Re: Cantor Confusion
- From: "MoeBlee" <jazzmobe@xxxxxxxxxxx>
- Date: 6 Nov 2006 15:04:32 -0800
mueckenh@xxxxxxxxxxxxxxxxx wrote:
MoeBlee schrieb:
mueckenh@xxxxxxxxxxxxxxxxx wrote:
MoeBlee schrieb:
For ordinals,
x<y <-> xey
where 'e' is the epsilon membership symbol.
That is identical with my definition. Take for instance Zermelo's
definition of the naturals or Cantor's own (Collected works, p.
289-290), then you can see it. Your definition has been simply
translated from Cantor's. Please don't conclude from your own ignorance
on mine.
You wrote:
"My question is : Do you maintain omega > n for all n e N? I know that
modern set theory says so. If something can be larger than a number,
then it must be a number."
Given that "n < omega" stands for "n e omega", and given that N is
omega, what point is there in asking anyone whether they maintain that
n < omega for every n e N? You're asking someone whether he or she
maintains that n e omega for every n e omega. What is the point of
asking that?
If omega > n then we cannot have a diagonal with omega digits in a
matrix the lines of which have only n (= less than omega) digits,
because the digits of the diagonal are digits of the lines.
That's just another reenactment of your dogmatic pronouncements. I gave
you a proof in set theory, from axioms using only first order logic.
You give an argument that does not use any recognizable logic and from
no stated axioms.
Please just show any step of my proof that is not justified by first
order logic applied to the axioms, or show a proof of a sentence P and
~P in the language of set theory such that both sentences are theorems
of set theory.
If it cannot be a fraction because ZF does
not yet know how to divide elements,
In ZF we define various operations of division. As far as I know, there
is not a dvision operation for sets in general.
By the way, I do see that there are definitions of subtraction,
division, and logarithms for ordinals. But they are not anything like
your non-definition of ordinal division.
Why then do you not understand how an edge of the binary tree can be
divided?
Haven't I asked you a few times already not to effectively put words in
mouth? I never said I don't understand how an edge can be divided. I
never said anything about "divided edges".
So do you understand it now, how an edge can be divided in two pieces
and each one can be related to a path? So do you understand now, how
half an edge can be divided in two pieces and each one can be related
to a path? So do you understand now, how a fraction of an edge can be
divided in two pieces and each one can be related to a path?
If you want to define a particular relation on a particular object that
is proven to exist in set theory, then I'm all ears.
You are ignorant of the basics of set theory.
I know its basics. That does not mean that I have to accept them.
No, you don't know basic set theory. You don't even have a SENSE (let
alone a precise understanding) of mathematical definition.
MoeBlee
.
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