Having a universal set [in naive set theory] is a sufficient condition for Russel's paradox. See Naive Set Theory[1] bottom of page 6. This is why we can have no Universe in any consistent set theory.

Edit: @mafribe makes the point that there are some set theories that can still have universal sets by culling other features that ZF-style set theories have. I was mostly referring to ZF-style set theory (hence my citation). Indeed, one could even make a ZF-style set theory paraconsistent and still have Universal sets.

[1] http://sistemas.fciencias.unam.mx/~lokylog/images/stories/Al...

   Having a universal set is a sufficient condition for Russel's paradox.

That's not true. There are set-theories, e.g. Quine's NF [1] which allow universal sets, and other things like the set of all ordinals, that are forbidden in ZF-style set-theories. The problem in ZF is caused by unlimited comprehension. NF circumvents this by restricting comprehension. Tom Forster [2] has written a great deal about set theories with universal sets, including the wonderful [3]. He makes the historical point that set theory was born with universal sets.

[1] http://plato.stanford.edu/entries/quine-nf/

[2] https://www.dpmms.cam.ac.uk/~tf/

[3] T. E. Forster, Set Theory with a Universal Set. http://ukcatalogue.oup.com/product/9780198514770.do

True, I was mostly referring to ZF-style set theories (which is what the thread is mainly about). Your point could even be extended by saying that there are proofs for a paraconsistent ZF with a universal set.

Your [3] link doesn't work by the way, I'm interested in reading Forster!

[3] works on my browser. Anyway, the link was to the publisher's page for the book. Here is another one: http://www.amazon.co.uk/Set-Theory-Universal-Exploring-Unive... .

You can fix the Russel's Paradox in ZF as well.

I have no idea where you got this from but your statements don't follow any form of logic I'm familiar with.

If V = {x | x not in x}, then if V contains itself, V is not in V (and vice versa) is an obvious contradiction. Your new set S doesn't help in the slightest.

Fixing it is emphatically "not easy" and mathematicians generally rely on the ZFC axiomation (although several other possibilities were proposed).

It's no big deal if you don't admit universal set or anything other than ZFC.

It took 16 years (1901-1917) to get from russell's paradox to a set theory which didn't allow it, but which was able to create a lot of interesting and useful sets (ZF). So it seems like a big deal. And we still can't talk about "the set/collection/whatever of all sets" in the language of ZFC, so we're still missing out.

> universal set

But now you're talking about non-standard set theory[1] which is fine but you are kind of side-stepping the issue.

[1] https://en.wikipedia.org/wiki/Axiom_of_regularity

Well, it's not that difficult to fix it in ZF.

If you define S to be S = {x in U: x not in x}, then it simply means S is not in U.

There is no U in ZF so I really don't understand what you are saying.

You can call U anything you want. It's by axiom schema of specification: for any set A, there some set B with a set C in B iff C in A.

Why is this downvoted? Is the statement incorrect?

Yes

By the way, I forgot to ask do you object to my post about "killing" the Russel set because you don't accept the existence of universal set or because you don't understand how {x in S| x not x} helps here? If it's the former, can you assume that U exists and explain how {x in S| x not x} solves the problem? So that I know you're not simply "saying things".

Ah, I see you explained. Thanks. I love the subject, but I'm still a neophyte.