A note on surjective cardinals
Abstract
For cardinals $\mathfrak{a}$ and $\mathfrak{b}$, we write $\mathfrak{a}=^\ast\mathfrak{b}$ if there are sets $A$ and $B$ of cardinalities $\mathfrak{a}$ and $\mathfrak{b}$, respectively, such that there are partial surjections from $A$ onto $B$ and from $B$ onto $A$. $=^\ast$-equivalence classes are called surjective cardinals. In this article, we show that $\mathsf{ZF}+\mathsf{DC}_\kappa$, where $\kappa$ is a fixed aleph, cannot prove that surjective cardinals form a cardinal algebra, which gives a negative solution to a question proposed by Truss [J. Truss, Ann. Pure Appl. Logic 27, 165--207 (1984)]. Nevertheless, we show that surjective cardinals form a ``surjective cardinal algebra'', whose postulates are almost the same as those of a cardinal algebra, except that the refinement postulate is replaced by the finite refinement postulate. This yields a smoother proof of the cancellation law for surjective cardinals, which states that $m\cdot\mathfrak{a}=^\ast m\cdot\mathfrak{b}$ implies $\mathfrak{a}=^\ast\mathfrak{b}$ for all cardinals $\mathfrak{a},\mathfrak{b}$ and all nonzero natural numbers $m$.
- Publication:
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arXiv e-prints
- Pub Date:
- August 2024
- DOI:
- 10.48550/arXiv.2408.04287
- arXiv:
- arXiv:2408.04287
- Bibcode:
- 2024arXiv240804287J
- Keywords:
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- Mathematics - Logic;
- Primary 03E10;
- Secondary 03E25
- E-Print:
- 14 pages