Consequences of arithmetic for set theory

Abstract

Abstract. 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 ^ and 31, either 8° < 3 or 9! < f. However, in ZF this is no longer so. For a given infinite set A consider seq 1 " 1 (A), the set of all sequences of A without repetition. We compare | seq'"'(/4)|, the cardinality of this set. to |^( J / ) | , 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 h VX(| seq, and we show that this is the best possible result. Furthermore, it is provable in ZF that if B is an infinite set, then | fin(B)| < |^(B)| even though the existence for some infinite set B* of a function / from fin(5*) onto 9>{,B*) is consistent with ZF. §0. Introduction, definitions, and basic theorems. Introduction. In ZFC the cardinality of ordinal numbers plays an important role, since by AC each set has the cardinality of some ordinal.We use "alephs" for the cardinalities of ordinals. Thus in ZFC each cardinal number is an aleph. However, this need not be the case in ZF.If we have a model M of ZF in which the axiom of choice fails, then we have more cardinals in M than in a model V of ZFC, even if we have fewer sets in M than in V. (This occurs when the choice-functions are not all in M.) This is because the ordinals are in M and, hence, the alephs as well.In this paper we are interested in the relation between three cardinals arising in connection with a set S, namely,(1) the cardinality of the power set of S, (2) the cardinality of the finite subsets of S, (3) the cardinality of the finite sequences without repetition of S. This section contains definitions and basic theorems provable in ZF. In the next section we present two relative consistency proofs illustrating possible relations between these cardinals.The last two sections contain three results provable in ZF. The proofs of these are based on the same idea originally from E. Specker, who used it to prove that the axiom of choice follows from the generalised continuum hypothesis [Spl]. Assuming the existence of a function we derive a contradiction to Hartogs's Theorem.

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