The discussion revolves around how Zermelo-Fraenkel set theory with the Axiom of Choice (ZFC) addresses Russell's paradox, particularly focusing on the implications of its axioms for the existence of certain sets. Participants explore the nature of sets, predicates, and the axioms of ZFC in relation to the paradox.
Participants generally agree that ZFC provides a framework that avoids Russell's paradox, but there are multiple competing views on how this is achieved and the implications of the axioms involved. The discussion remains unresolved regarding the clarity of certain axioms and their interpretations.
Some participants express uncertainty about the axioms of ZFC, particularly the axiom of Regularity/Foundation, and how they relate to the definition of sets. There are also discussions about the complexity of demonstrating whether something is or isn't a set within different models.
it's not clear how to prove Russell's paradox is impossible.
zefram_c said:Basically, what ZFC does NOT have is an axiom that says every predicate can define a set. What it does have in its stead is an axiom that every predicate can define a subset of an existing set, plus a few other axioms to make up for the lost functionality. Hence there is no way to define Russell's set through the axioms of ZFC.
I agree. Doesn't that make Russell's paradox kind of trivial?Hurkyl said:Well, in ZFC, Russel's paradox is a proof that the class of sets that do not contain themselves is not, itself, a set. (because if it was a set, then we get a contradiction)
Euclid said:How does ZFC manage to block Russell's paradox? I've read through the axioms extensively, and it's not clear how to prove Russell's paradox is impossible.
In particular, I'm talking about Russell's paradox that shows {x| x not in x} is not a well-defined set.
matt grime said:Deciding whether something is or isn't a set depends on the model ZF you use.
A model of ZF consists of a S** of things, where S** means a naive collection of objects. This collection must satisfy the axioms of ZF. The things in the collection are called the sets in the model.
You might have seen this idea before if you've studied groups or vector spaces.
A vector space is a set satisfying some axioms, elements in the vector space are vectors. And just as the vector space is not a vector neither is the S** a set.
A common way to show something isn't a set is to demonstrate that it doesn't satisfy some axiom or other, or has cardinality strictly greater than any set cardinal.
Let's do 3 analogies, since they are easier:
Is the 2x2 matrix whose entries are a_11 = 1 a_12 = 1 a_21=0 a_22 = 1 an element of a group? Well, we find a group of which it is an element, which is elementary (it is a member of the group of all two by two invertible matrices under multiplication).
The matrix given by 1,0,0,0 resp in that list can never be a member of a matrix group where the operation is matrix multiplication, but it might be an element in some other model, and indeed is.
Can the elements a,b,c be elements of a group where a,b,c are all distinct and ab=a=ac? No, this can never happen since then b and c would both be distinct identity elements, so no model of GROUP can have elements that behave like that.
Deciding if something is or isn't a set in some model, or if there is a model in which it is a set is very hard, don't expect there to be easy examples.
Note this post owes a lot to an article I read by Tim Gowers.
I looked at that axiom repeatedly and never been able to make sense of it:CrankFan said:Did you take a look at the axiom of Regularity/Foundation ?
No set (of ZF) contains itself as a member, thus the "set of all sets not containing themselves as members" is the class of all sets; which isn't a set (in ZF).
Euclid said:Every non-empty set x contains some element y such that x and y are disjoint sets.
Hurkyl said:I presume you see why Russel's construction isn't directly applicable: because the axiom of extensionality has been replaced with the axiom of subsets (and a few other axioms denoting how to "safely" construct sets).