Can a set be member of itself?

[jambaugh]

See Finsler Set Theory. Finsler played with the idea of self containing sets. Basically you construct sets as nodes of directed networks. To avoid Russel paradoxes you must be absolutely constructive in defining these sets. However you have more flexibility in that you start with directed nets and see if a node satisfies the requirements of being a Finsler Set.

 

[maverick280857]

Ok, now you got me more confused.

Sorry about that. I've been away a few days so I couldn't get a chance to reply. Anyway, I hope your confusion has been removed.

 

[DeaconJohn]

Consider, is this a question of intuition or how set theory is usually defined? And the further question why set theory is so defined, is something you and I have not gone into.

Consider the Axiom of the Empty Set and the Axiom of the Power Set.

Also, take the matter of whether a number N is considered a divisor of itself.

Robert,
Very thought provoking comment about the relationship of the "matter of whether a number N is considered a divisor of itself" and the status of self-referencing statements in mathematics.
Let's see if I can make a first step at formalizing your "matter." (Well put! by the way.)
Let S be the set of all natural numbers "n" such that "n divides n."
OK, that's a shot. If "n" is a divisor of itself (obviously just a matter of definition in this case), then S = N. (N = {1, 2, 3, ... } = the set of natural numbers.
Note that Russell's paradox works with the set of all sets that are NOT elements of themselves. It's not the "matter" of sets that are elements of themselves that causes the problem, it's the self-referencing nature of the matter.
If I remember correctly, the nature of self-referencing statements are the essence of Godel's proof of his theorem.
For a "popular" example, consider the statement p = "This statement is false." Ask the question whether p is true or false. If p is true then it is false. If p is false then it is true.
Hmmmmmmmmmmmmmmmmmmmm.
Awesome are the mysteries of the universe that we live in.
DeaconJohn

 

[dream runner]

Alright, it seems logical that in ZFC set theory, a set cannot be a member of itself. My question pertains to the definition of ordinals. According to Thomas Jech's edition of set theory, a set is ordinal if it is both transitive and well ordered by membership. I've been poking around trying to find an example of a set which is transitive and not well ordered by membership and only two possibilities seem to arise:

1: By definition, a set A is not well ordered by membership if there exists some subset B of A does not contain a least member. Thus every element b within B implies the existence of another element c within B which is also an element of b. This seems recursively to lead to an infinite set B.

2: A common counterexample I've seen is when proving that the class Ord of all ordinals is a proper class for else it would contain a member [alpha] which is an element of itself, and thus not well ordered. Furthermore, by our above discussion, no set exists such that it is a member of itself and thus no set exists containing a set which is a member of itself, as this flies in the face of the Axiom Schema of Seperation.

It seems that both cases pose a stark contradiction and thus imply that no such set exists. However, absurdity is not a requisite for truth, thus I would like to know, since all evidence seems to point towards the contrary, whether it is even possible for any set A to be transitive by the definition that every element of A is also a subset of A, and yet not be well ordered by membership.


P.S. My idea of a set here does not include any urelements, which seems rational, for consider the set

A = {b, {b}}
The only way this set can possibly be transitive is if b is a subset of A, and the only non set element which inherently a member and a subset is the null element, thus
A = {null, {null}} is transitive, and adding another element would require that element be the successor {{null}}. Much like a russian nesting doll.

 

[Hurkyl]

1: By definition, a set A is not well ordered by membership if there exists some subset B of A does not contain a least member. Thus every element b within B implies the existence of another element c within B which is also an element of b. This seems recursively to lead to an infinite set B.

You have an omission here -- you're implicitly assuming that membership is an order relation. However, that need not be true, even for transitive sets. For example, define:

A = {}
B = {A} = {{}}
C = {B} = {{{}}}
D = {A, B, C} = {{}, {{}}, {{{}}}}​

D is a transitive set. However, membership does not satisfy the trichotomy law: all three of the following statements are false:

A is a member of C
C is a member of A
A = C​

 

[dream runner]

Aha! Thank you so much. I was spending too much time studying ordinals, in which membership is a linear ordering before thinking of this question.

Sometimes thinking too deeply about another topic can lead me into a quagmire such as these.

 

[dream runner]

And I haven't gotten to the trichotomy axiom quite yet, and how it applies to binary relations. So I'm assuming in order for the membership relation to be considered as an order relation it must be trichotomous, which makes sense. Very interesting.