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 $\mathsf{ZF}+{\mathsf{DC}}_{\hat{I}}º$, where $\hat{I}º$ 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 $m$.

13 pages