Re: Cantor Confusion

In article <1162828228.538226.143300@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:


There is a mapping of the diagonal on a (each) column.
There is no mapping of the diagonal on any line.
The diagonal cannot have more elements than the width of the matrix is.

For every n in N, there is a line longer than n,
For every n in N, there is a line number larger than n.
There is no last line.
here is no last position in the diagonal.

The number of elements of the diagonal is assumed to be omega.
That is wrong, because only the supremum is omega.

WM assumes that omega must be a member of omega, which is false.
To say that the number of elements in a well- ordered set is omega
prohibits any of those members, taken in order, from corresponding to
omega.

To insist otherwise is to require omega to be a member of itself.
.

Relevant Pages

  • Re: Implementable Set Theory and Consistency of ZFC
    ... By replacement on the unusual omega, ... Let us define in the meta-theory, an unconventional rank function ... Then each successor step only increases rank by 1, so every member ...
    (sci.math)
  • Re: A dilemma with Z set theory
    ... any list of subsets of omega a subset of omega ... atom is a member of P, and we are sure that T is a member of P, ... extension of x "in the real world" does there exist an object y in M ... undefinable subsets of Omega IS *ALL* undefinable subsets of Omega, ...
    (sci.logic)
  • Re: Review of Mueckenheims book.
    ... omega is a mapping from omega into the power set of omega. ... claimed that no natural number is a member of the power set of ...
    (sci.math)
  • Re: Why? [was Re: Cantor`s powerset theorem is false?]
    ... were a member of omega, then there would be a member of omega that is a ... I do not think we can start a proof by assuming a contradiction to ... We prove the principle of mathematical induction on omega. ...
    (sci.logic)
  • Re: Why? [was Re: Cantor`s powerset theorem is false?]
    ... were a member of omega, then there would be a member of omega that is a ... I do not think we can start a proof by assuming a contradiction to ... We prove the principle of mathematical induction on omega. ...
    (sci.logic)
Loading