From: utzoo!decvax!ucbvax!arens@UCBKIM
Newsgroups: net.math
Title: Re: I am trying to comprehend
Article-I.D.: ucbvax.812
Posted: Mon Feb  7 15:14:13 1983
Received: Wed Feb  9 02:55:38 1983

From: arens@UCBKIM (Yigal Arens)
Received: from UCBKIM.BERKELEY.ARPA by UCBVAX.BERKELEY.ARPA (3.300 [1/17/83])
	id AA04646; 7 Feb 83 15:13:26 PST (Mon)
To: net-math@ucbvax, harpo!ihnp4!ihuxr!lew@decvax.UUCP
In-Reply-To: Your message of  7 Feb 1983 1413-PST (Monday)


What the comprehension axiom says is that every subcollection of elements of
a set defined as those elements of the set that satisfy some formula -- is
also a set.  In the formulation you gave,

    Ey Ax (x < y <-> x < z ^ phi) where phi is a formula NOT INVOLVING y.

z is the original set, phi is the defining formula, and y is the new set
this axiom claims exists.

This can be used to prove the non-existence of a universal set in the
following way (using Russel's paradox).

We prove by assuming a universal set ("the set of all sets") exists, and
proving a contradiction.

Let's assume the set of all sets exists and call it U.  We'll now use the
axiom of comprehension with phi = x ~< x   (i.e. "x doesn't belong to x"),
and z = U.

The axiom in this particular case will read:

    Ey Ax (x < y <-> x < U ^ x ~< x) 

(i.e. there exists a set such that it is made up exactly of all sets in U
which do not contain themselves).  Let's call this set that the axiom tells
us exists, P.

So,   Ax (x < P <-> x < U ^ x ~< x).  This is a universal statement, stated
for ALL x, and so must be true for P too.  In that particular case it will
say,  
      P < P <-> P < U ^ P ~< P.

A contradiction can be gotten easily from this, since trivially P < U.

Therefore U doesn't exist.

This should take care of what Kunen is doing.  It isn't clear to me from
your description what Monk is trying to do in his book.

The notation  E!x  is read as "there exists a unique x".

Yigal Arens
UC Berkeley