# Thread: This is making me go 1N54N3!!

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

2. Originally Posted by ThePerfectHacker
Are you saying the the set of all finite sets fails to satisfy ZF axioms?
No, but I can proove it from what I did say.

RonL

3. Originally Posted by CaptainBlack
No, but I can proove it from what I did say.

RonL
Is the proof simple because all I know about set theory is from wikipedia so you would have to keep it simple.

4. Originally Posted by ThePerfectHacker
Is the proof simple because all I know about set theory is from wikipedia so you would have to keep it simple.
Let S denote the "set of all sets", then there is a one-to-one mapping from
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

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