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You are watching: Find two sets a and b such that a ∈ b and a ⊆ b

Would this average A and B are equal due to the fact that A is an element of B and a subset?Or favor thisA = 1,2,3b = 1,2,3,4,5,6would the work?
B deserve to never it is in an facet of B, for this reason no.
It depends... If you specify numbers as well-ordered sets, then this works... If you have actually no idea what I just said, then your prof most likely won"t accept that solution. :roll: The set 1,2,3 is not an facet of 1,2,3,4,5,6, so that solution doesn"t work (unless friend encode organic numbers a particular way).B can contain any other set as one element, so pick A-- choose your favorite set. (If girlfriend don"t have actually one, discover one :mrgreen:.)Now take the collection whose only facet is A: A, this set contains A as an element, yet unfortunately the doesn"t contain A together a subset unless A= (Its only subsets space A and , but A?A...)So, we need to include a bunch of aspects to this collection to get the B us want... Have the right to you figure out which elements?

Or, adhering to Azcel, pick $$\displaystyle \Omega$$, which has $$\displaystyle \Omega=\\Omega\$$. Of food that"s not permitted in ZF....

Not at all. Why would certainly leaving an axiom out cause contradictions? You get contradictions by including too plenty of axioms, not as well few. (In this case, I"m pretty sure ZF and also ZF - structure are in reality equiconsistent, for this reason there"s no "danger" to allowing Foundation. But that doesn"t typical I have to like it!)
I was largely being cheeky...Anyway; the just arisen to me that we haven"t actually discussed structure at every in my collection theory class... He noted most the the axioms on work one, added replacement later, and introduced selection a little bit later... But foundation has no been discussed... I will have to ask mine professor about this.

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