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Homework Statement
Let A, B, and C be sets. Assume the standard ZFC axioms.
Please see below for my updated question.
Thanks.
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Yes I understand this. But how does the chain of set inclusions I mention lead to the fact that C ε C?Pick A={a}. Pick B={A}={{a}}. So A ε B. Correct so far? Pick C={B}={{{a}}}. So B ε C. Is A ε C?? This is a little tricky. a is not equal to the set consisting of a.
You seem to have deleted the question. I thought I suggested you try to prove that A ε B and B ε C doesn't necessarily imply A ε C? Since inclusion is not transitive?Yes I understand this. But how does the chain of set inclusions I mention lead to the fact that C ε C?
(Thanks for all your help!)
I agree. I am now asking modified question: How then can I show that if A ε B and B ε C and C ε A, that C ε C (yields the contradiction I require).You seem to have deleted the question. I thought I suggested you try to prove that A ε B and B ε C doesn't necessarily imply A ε C? Since inclusion is not transitive?
I think you are reaching way too far for a contradiction. C ε C puts you squarely in Bertrand-Russell paradox territory. A simple example of A ε B and B ε C with simple sets and A not an element of C will serve nicely. I basically gave you one. Follow it up. It shows inclusion isn't transitive. That's all you need.I agree. I am now asking modified question: How then can I show that if A ε B and B ε C and C ε A, that C ε C (yields the contradiction I require).
Apologies for the confusion.
This is a separate question. I absolutely want to be in such territory. I need to show that the aforementioned set inclusions yield C ε C and then I can invoke the axiom of regularity/foundation to show it is a contradiction.I think you are reaching way too far for a contradiction. C ε C puts you squarely in Bertrand-Russell paradox territory. A simple example of A ε B and B ε C with simple sets and A not an element of C will serve nicely. I basically gave you one. Follow it up.
I don't think so. Inclusion ISN'T transitive. You can't say A ε B and B ε C implies A ε C. At all.This is a separate question. I absolutely want to be in such territory. I need to show that the aforementioned set inclusions yield C ε C and then I can invoke the axiom of regularity/foundation to show it is a contradiction.
Hmm....I don't think so. Inclusion ISN'T transitive. You can't say A ε B and B ε C implies A ε C. At all.
C ε C violates regularity. But the statement C ε C doesn't follow from anything you've said before because inclusion isn't transitive.Hmm....
Then how do I go about using the axiom of regularity to prove that no set membership loops like that I described exist?
Yes but I want to show that A ε B and B ε C and C ε A violates regularity.C ε C already violates regularity. The statement C ε C doesn't follow from anything you've said before because inclusion isn't transitive.
Mmm. I'm not all that hot with set axiomatics. But you can simplify that. Suppose A ε B and B ε A, can you show that violates regularity?? Like I say, I don't have ZFC axioms at my fingertips.Yes but I want to show that A ε B and B ε C and C ε A violates regularity.
Then I can, but that doesn't help me.Mmm. I'm not all that hot with set axiomatics. But you can simplify that. Suppose A ε B and B ε A, can you show that violates regularity?? Like I say, I don't have ZFC axioms at my fingertips.