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

apoorv wrote:
MoeBlee wrote:
But since the
existence of omega also proves the axiom of infinity, indeed, your 1A
is equivalent with the axiom of infinity.
and
No ordinal is a member
of itself (provable without the axiom of regularity) but there are
ordinals that are not members of omega.
What in the world? If omega is an ordinal (which it is), then omega
does not belong to itself. It does not follow that if omega is not a
member of itself then omega is a member of itself.


Let us agree, as you say ,that the axiom of infinity is equivalent to
1A i.e the set N exists.So we have the null set, its successors and
the set N. We want to deduce the existence of a set containing N in a
situation where we do not have powerset and regularity.

I take it that 'containing' means 'have as a member'. If so, we don't
need the power set axiom nor the axiom of regularity to prove that
there is a set s such that omega is a member of s, since omega is a
member of {omega}.

Now N does not
belong to N if and only if S(N)=NU{N} is different from N . To assert
that omega does not belong to omega is equivalent to asserting the
existence of the successor of omega---that is to add that as an
additional axiom to our system.

We don't need an additional axiom. The existence of successors comes
from the pairing axiom and the union axiom. Also, we can prove that
omega is not a member of itself without even mentioning the successor
of omega.

If we confine ourselves to the axiom 1A, then we have the set N as the
set of all ordinals that exist in that system;(these ordinals are
different from their successors, or equivalently do not belong to
themselves).

No, I already told you that the axiom of infinity does not entail that
omega is the set of all ordinals and not the set of all ordinals that
are not members of themselves. In fact, we can prove, without the power
set axiom or the axiom of regularity, that omega is not the set of all
ordinals nor the set of all ordinals that are not members of
themselves.

So, in this system N is the set of all ordinals that do not
belong to themselves.

In what system? ZF with pairing but without power set and without
regularity? If so, then you are incorrect. In that system, omega is NOT
the set of all ordinals that are not members of themselves.

If N is an ordinal, then in this system, N e N
implies N ~eN and vice versa.

That is is incorrect. In the axiomatization you've mentioned, omega not
in omega does not imply omega in omega. That is because, in the
axiomatization that you mentioned, omega is NOT the set of all ordinals
that are not members of themselves.

I am asking sincerely: what set theory textbooks have you studied? I'm
curious about that because you have some very fundamental
misunderstandings and I'm wondering whether they can be cleared up by
going back to the appropriate sections in the books that you've
misunderstood.

MoeBlee

.

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