Re: Request for Peer Review - Refutation of Cantor Theorem Conclusion

Hi All:

Thanks for the feedback.

Okay. First, IF you could show that the diagonal set is not an element
of the power set, this would NOT establish that "f is [...] a
bijection." The proof begins with an arbitrary function f:S->P(S), so
at best it would establish that f ->could be<- a bijection.

Assume the following:
1. D is not an element of the powerset.
2. A mapping function f:N->P(N).
What other things are necessary to show f is a bijection between N and
P(N)?

concentrate on demonstrating why Cantor's diagonal
argument does not prove what it appears to prove. That is presumably a
smaller piece, easier to explain and evaluate, and so requiring a
smaller investment on the part of reviewers.

I will take your suggestion and reduce the paper to the necesary proof.

Can you explain how the diagonal avoids being in the power set? Do you
regard it as not being a set, or do you think it contains some element
that is not in the base set?

If D is an element of the powerset, then the powerset/set-theory is
inconsistency. (Its a proof by contradiction.) Therefore, D does not
exist. Cantor's Theorem states D is not in the image of f. Therefore,
whether D exists or not is immaterial to the Cantor's proof of D not in
the image of f, just the conclusion; namely, card(N) < card(P(N)).
Since D does not exist and there exists a mapping function f:N->P(N)
and N and P(N) are equinumermous, then card(N) = card(P(N)).

I just posted this, because maybe,
the linear ordering appeals more
to intuition, than the ordering of
sets. Diagonalization is more abstractly
based on sets.

Thanks for the discussion. I briefly looked at it. While I believe it
can be solved, I would like to focus on the diagonal/set-theory
approach because that is the easiest to understand.

Thus, for each element y of S, "y
is an element of f(y)" is either true or false and not both

But it is by *definition* (at least in ZFC):
D := {x e S : x !e f(x)}.

Tangential Thought: Under Cantor's Theorem, why is the membership
function for D considered a valid predicate when it tries to return
both true and false (or neither)?

Content yourself with people who'd show you the errors of your ways
were to have the courage to post your conjectured proof here.

Fair enough. Let me update with Patricia's suggestion and then I will
post it.

If you have some other
collection of axioms for set theory in mind you need to
start by saying exactly what the axioms are.

Nope, standard axioms.

.

Relevant Pages

  • Re: SUBTHREAD : fixed point
    ... There is a natural extension of the powerset ... ideas for his proposed set theory. ... here are the aleph's without ordinals (under some standard axioms) ... is more like mereology than set theory. ...
    (sci.math)
  • Re: Induction, Kant, Infinity
    ... RF> If you use the notion of proper classes in ZF ... THERE IS NO SUCH THING as "the powerset ... theorem that infinite sets aren't equivalent". ... You can't have a set theory without axioms, ...
    (sci.logic)
  • Re: Cantors diagonal proof wrong?
    ... DAMN, you're stupid. ... Then there ARE AXIOMS, TOO, dumbass. ... > About the multiple representation of ordinals as sets, ... > powerset of a set X, P, with the script P to delineate it from the ...
    (sci.math)
  • Re: Cantors diagonal proof wrong?
    ... DAMN, you're stupid. ... Then there ARE AXIOMS, TOO, dumbass. ... > About the multiple representation of ordinals as sets, ... > powerset of a set X, P, with the script P to delineate it from the ...
    (sci.logic)