# Read e-book online Abstract Set Theory PDF

By Thoralf Skolem

ISBN-10: 026800000X

ISBN-13: 9780268000004

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**Extra resources for Abstract Set Theory**

**Example text**

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. K this condition is fulfilled, we have according to Theorem 54, that x and y exist such that (a,b,x)eP & (a,b,y)eP together with (x,a,Ci)eS & (y,a,c 2 )eS. Because of the validity of S(b) this yields first x = y, whence GI = c2 by Theorem 51. Theorem 56. (x)(y)(Ez) ((x,y,z)e P). Proof. Let S(b) here be the statement (x)(Ez) ((x,b,z) e P).

Furthermore, for every yex 0 , where 0=1= y, there is a y ^ e x o such that y = y-i + g(y-i). Otherwise x0- {y}would still possess the properties 1) and 2) which is contrary to the definition of XQ. Then we may define a mapping of u on a proper part of u as follows. We let u - Sxo be mapped identically onto itself while every g(y), where yex 0 , shall be the image of g(y-i) for the corresponding y _ j . This provides a mapping of Sxo onto the proper part Sxo - (g(0)}. Indeed every zeSxo must be a g(y) for some yexo, because otherwise we could remove all elements y containing the element z from XQ and still have a subset x with the properties 1) and 2).

Then, if BeC, neither A = B nor AeB, because of the transitivity of C. Therefore BeA and thus CEA because BeC yields BeA for all B. Since AEM and A H(M - C) = 0, it follows that A E C , whence A = C, whence CeM. Thus I have proved that every transitive proper subset of M is element of M. Remark 3. It is clear according to this that every element of an R - ordinal is an R - ordinal. 36 LECTURES ON SET THEORY Theorem 32. If A and B are R-ordinals, AeB-*--AC B. Proof. AeB yields, because of the transitivity of B, A EB, but A = B is excluded.

### Abstract Set Theory by Thoralf Skolem

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