- #1
Old Monk
- 8
- 0
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.
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.