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)

It is shown in ZF (without choice) that if there is a finite-to-one map P(X) → X, then X is finite.