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

apoorv wrote:
MoeBlee wrote:

I use standard weak mathematical induction on omega:

0 not in 0.
Suppose, for an arbitrary n in omega, n not in n. It follows that S(n)
not in S(n).
Therefore, for all n in omega, n not in n.

Please note that we restricted ourselves to 1A part of the axiom of
infinity. So, the domain of discussion is omega.

I don't know what you think a 'domain of discussion' is, but if you
mean a domain of discourse, then that is not part of the theory but
rather is part of the structure for the language of the theory.

So, your argument is equivalent to:
0 not in 0.
Suppose, for an arbitrary n , n not in n. It follows that S(n)
not in S(n).
Therefore, for all n , n not in n.
And So
Omega is all n not in n and we run into Russell's antinomy.

No, since you have not derived the formula: Ax x in omega.

Either you're pretending not to understand the most basic things about
formal theories such as set theory, or you really don't understand
them. If the former, then I can only say that the laugh's on me for
taking this conversation with you seriously. If the latter, then again
I ask what textbook you've been studying, so that perhaps I can refer
you to the section you need to review.

To put it somewhat differently, our restricted axiom set is:
1)There exists a null set.
2)Given x, there exists S(x)=xU{x} ~=x (pairing and finite union)
3)There exists N (omega) containing all x.(Infinity)

No, you said that we are considering ZF without the power set axiom and
without the axiom of regularity but with the pairing axiom. That theory
is:

first order logic with identity and:

axiom of extensionality
axiom schema of replacement
union axiom
pairing axiom
axiom of infinity

We don't need an empty set axiom, since the existence of an empty set
is derivable from the axiom schema of replacement, and uniqueness is
derivable from the axiom of extensionality.

The union axiom is not limited to finite unions.

The axiom of infinity has no clause about omega "containing all x"

Clearly, here the axiom of infinity is equivalent to asserting the
existence of a Universal Set and we run into Russell' antinomy.

Okay, you are joking, right? The axiom of infinity is not equivalent to
asserting the existence of a universal set. If you propose a DIFFERENT
axiom that you call 'the axiom of infinity', then maybe that different
axiom is equivalent to asserting the existence of a universal set. But
so what? All you've done then is assert an axiom that entails an
inconsistent theory.

If we use , instead of 3) the following:
3A) There exists N (omega) containing all x other than itself
(Infinity)

Then we have an inconsistent theory, since N u {N} would be a universal
set. But so what? What is the point here of intentionally axiomatizing
an inconsistent theory?

Then 3A) is effectively 2 axioms :
1) N contains all x ~=N
2)N does not contain itself <----> S(N)~=N exists.
or, We are asserting the simultaneous existence of two different
infinite sets.

Actually, since 3A causes an inconsistent theory, you don't have to
limit to clauses 1) and 2) above, but rather you can list any formula
in the language as a theorem.

If you're postings in this discussion are an intentional joke, then
consider yourself successful in having fooled me with the prank. But if
you're not joking and you think you've found some problem with set
theory as opposed to having just made your own axiomatization of an
inconsistent theory, then please take a break from posting nonsense
while you review a good textbook.

MoeBlee

.

Relevant Pages

  • Re: Inconsistent sets
    ... The axiom of infinity ... > It does not guarantee an infinite set. ... > 2) the axiom creates or guarantees only natural numbers (never omega). ... The natural numbers don't need the axiom of infinity. ...
    (sci.math)
  • Re: Cantor Confusion
    ... Only in WM's world is omega itself needed to make things not finite. ... axiom of infinity, because that guarantees that you can even talk about ...
    (sci.math)
  • Re: Why? [was Re: Cantor`s powerset theorem is false?]
    ... The axiom of infinity states that there is a set closed under the ... simply that of having a bijection with omega or with a member of omega. ...
    (sci.logic)
  • Re: Why? [was Re: Cantor`s powerset theorem is false?]
    ... are and are not certain theorems of certain axiom sets. ... ZF has a model in which the domain of discourse is omega. ... ZF without regularity and power set proves ...
    (sci.logic)
  • Re: Why? [was Re: Cantor`s powerset theorem is false?]
    ... The axiom of infinity ALREADY says and does that. ... that omega does not belong to omega is equivalent to asserting the ... That the successor of EACH AND EVERY ...
    (sci.logic)