Regularity and self containing sets

  • Thread starter Old Monk
  • Start date
  • #1
8
0

Main Question or Discussion Point

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 [itex]\notin[/itex] 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 [itex]\cap[/itex] 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 [itex]\cap[/itex] 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.
 

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Realize that to disprove the existence of a solution to [itex]A \in A[/itex], 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
8
0
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
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
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 [itex]A \in A[/itex]. What element of {A} is disjoint from {A}?
 
  • #5
8
0
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 [itex]\cap[/itex] 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 [itex]\cap[/itex] A ≠ ∅
I've made an error in the symbols used and didn't re-check carefully enough. I meant to write,

a [itex]\cap[/itex] A = ∅

and

1 [itex]\cap[/itex] 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 [itex]A \in A[/itex]. 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.
 
  • #6
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
No element of {A} is disjoint from {A}. But how would that resolve the problem?
Using the assumption [itex]A \in A[/itex], we derived a contradiction: the set {A} violates the regularity axiom. Therefore, for all A, [itex]A \notin A[/itex].

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 [itex]A \in A[/itex], then [itex]{1,A} \neq {A}[/itex].
 

Related Threads on Regularity and self containing sets

Replies
2
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
4
Views
1K
Replies
20
Views
2K
Replies
2
Views
9K
Replies
1
Views
717
Replies
25
Views
12K
Replies
16
Views
906
Replies
18
Views
3K
Top