# Abstract Set Theory by Thoralf Skolem

By Thoralf Skolem

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This is impossible, however, because in such a case we should have (a,b,y) e P and (y,a,c)e S. Thus P1 would also possess the property 2), and that is absurd. Theorem 55. (x)(y)(z)(u) ((x,y,z) e P. & (x,y,u) e P -> (z = u)). Proof. Let S (b) denote the statement (x)(z)(u) ((x,b,z)eP & (x,b,u)eP -» (z = u)). Then S(0) is true because (x,0,z)eP -*(z = 0) and (x,0,u)eP -»(u = 0) (see Theorem 53). Let us assume that S(b) is true, and let us look at the conjunction (a,b f ,Ci)e P & (a,b f ,c 2 )e P.

Thus the cardinal numbers are also well-ordered by the relation < . More exactly expressed: All cardinals = a given cardinal constitute a wellordered sequence according to their magnitude. The least of the transfinite ones, the cardinal of the denumerable sets, we denote, as Cantor did, by N0, the following by NI , and so on. e. w= a, then we have 1 + a = a, because we may write a = w+ j3, whence l + a = l + ( w + |8) = (l + w) + j3 = co + /3 = a. More generally we have of course n + a = a, n finite.

5, p. 40) the axiom of infinity can be put into defining form. The easiest way of doing that is to use the notion of ordinal set introduced in § 8. We may define a finite ordinal as an ordinal set M such that (Ex) (xeM) & (M = x*) & (y)(y eM -* (Ez)(zey & y = z*). Here x* means x U{x}. Then the axiom of infinity can be expressed by saying that the finite ordinals constitute a set. The axiom of choice has given rise to many discussions. The reason for this is of course its non-constructive character.