From: Torkel Franzen <torkel@sm.luth.se>
Subject: Re: Axiom of extension
Date: 31 Jul 2000 12:20:33 +0200
Newsgroups: sci.math
Summary: [missing]

"Clive Tooth" <clive@pisquaredoversix.force9.co.uk> writes:

 > (Ax)(Ay)((x=y) iff (Az)((z%x) iff (z%y)))
 > 
 > appears to be a definition of "=".
 > 
 > Consider two sets F: {f} and G: {g}. Do we have F=G? We need (Az)((z%F) iff
 > (z%G))). Let z be f, the question arises "Do we have f%G?". To resolve this
 > we need to know "Do we have f=g?". We appear to be in a state of (possibly
 > infinite) regression.

  The axiom of extensionality is just a *statement*: sets X and Y are
identical if and only if they have the same members. The regress that
you comment on has to do with using a certain *procedure* to determine
whether sets X and Y are identical. If (for example) a={a} and b={b},
the procedure you describe does indeed result in an infinite
regress. This however is not a problem with the axiom of
extensionality. Accepting non-well-founded sets is no obstacle to
accepting the axiom of extensionality. We just can't use the axiom of
extensionality to decide whether there are many sets a such that
a={a}, but must decide this on other grounds.
==============================================================================

From: hrubin@odds.stat.purdue.edu (Herman Rubin)
Subject: Re: Axiom of extension
Date: 31 Jul 2000 20:02:27 -0500
Newsgroups: sci.math

In article <8m3i6q$klv$1@mail.pl.unisys.com>,
Clive Tooth <clive@pisquaredoversix.force9.co.uk> wrote:
>I use % for "is an element of".

>The axiom of extension:

>(Ax)(Ay)((x=y) iff (Az)((z%x) iff (z%y)))

>appears to be a definition of "=".

In set theory with equality, it is a true axiom.  In
Fraenkel-Mostowski models, one starts with a set of
individuals, which have no elements, but are not equal
to the empty set.

Both the axiom and its negation are consistent.

>Consider two sets F: {f} and G: {g}. Do we have F=G? We need (Az)((z%F) iff
>(z%G))). Let z be f, the question arises "Do we have f%G?". To resolve this
>we need to know "Do we have f=g?". We appear to be in a state of (possibly
>infinite) regression.

>(1) Does the axiom of foundation, (Ax)((x/={})->((Ey)((y%x) and (x intersect
>y={})))), prevent this regression from being infinite?

With choice, yes.  I am not sure without.

>(2) If the answer to (1) is yes: How is it possible to formulate an
>effective axiom of extension in ZF without foundation?

It might not be effective, but it certainly can be done.
The Specker models replace the individuals in the
Fraenkel-Mostowski models by non-regular sets which are
their own singletons.  Less trivial versions have also
been produced.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558