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