑
2016-09-07 v1
摘要
We prove for any mu = mu^{< mu}< theta < lambda, lambda large enough (just strongly inaccessible Mahlo) the consistency of 2^mu = lambda-> [theta]^2_3 and even 2^mu = lambda-> [theta]^2_{sigma,2} for sigma < mu . The new point is that possibly theta > mu^+ .
引用
@article{arxiv.math/9712282,
title = {Was Sierpinski right? IV},
author = {Saharon Shelah},
journal= {arXiv preprint arXiv:math/9712282},
year = {2016}
}