Fun with Russell's paradox

[suszterpatt]

I was taught in my Intro to Math I class that what Russell's paradox concludes is that there is no such S set that contains absolutely every set. This is because if it existed, it could be divided into two subsets: N would be the subset that contains those sets that contain themselves, and M would be the subset of those sets that don't contain themselves. Obviously, any set in S must be in either N or M, but can't be in both at once. But N and M are also sets, and as such must be in either N or M. Now the question is, where is M? If M is in N, then it doesn't contain itself, and should thus be in M. Adversely, if it's in M, then it contains itself, and should be in N. Since neither scenario is possible, and there are no more scenarios, our initial assumption that S exists must be false.


BUT!

I recently checked Wikipedia, and according to that, all Russell's paradox says is that the above mentioned M can't exist, it doesn't say anything about S or N.


This got me thinking.


One of the axioms of set theory is that for every a element and B set, one can unequivocally tell whether a is an element of B. This is why M can't exist: supposing that M either contains or does not contain itself, we can conclude the opposite. However, the above mentioned N can't exist for a similar reason! Suppose that N is not an element of N: this is possible. Now suppose that N is an element of N: this is possible too! Therefore, you can't tell if N contains N, which contradicts the aforementioned axiom. This way, you get that N can't exist either.


But if neither N or M exists, then what's to stop S from existing? We used N and M to disprove the existence of S, but apparently they don't exist.



So I guess my final question is: does S exist, and how do you prove/disprove its existence?

 

[CycloWizard]

Doesn't sound like a very introductory math class to me...

I will just throw in my two cents: if neither M nor N exist, and S encompasses both, then S either does not exist or must be empty. Or, since your definition of N is very similar to your definition for S, disproving N as you did would disprove S.

Of course, I'm probably wrong, which is why I just audit analysis classes instead of actually enrolling.

 

[suszterpatt]

Well, no set can contain stuff that doesn't exist, that's the whole point. If neither N or M exists, then S can't encompass them, and you can't divide S into those two subsets. I sense a contradiction here, what with being unable to take a certain subset of a set, but I'm not really sure.

Or, since your definition of N is very similar to your definition for S, disproving N as you did would disprove S.

Nah, that doesn't work. It's obvious that S contains itself. If it didn't, then it wouldn't be the set that contains every set, now would it?