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)

With the Axiom of Choice, for any infinite cardinal but, without, we cannot conclude any relationship between and for an arbitrary infinite cardinal. In this paper, we give some properties of in the absence of and compare them to those of for an infinite cardinal. Among our results, we show that “ for any infinite cardinal and any natural number n” is provable in although “ for any infinite cardinal ” is not.

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)