# Transitivity of Element of

## Homework Statement

Let A, B, and C be sets. Assume the standard ZFC axioms.

Please see below for my updated question.

Thanks.

Last edited:

Dick
Homework Helper
A ε B means the set A is a element of the set B, right? That's not the same thing as A being a subset of B. You have sets that are elements of other sets. Try to construct a counterexample.

That's what I was afraid of. 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).

Dick
Homework Helper
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.

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.
Yes I understand this. But how does the chain of set inclusions I mention lead to the fact that C ε C?

Dick
Homework Helper
Yes I understand this. But how does the chain of set inclusions I mention lead to the fact that C ε C?

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?

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 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.

Dick
Homework Helper
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.
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 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.
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.

Dick
Homework Helper
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 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.

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.
Hmm....

Then how do I go about using the axiom of regularity to prove that no set membership loops like that I described exist?

Dick
Homework Helper
Hmm....

Then how do I go about using the axiom of regularity to prove that no set membership loops like that I described exist?
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.

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.
Yes but I want to show that A ε B and B ε C and C ε A violates regularity.

Dick
Homework Helper
Yes but I want to show that A ε B and B ε C and C ε A violates regularity.
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.

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.
Then I can, but that doesn't help me.

See http://en.wikipedia.org/wiki/Axiom_of_regularity

It states:

Every non-empty set A contains an element B which is disjoint from A.

I appreciate all of your help.