Example of transitive but not well ordered set needed

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The discussion revolves around the search for an example of a set that is transitive but not well-ordered by membership. It highlights that a set is not well-ordered if it contains a subset without a least member, leading to the possibility of infinite descending chains. The conversation references the class of ordinals, emphasizing that no set can contain itself as a member, which complicates the existence of such a set. An example provided is the set {0, {0}, {{0}}}, which is transitive yet not well-ordered. The conclusion suggests that while the existence of such sets seems contradictory, it remains a topic of inquiry in set theory.
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Example of transitive but not well ordered set needed!

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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In the first example, I am not implying in any way that a transitive set cannot be infinite, it is the way in which it is infinite that seems counter-intuitive.

Let B be a subset of A having no least membered element.

Then if b is an element of B, then there exists c in B such that c is also in b, thus
b = {...,c}
but then c cannot be a least element, thus there must exist d in B such that d is in c
c = {...,d}
thus b = {...,{...,d}}
This continues on to create an infinite set B, but B cannot be inductive, or else null would be in B and by definition null is a least membered element. Thus B need be a non-inductive yet infinite set, containing some kind of infinite cascade of subset elements.

The situation seems comparable to an open interval of the real line. Imagine an infinite collection of Mitroshka dolls (the russian nesting dolls). There may exist a largest Mitroshka, but there are infinitely many smaller metroshka's within it. Much like an interval (a, b] of the real line which is also not well ordered in the regular sense of the word.

Is such a set possible?
 


Not well-ordered by membership means that either two elements are not comparable or an infinite descending chain, the latter is forbidden by the axiom of regularity, assuming we're talking about ZFC. { 0, { 0 }, { { 0 } } } is an example of a transitive set that is not well-ordered.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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