If we assume the axiom of choice, then every two cardinal numbers are comparable, In the absence of the axiom of choice, this is no longer so. For a few cardinalities related to an arbitrary infinite set, we will give all the possible relationships between them, where possible means that the relationship is consistent with the axioms of set theory. Further we investigate the relationships between some other cardinal numbers in specific permutation models and give some results provable without using (...) the axiom of choice. (shrink)

In this paper, we consider certain cardinals in ZF (set theory without AC, the axiom of choice). In ZFC (set theory with AC), given any cardinals C and D, either C ≤ D or D ≤ C. However, in ZF this is no longer so. For a given infinite set A consider $\operatorname{seq}^{1 - 1}(A)$ , the set of all sequences of A without repetition. We compare $|\operatorname{seq}^{1 - 1}(A)|$ , the cardinality of this set, to |P(A)|, the cardinality of (...) the power set of A. What is provable about these two cardinals in ZF? The main result of this paper is that $ZF \vdash \forall A(|\operatorname{seq}^{1 - 1}(A)| \neq|\mathscr{P}(\mathscr{A})|)$ , and we show that this is the best possible result. Furthermore, it is provable in ZF that if B is an infinite set, then $|\operatorname{fin}(B)| <|\mathscr{P}(B)|$ even though the existence for some infinite set B* of a function f from $\operatorname{fin}(B^\ast)$ onto P(B*) is consistent with ZF. (shrink)