Can you take Toernquist's inaccessible away?
英文摘要：
We prove that ZF+DC+"There are no mad families" is equiconsistent with ZFC.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
继续探索
相关论文
你能去掉Komjath的不可达基数吗？
Can You Take Komjath's Inaccessible Away?
本文比较Kurepa树与Aronszajn树，通过序数步方法将相关相容性结论的大基数假设从两个不可达基数降至一个，还得到无需大基数的相容性结果及MA相关推论，并证明了Todorcevic ρ函数的新引理。
math.LO
Can you take Akemann--Weaver's $\diamondsuit_{\aleph_1}$ away?
By Glimm's dichotomy, a separable, simple $\textrm{C}^*$-algebra has continuum-many unitarily inequivalent irreducible representations if, and only if, it is non-type I while all of its irreducible representations are unitarily equivalent if, and only if, it is type I. Naimark asked whether the latter equivalence holds for all $\textrm{C}^*$-algebras. In 2004, Akemann and Weaver gave a negative answer to Naimark's problem, using Jensen's diamond axiom $\diamondsuit_{\aleph_1}$, a powerful diagonalization principle that implies the Continuum Hypothesis ($\mathsf{CH}$). By a result of Rosenberg,…
math.LOmath.OA
