universal set paradox - Tilted Forum Project Discussion Community

noahfor · 02-22-2005, 02:24 PM

"Try to take the power set of the universal set. You will find that the universal set must be a member of itself and not a member of itself

Russell does use the Russell set and hierarchy to solve this problem."

I understand the contradiction of russell set, but not of the universal set. Why must the universal set not be a member of itself. How does russell use the russell set to solve the problem. I understnd the theory of types, the hierarchy, but not why russell's set is needed to solve the problem. I also understand the contradiction that the power set must be bigger and the same size as universal set but not the contradiction noted above. Thanks.

asaris · 02-22-2005, 02:27 PM

It's been a while since I've studied set theory, but in a nutshell, every set has to have a domain, so there is no such thing as the universal set.

noahfor · 02-23-2005, 12:08 PM

What if there were a set whose members are all the members of the power set of this set, so x is a member of S if x is a member of P(S). Doesn't this violate Cantor's theorem, and also is not able to be resolved by Russell's hierarchy?

I came up with that myself.

asaris · 02-23-2005, 12:53 PM

Do you mean x is a member of S if and only if x is a member of P(S)? I think you could use Cantor's theorem to show that such a set is impossible -- take f(x) -> x. Since this function is one-to-one (forgive me if I'm forgetting terminology here), S = P(S). But this is impossible, so your definition of a set defines a different set.

I think that's just an impossible set, and so a bad definition of a set. But I'd really feel more confident if someone who's had more set theory could confirm this. Maybe if I have time today or tonight, I'll try to refresh my memory a bit better.

02-22-2005, 02:24 PM	#1 (permalink)
noahfor Addict	universal set paradox "Try to take the power set of the universal set. You will find that the universal set must be a member of itself and not a member of itself Russell does use the Russell set and hierarchy to solve this problem." I understand the contradiction of russell set, but not of the universal set. Why must the universal set not be a member of itself. How does russell use the russell set to solve the problem. I understnd the theory of types, the hierarchy, but not why russell's set is needed to solve the problem. I also understand the contradiction that the power set must be bigger and the same size as universal set but not the contradiction noted above. Thanks.

02-22-2005, 02:27 PM	#2 (permalink)
asaris Mad Philosopher Location: Washington, DC	It's been a while since I've studied set theory, but in a nutshell, every set has to have a domain, so there is no such thing as the universal set. __________________ "Die Deutschen meinen, daß die Kraft sich in Härte und Grausamkeit offenbaren müsse, sie unterwerfen sich dann gerne und mit Bewunderung:[...]. Daß es Kraft giebt in der Milde und Stille, das glauben sie nicht leicht." "The Germans believe that power must reveal itself in hardness and cruelty and then submit themselves gladly and with admiration[...]. They do not believe readily that there is power in meekness and calm." -- Friedrich Nietzsche

02-23-2005, 12:53 PM	#4 (permalink)
asaris Mad Philosopher Location: Washington, DC	Do you mean x is a member of S if and only if x is a member of P(S)? I think you could use Cantor's theorem to show that such a set is impossible -- take f(x) -> x. Since this function is one-to-one (forgive me if I'm forgetting terminology here), S = P(S). But this is impossible, so your definition of a set defines a different set. I think that's just an impossible set, and so a bad definition of a set. But I'd really feel more confident if someone who's had more set theory could confirm this. Maybe if I have time today or tonight, I'll try to refresh my memory a bit better. __________________ "Die Deutschen meinen, daß die Kraft sich in Härte und Grausamkeit offenbaren müsse, sie unterwerfen sich dann gerne und mit Bewunderung:[...]. Daß es Kraft giebt in der Milde und Stille, das glauben sie nicht leicht." "The Germans believe that power must reveal itself in hardness and cruelty and then submit themselves gladly and with admiration[...]. They do not believe readily that there is power in meekness and calm." -- Friedrich Nietzsche

02-23-2005, 12:08 PM	#3 (permalink)
noahfor Addict	What if there were a set whose members are all the members of the power set of this set, so x is a member of S if x is a member of P(S). Doesn't this violate Cantor's theorem, and also is not able to be resolved by Russell's hierarchy? I came up with that myself.