Re: bXc = b -> b=0 (with regularity but not infinity)
- From: "Rupert" <rupertmccallum@xxxxxxxxx>
- Date: 26 Aug 2006 01:55:10 -0700
MoeBlee wrote:
Here's one that has me stumped:
In ZR(-I) (that's Z set theory with the axiom of regularity but not the
axiom of infinity), prove:
bXc = b -> b=0
where 'X' stands for the Cartesian product.
You may not use the 'infinite sequence of nested members' version of
regularity. But the following previous theorems are available (where
'e' stands for 'is a member of') along with the usual "first chapter"
theorems of set theory:
~SeS
~SeTeS
~SeTeVeS
S subset SXS -> S=0
If bXc=b, then (bXc)Xc=b, so b=bXc=bX(cXc), and ran(b)=c=cXc=0, so b=0.
But I still have no idea how you prove this result with regularity but
without infinity.
Source: 'Axiomatic Set Theory' by Suppes, pg. 56, exercise 2 (Dover
edition).
.
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