Re: Two results of set geometry

On 25 Sep., 22:46, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On Sep 25, 4:09 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:





On 25 Sep., 17:48, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Sep 25, 11:00 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On 25 Sep., 15:00, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Sep 25, 8:42 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On 25 Sep., 13:36, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Sep 25, 6:55 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On 24 Sep., 22:37, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Sep 24, 4:09 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On 24 Sep., 20:39, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

<snip>

The only way to get the finite segments without the complete
column is to write each finite segment in its own column.

That can be done. And then these segments can be projected in one
column.

And note that you do not change the finite segments in any way.
So any bijection that exists when each finite segment is written
in its own column still exists when all finite segments are written
in the same column.

When each of the finite segments is written in its own column there
is a bijection from the finite segments to N.

Yes.

When each of the finite segments is written in the same column,
there is a bijection from the finite segments to N.

Yes.

Therefore there are as many elements in N as there
are finite segments. From

WM: There are omega finite segments 1, 11, 111, ...

we conclude there are omega elements in N.

Without the complete column. So the complete column is not more and
not less - it is simply not existing?

No, the complete column is the set of all finite segments.
It certainly exists. However, it is a *set* of finite
segments (or equivalently, a *set* of indices).
It is not a finite segment.

As you acknowledge, "there is a bijection from the finite
segments to N". As you point out
there are omega finite segments. So there
are omega elements in N.

Not without N.

N is not an element of N.
A bijection to a set B is to the elements of B.
The bijection is between the finite
segments and the elements of N.
We have: "there is a bijection from the finite
segments to N". There are omega finite segments. So
there are omega elements in N.

<snip discussion of the properties of omega.
The only important fact here is that there
are omega finite segments.>

You have agreed that there is a bijection from
the finite segments to the elements of N
and that there are omega finite segments.

omega is double valued. omega is the number of finite elements (rows)
which is not completed, and omega is the completed infinite set.

Do you also agree that N has omega elements?

Here the double valuedness shows up again. N has omega elements and N
is omega elements. N is a heap? Not more than its elements?

Regards, WM

.

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