Re: Two results of set geometry

In article <1188660241.121693.309500@xxxxxxxxxxxxxxxxxxxxxxxxxxx> WM <mueckenh@xxxxxxxxxxxxxxxxx> writes:
On 31 Aug., 04:07, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> Only the fact that the first column has a lot of finite initial
> segments and an infinite initial segment while the rows are all finite
> may be raising some suspicion.

What this is stating, precisely is:
That the ordered set of natural numbers has a lot of finite initial
segments and one infinite initial segment, while the set of natural
numbers is infinite may be raising some suspicion.
Yes, I knew all along that you distrust the axiom of infinity. Well,
reject it, and go on.

I do not reject the axiom of infinity. I assume that every initial
segment of the first column (including the complete first column)
corresponds to (is in bijection with) a natural number.

Now you are using (again) your very own terminology. There is a bijection
between the initial segments (including the complete first column) with the
natural numbers. There are lots of them. But you are not talking about
bijections, but about oder-preserving bijections between ordered sets. And,
no, there is no order-preserving bijection between the ordered set of initial
segments and the ordered set of natural numbers.

> That is incorrect. Each partner on the one side is a number of natural
> numbers and on the other side each partner is a natural number. I
> tried to express that by the variables {1,2,3,...,n} <--> n.

Again, you lost me here. The partners where both the set of natural
numbers.

The partners include {1,2,3,...}. That is not a natural number.

I disrember the start. As far as I remember the partners where the set of
natural numbers. {1, 2, 3, ...} is not in the set of {1, 2, 3, ..., n}.

> In fact I do not add something to the set the column is bijected witd,
> but I do neither add something to the column. The initial segments on
> the "left hand side" change from
> 1, 11, 111, ..., 111... to
> 11, 111, ..., 111..., 111...1.

There is no last row, so where does the 111...1 come from?

There are omega initial segments of the first *column*. Adding one 1
to every initial segment supplies 11, 111, ..., 111..., 111...1, the
last one having ordinal omega + 1.

Again, adding again a one to the initial segment that is the first column.
Which row has that complete initial segment?

> If they rermain in bijection with the sequences in the rows, then the
> set {2, 3, 4, ...} has ordinal omega + 1.

It is based on your assumption that there is a last row. There is none,
so now what?

No. There is no last row. But every initial segment of the first
column is said to have a partner in the set of sequences in the rows.

Who said that? There is not such partner for the *complete* first column.
That could only be true it there *was* a last row.

> We have a bijection between natural numbers and ordinal numbers of
> sets of natural numbers.

Oh. I did not know that. As far as I know, in this part we were talking
about matrices and of bijections between lines and columns. You switch
point of view from one to another without consistency.

The natural numbers are represented by the sequences in the rows. The
ordinal numbers are represented by the initial segments of the first
column.

O. Which initial segment represents 0? And to which line does it
correspond?

I said *replaced*. But let's see whether I do understand you:
a natural number n has the set of second indices {1, 2, 3, ..., n}
a set of natural numbers {n1, ..., nk} has the set of second indices
U{l = 1, ..., k} {1, 2, 3, ..., l} = {1, 2, 3, ..., k}. The set of
natural numbers has as second indices {1, 2, 3, ...} = N. If we add
that "infinite number" 111..., we have that that number has the set
of indices {1, 2, 3, ...} = N. So, yes, the sets of second indices
is the same.

Initially you were of opposite opinion.

Perhaps. But as you do not clearly state what you mean it is easy
to be set on a wrong footing.

I do not see why that should be a problem.

If you add a sequence which is larger than any finite sequence, then
you do not add any index?

Why is that a problem?

That hints to the fact that you add nothing.
The infinite sequence does not provably existt, because there is no
indication (no piece of circumstantial evidence) that you have done
anything at all.

You have done something. Namely adding a line with *all* indices.
But of course you can not prove that an infinite line exists. It is
there due to the axiom of infinity (which you consistently reject,
although you state you do not).

Again the negation of the axiom of infinity. There is no problem once
you are able to distinguish between natural numbers and ordinal numbers.
Properly speaking, the first ordinal number is 0, not 1. So each
ordinal number is the order number of the ordered set of all preceding
ordinal numbers. And once you see that, there is no problem at all.

Wrong. Repeat twice adding a 1 to all partners of the bijection. Then
you see the problem again.

Your context is so far away that I do not even understand what you mean.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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