Talk:Set of all Sets

From ProofWiki
Jump to navigation Jump to search

This proof is philosophically circular. It was because the concept of the set of all sets produced a contradiction (namely Russell's Paradox) that the Zermelo Axioms (and hence ZF, and in particular the Axiom of Foundation) were designed in the first place. So this proof says: "The set of all sets is a contradiction because it can't be created using ZF". To keep it honest we must now link to the result which says "Any set which can not be created by the ZF axioms is inherently contradictory" which I don't believe we can do. --prime mover 10:25, 15 May 2011 (CDT)

Possibly...my understanding is "a thing which doesn't satisfy ZF isn't a set". But set theory is one of those things I learnt a (very) little about as an undergraduate, but never read anything serious on the subject. By all means amend/delete as you see fit. --Linus44 11:07, 15 May 2011 (CDT)
Okay - won't be immediate. This is stuff I've taught myself so my thinking may be biased from the limited selection of books I've managed to unearth. My take is: while ZF provides an appropriate axiomatic framework, we need to hammer home to the casual learner and beginning student why we can't define "the set of all sets" as a "set", before we are able to say: "... and so because we can't trust you to behave yourselves, we need to give you a strict list of stuff you're allowed to do."
As I say, it won't be immediate, I'll put a flag in place. I have a feeling that all we need to do is point to the comprehension principle, and/or just go straight to Russell's Paradox. I want to get Steen and Seebach done first. --prime mover 12:14, 15 May 2011 (CDT)
I don't think the proof is philosophically circular in the way you've suggested. It might seem that way because of the level of formality of the statement, though. We're saying that "The existence of a set which contains all other sets leads to a contradiction". But when you chase out the definitions and context that are likely being assumed, what we're really trying to do is use natural language to say something like "In the context of first-order logic, $\operatorname{ZFC} \vdash (\exists x \forall y (y\in x))\to (\phi \wedge \neg\phi)$". The informal statement is loaded with a lot of background information. By "set" we might mean something like "variable in the first-order language of sets $\mathcal{L}=\{\in\}$" or "element of an interpretation of ZFC", by "contains all other sets" we mean some statement involving a specific relation $\in$ which we have assumptions about, by "leads to" we mean "using that as an assumption, we can prove... (in some particular system)".
Even if you don't want to write this theorem from the perspective of some formal system, that break-down still has to occur implicitly. All of those words have to have meanings. It sounds like you want to be able to say that the set of all sets is inherently contradictory, without regard to a particular system's idea of what a set is. (I'm inferring this since you seem unhappy with showing that it's contradictory according to ZFC.) But we can't say "a set of all sets cannot exist" in a vacuum though. The statement is meaningless without definitions or rules for how "set" and "containment" behave.
Similarly, it's not quite right to say "the concept of a set of all sets produced a contradiction (namely Russell's Paradox)" in a vacuum. Russel showed a contradiction in, e.g., naive set theory, where "set" has a different meaning than the word "set" does when we're talking about ZFC. That is, when Russel wrote "set", he meant something different than when we write "set" today usually. Moreover, there's nothing wrong (in absolute terms) with having some kind of notion of "set" where "a set of all sets" exists. Some non-standard set theories do this (e.g.: Quine's New Foundations). But again this involves using a different notion of "set" than the default notion used in modern mathematics (ZFC's).
My point, which hopefully wasn't lost in all of that, is that statements like this depend on what we mean when we say "set". We have to assume some context. You can't hope for a proof which establishes something about "sets" without respect to a system that tells you what a "set" is.
That said, I'd be okay with calling attention to the fact that we're using ZFC, and I'd encourage adding a proof that used Axiom:Axiom_of_Replacement to define Russel's set and derive the historic contradiction.
Qedetc 17:50, 8 June 2011 (CDT)
Well first things first you've got to provide links to your definitions. Otherwise the entire point of what your trying to do has been lost. To quote you: "... statements like this depend on what we mean when we say "set". We have to assume some context. You can't hope for a proof which establishes something about "sets" without respect to a system that tells you what a "set" is."
You wrote the proof - you provide the context. --prime mover 00:29, 9 June 2011 (CDT)


I didn't create this entry. Also, I'm not sure how you took my last comment, but it doesn't feel like you took it how I intended it to be taken. You seemed to be asking for something impossible and I was trying to explain why it can't be done, because I think it is worthwhile to think about. I wanted to draw attention to some of the meta-math-y issues going on here. I think the stuff you said deals with interesting questions/points that I know I didn't really think about until a while after having taken some set theory / logic courses, trying to explain Russel's paradox to other non-math-people. I think there's some value in what I said, independent of what it means for this particular page. I don't know exactly how you read it before, but I guess I'm encouraging you to re-read it as "hopefully interesting philosophy-of-math side-discussion" rather than "critique of page or earlier comment".
Qedetc 01:07, 9 June 2011 (CDT)
Seriously beg your pardon - so you didn't. Apologies, pretend I never wrote that. I will give this a think in a bit - there's a ton of matters arising from entries from the last few hours or so. --prime mover 01:26, 9 June 2011 (CDT)

It would be better to prove that this set is impossible using Cantor's Theorem. Because if the universal set exists, then its powerset is itself. But the identity map is bijective from $V$ onto $\mathcal{P} V$, hence, we have a contradiction. I may write it up later. --Andrew Salmon 20:15, 25 July 2012 (UTC)