Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:
MoeBlee schrieb:



As I said, I don't care about (3).

My mathematical notions include the fact that the diagonal of a matrix
cannot have more elements than every line. If in your opinion this
position is not valid in ZFC or in your logic, then we should stop
here.

Then why did you even START to comment on my proof and raise all kinds
of specious objections when all your objections only boil down to the
fact that the axiom of infinity is used? I said it is a proof in Z set
theory, nothing more and nothing less. If you had simply STARTED by
saying, "But your proof uses the axiom of infinity", then I would have
said, "Yes, of course it does, just as I said AT THE OUTSET that my
proof is in Z set theory" and that would be that.

The diagonal is assumed to exist such that each of its digits exists,
actually. This is established by the mapping on a column.

No it is not.

But it cannot
be established by the mapping on any line. You should recognize that
the following bijection between columns and lines shows a
contradiction, because one element is missing:

1 <--> 1
1,2 <--> 2
1,2,3 <--> 3
...
1,2,3,...n <--> n
...
1,2,3,... <--> omega

1, 2, 3 ... is NOT a line.

But it is the whole column which would correspond to the line omega
(which does not exist).

And it is NOT a contradiction that 1, 2, 3, .. omega is equinumerous
with omega+1.

Of course, the set 1,2,3... is equinumerous to the set 2,3,4,..., 1.
Nevertheless the diagonal of a matrix cannot have more elements than
every line. If in your opinion this position is not valid in ZFC or in
your logic, then we should stop here.

Again, NOW you say that, after I already said so many times that my
proof is in Z set theory.

I answered that already. And the cardinality of the diagonal is not
assumed, but is proven, to be omega.

It is *assumed* by stating the axiom of infinity. Without this
assumption the length was not omega.

It is PROVEN from the axiom of infinity.

But the existence of this axiom is assumed.

The existence of the AXIOM is assumed? The axiom exists, obviously. You
can read it for yourself. It exists as a formula, which is what it is -
a certain kind of formula. You can see a typographical representation
of the formula whenever you like. Are you now saying we should doubt
the existence of formulas? If that is not what you're saying, then in
what sense is relevent to say that the existence of the axiom is
assumed?

since I would not have bothered to even post
a proof about denumerable sequences and talk with you about it for so
long if I accepted any condition that I can't use the axiom of
infinity.

That is *not my condition* but it is the result which follows from the
fact that the diagonal cannot be longer than every line.

If it is not a condition, then what is the sense of your objection? OF
COURSE I used the axiom of infinity. I said my proof is in Z set
theory.

If I have the
choice either to accept the axiom of infinity with the condition that a
diagonal can be longer than every line, or to drop both notions, then I
choose the second.

Fine! Don't accept the axiom of infinity. But that's irrelevent to my
claim, which is that my proof is a proof, in Z set theory, of the
denumerability of the diagonal.

If you can live with the contrary, then try to do
it. I wish you nice dreams.

It's not a matter of whether I accept the axiom of infinity. I said
mine is a proof in Z set theory. And mine is a proof in Z set theory,
whether you or even *I* accept the axiom of infinity or not.

MoeBlee

.

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