# Regularity and self containing sets

1. Aug 26, 2012

### 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.

2. Aug 26, 2012

### Hurkyl

Staff Emeritus
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}.

3. Aug 26, 2012

### 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.

4. Aug 26, 2012

### Hurkyl

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

5. Aug 27, 2012

### Old Monk

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?

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.

6. Aug 27, 2012

### Hurkyl

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

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