Re: Two results of set geometry

On Sep 25, 5:19 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

Or do yuou claim that the complete diagonal, a subset of the union of
rows, can be longer than the rows?

The diagonal is not just a subset of the union of the set of rows; the
diagonal IS the union of the set of rows. And I could not have made it
more clear already for you. Again:

The diagonal is longer than any single row, but the diagonal is the
union of the set of rows.

For every proper initial segment of the diagonal, there is a row that
equals that proper initial segment. But there is no row that is as
long as every proper initial segment.

Again:

For every p there is an r.

It's not the case that there is a single r for all p's.

That is not a contradiction.

Even more basically:

If you have a principle "if every proper initial segement of a
sequence is fintite, then the sequence is finite", then there's no
reason for you to do proofs regarding set theory. With your principle,
you will derive a contradiction with set theory just about
immediately.

Even more basically, if you have a principle "there are no sets that
can't be put in bijection with a natural number", then there is no
reason for you to have to do proofs about set theory. You've
immediately contradicted set theory already with your principle. All
the stuff about trees and whatnot is superfluous. With your principle,
you point blank contradict set theory.

Every row is finite. The diagonal is infinite. A contradiction in my
eyes.

Right, in YOUR EYES, since it contradicts your principle that a
sequence can't be infinite if every proper intitial segment is finite.

But why even bother with that? You have a principle that there do not
exist sets that can't be bijected with a natural number. And set
theory is swimming with contradictions of that. So, fine, we know that
set theory is in contradiction with your principles. No need for
futher demonstrations in that regard.

The diagonal is a proper subset of the finite segments of 1's in the
rows.

No it's not. Every member of the diagonal is a member of some row.
The diagonal itself is not a proper subset of any row nor even of the
union of the rows. Rather the diagonal IS the union of the rows.

It is the union of rows and as such it is a subset of the union of
rows. May be improper,

The diagonal is the union, therefore is a subset of the union.
Correct. But it is not a PROPER subset. That is trivial from the
defintion of 'proper subset'. If x=y, then x subset of y and NOT the
case that x propersubsetof y.

but in any case the union of rows does not
contain an infinite element.

Right.

If you want to argue that the infinite
number of rows makes their union infinitely long, then you are wrong.
You argument would also apply to an infinite union of rows of length
1.

No, if the lengths were all 1, then all the rows are identical (a row
is a sequence, and a sequence is determined by its length and its
values at each position, so for any sequences r and s, if the length
of r is the length of s and both have the value 1 at the first
position, then r=s.

Why would you try such a silly argument?

I don't.

The proof that the diagonal is infinte is trivial. It comes from
applying first order logic to the axioms of set theory (and even
informally from basic notions about sequences and such). I've already
posted that proof for you a few times; it is utterly trivial.

Meanwhile, you have certain principles that contradict those axioms.
Okay, you're entitled to adopt whatever principles you want. And I
agree that the set theory axioms contradict your principles. But that
doesn't change the fact that there exists an exact finite sequence (a
proof) of finite sequences (formulas) such that every member of the
proof sequence is an axiom of set theory or axiom of first order logic
with identity or follows by modus ponens from previous entries in the
sequence, and the final line of that proof sequence is the theorem
that the diagonal we've been talking about is infinite. That's simply
an incontrovertible fact about finite sequences (the formalized
version of the statement "the diagonal is infinite" is itself - the
STATEMENT - a finite sequence of formal symbols and the proof is a
fintite sequence of finite sequences).

If there is no infinite row, then there cannot be an infinite
diagonal.

You've never proven any such thing.

It is obvious. Take infinitely many rows of height between 1 and 2. Do
you think it must be proved that the union i less than 3?

You've not proven it from the axioms of set theory.

You've not proven if from any axioms of your own.

Instead, you claim, upon some other "experimental" principle of
yours that it is true of "reality".

No one disputes that "If there is no infinite row, then there cannot
be an infinite diagonal" contradicts set theory.

So, since you take it as a principle (or a "truth of reality" or
whatever you want to call it), then I don't see why you waste any time
at all trying make proofs about set theory. Just invoke your principle
and set theory is contradicted - we AGREE that set theory is thus
contradicted - nothing more is required simply to contradict set
theory.

If you really argue that an infinite set of finite heights results in
an infinite union height, then we have no dissent.

I haven't argued about heights (except that the first column is
infinite; actually all the columns are).

I've given you renderings of formal proofs that the diagonal is
infinite. And I've said that I agree that the conclusion of the proof
(not the existence of the proof itself, which is incontrovertible)
contradicts your principles. No need for any argument about trees and
whatnot to convince me of what I already know: set theory contradicts
your principles.

MoeBlee

.

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