Re: Why? [was Re: Cantor`s powerset theorem is false?]

apoorv wrote:
[Moeblee has] said 'However, in Z set theory, if we define 'z is a set' by
'z = 0 v Es zes' then it's trivial to prove Az z is a set'. I am afraid this is not very clear to me.

MoeBlee's definition is a bit obscure; he probably meant 'z is the empty set or there is a s such that s e z'. What is meant by saying that the axioms of Zermelo set theory - in its modern form - imply that everything is a set can be explained by the following observations. One of the axioms of Zermelo set theory states that the identity of an object is determined solely by its members. This is usually taken to be the characteristic feature of a set. From this axiom it follows that there can be only one object with no members, which is taken to be the empty set. (The existence of an object with no members follows from the other axioms). If there were non-set objects, they would be memberless objects not identical to the empty set. Thus it follows that there are no non-set objects.

By this definition, it appears that z={z} would also be a set.

The axiom mentioned above does not rule out the existence of a z such that z = {z}. However, there is an axiom called the axiom of foundation in Zermelo set theory that prohibits such circular sets. Even if we drop this axiom and postulate the existence of circular sets the result about non-existence of non-sets still holds. It is thus somewhat unclear what the relevance of your remark is.

--
Aatu Koskensilta (aatu.koskensilta@xxxxxxxxx)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
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