A note on surjective cardinals
Fuente:
arXiv
Saved in:
| Main Authors: | , |
|---|---|
| Format: | Preprint |
| Published: |
2024
|
| Subjects: | |
| Online Access: | |
| Tags: |
Add Tag
No Tags, Be the first to tag this record!
|
| _version_ | 1866909777224794112 |
|---|---|
| author | Jin, Jiaheng Shen, Guozhen |
| author_facet | Jin, Jiaheng Shen, Guozhen |
| contents | 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 $\mathsf{ZF}+\mathsf{DC}_κ$, where $κ$ 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 $m$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2408_04287 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | A note on surjective cardinals Jin, Jiaheng Shen, Guozhen Logic Primary 03E10, Secondary 03E25, 03E35 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 $\mathsf{ZF}+\mathsf{DC}_κ$, where $κ$ 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 $m$. |
| title | A note on surjective cardinals |
| topic | Logic Primary 03E10, Secondary 03E25, 03E35 |
| url | https://arxiv.org/abs/2408.04287 |