Are you saying the the set of all finite sets fails to satisfy ZF axioms?

Printable View

- Feb 15th 2006, 01:41 PMThePerfectHacker
Are you saying the the set of all finite sets fails to satisfy ZF axioms?

- Feb 15th 2006, 07:44 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

RonL - Feb 16th 2006, 12:34 PMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

- Feb 16th 2006, 01:02 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

S into S1 the "set of all sets with one element", namely the map f which takes

s in S to {s} in S1. Now consider S1'=f(S). Then the map f is one-to-one and

onto form S to S1'.

Now f is also a one-to-one map from the power set P(S) of S onto the Power

Set of P(S1') of S1'. But if S1 is a set so is S1' as it is a subset of S1, so the

cardinality of P(S1') is strictly greater than the cardinality of S1'. But this

would imply that the cardinality of P(S) is strictly greater than that of S, but

its not so it aint.

RonL