Regularity and self containing sets

[Old Monk]

Hey all,

I was reading Terence Tao's text on analysis. After stating the axioms of pairing and regularity, he asks for proof of the statement that no set can be an element of itself, using the above two axioms. He has not defined any concepts like hierarchy or ranks.

I can see how,

A $\notin$ A

if A={A}, from the axiom of regularity.

But if the A were to contain itself any other set a, such that A={a, A}, where say a={1}, then we would have

a $\cap$ A ≠ ∅

Since A contains the set a and not the elements of a, the intersection would be disjoint. Alternately, if the set A were defined as A={1, A}, the intersection

1 $\cap$ A ≠ ∅

Isn't it possible to have such sets? What role does the axiom of pairing play in the preventing the existence of such sets?

Thanks.

 

[Hurkyl]

Realize that to disprove the existence of a solution to $A \in A$, we aren't required to apply the axiom of regularity to A specifically: any contradiction you can derive would suffice, such as considering the set {A}.

 

[Old Monk]

Wouldn't that contradiction disprove the existence of sets that have only themselves as elements? It'd have to be shown that any set that contains itself, contains only itself as an element. I'm assuming this is where the axiom of pairing is required to complete the proof.

 

[Hurkyl]

Wouldn't that contradiction disprove the existence of sets that have only themselves as elements? It'd have to be shown that any set that contains itself, contains only itself as an element. I'm assuming this is where the axiom of pairing is required to complete the proof.

Suppose $A \in A$. What element of {A} is disjoint from {A}?

 

[Old Monk]

But if the A were to contain itself any other set a, such that A={a, A}, where say a={1}, then we would have

a $\cap$ A ≠ ∅

Since A contains the set a and not the elements of a , the intersection would be disjoint. Alternately, if the setAwere defined as A={1, A}, the intersection

1 $\cap$ A ≠ ∅

I've made an error in the symbols used and didn't re-check carefully enough. I meant to write,

a $\cap$ A = ∅

and

1 $\cap$ A = ∅

not ≠ ∅.

Since the axiom of regularity requires the existence of atleast one element disjoint from the set itself, in either case we'd have an element disjoint from A itself. Wouldn't this leave scope for the existence of such sets?

Suppose $A \in A$. What element of {A} is disjoint from {A}?

No element of {A} is disjoint from {A}. But how would that resolve the problem? Still between {1,A} and {A}, we have an element 1, which belongs to one set but not the other. How can their equivalence be assumed?

My apologies for the errors in the original post.

 

[Hurkyl]

No element of {A} is disjoint from {A}. But how would that resolve the problem?

Using the assumption $A \in A$, we derived a contradiction: the set {A} violates the regularity axiom. Therefore, for all A, $A \notin A$.

Still between {1,A} and {A}, we have an element 1, which belongs to one set but not the other. How can their equivalence be assumed?

They aren't equal: if $A \in A$, then ${1,A} \neq {A}$.