# This is making me go 1N54N3!!

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• Feb 15th 2006, 01:41 PM
ThePerfectHacker
Are you saying the the set of all finite sets fails to satisfy ZF axioms?
• Feb 15th 2006, 07:44 PM
CaptainBlack
Quote:

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
• Feb 16th 2006, 12:34 PM
ThePerfectHacker
Quote:

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.
• Feb 16th 2006, 01:02 PM
CaptainBlack
Quote:

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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