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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 the forum discussion on this, 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.

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# Example of transitive but not well ordered set needed

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