William schrieb am Dienstag, 10. Mai 2022 um 15:40:05 UTC+2:
> On Tuesday, May 10, 2022 at 10:09:26 AM UTC-3, WM wrote:

> > > > There canot be infinitely many elements havin infinitely many successors.
> > > Piffle. Of course this is possible. The successors are not "used up" they are used more than once.
> > In a linear sequence like 1, 2, 3, ... there is no element used more than once.
> Piffle. 2 is used once as a successor (the successor of 1) 3 is used twice as a successor (the successor of 1 and the successor of 2), ... n is used n-1 times;. Note that the set of successors that are used more the once is infinite. The set of successors that is used more then twice is infinite. Given any n element of N_p, the set of successors that are used more than n times has cardinatlity aleph_0. There is no largest n.

What I meant is this: Divide the set 1, 2, 3, ... at any point you like. The lower set is finite. If you want to make the lower set infinite, then nothing remains for the upper set.

Regards, WM

Jim Burns schrieb am Dienstag, 10. Mai 2022 um 18:15:15 UTC+2:
> On 5/10/2022 9:14 AM, WM wrote:

> > Divide the set 1, 2, 3, ... at any point you like.
> It would need to be a point such that
> each non-empty collection =< that point
> contains a first and a last.
> ...because that is true of each natural.
> > The lower set is finite.
> Right.

> > If you want to make the lower set infinite,
> > then nothing remains for the upper set.
> This is why the intersection of all end segments
> is empty.

It is why there are finite endsegments.
>
> That would not be possible if
> each non-empty sub-collection of all naturals
> contains a first and a last.
> However, because
> some sub-collections of all naturals
> do not contain a last,
> it is possible.
> More: it is provable.

It is provably wrong. Inclusion monotony is a basic mathematical property, more basic than all of matheology. According to it the intersection of infinite endsegments is infinite.

REgards, WM

On Tuesday, May 10, 2022 at 5:34:38 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 10. Mai 2022 um 15:40:05 UTC+2:
> > On Tuesday, May 10, 2022 at 10:09:26 AM UTC-3, WM wrote:
>
> > > > > There canot be infinitely many elements havin infinitely many successors.
> > > > Piffle. Of course this is possible. The successors are not "used up" they are used more than once.
> > > In a linear sequence like 1, 2, 3, ... there is no element used more than once.
> > Piffle. 2 is used once as a successor (the successor of 1) 3 is used twice as a successor (the successor of 1 and the successor of 2), ... n is used n-1 times;. Note that the set of successors that are used more the once is infinite. The set of successors that is used more then twice is infinite.. Given any n element of N_p, the set of successors that are used more than n times has cardinatlity aleph_0. There is no largest n.
> What I meant is this: Divide the set 1, 2, 3, ... at any point you like. The lower set is finite.

And the upper set is infinite, You can do this at any point you like, so you can do this at each of an infinite number of elements. So there can be infinitely many elements each having infinitely many successors.

--
William Hughes

Jim Burns schrieb am Dienstag, 10. Mai 2022 um 18:38:58 UTC+2:
> On 5/10/2022 9:21 AM, WM wrote:

> > All fractions to be indexed are in this matrix:

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
....
>
> There is no last m. There is no last n.

But there are all m and n, according to set theory.

> > The indices are the positions of the first column.
> > At the beginning all fractions of all other columns
> > are not indexed.
> > Now we try to transfer all fractions, according to
> > Cantor's sequence, into the first column:
> The number of fractions m/n in the matrix for which
> _not all_ non-empty collections =< m or
> _not all_ non-empty collections =< n
> contain a first and a last
> is zero.

It is irrelevant.
>
> Assign the index m+(m+n-1)*(m+n-2)/2
> to the fraction m/n
>
> The number of fractions not-indexed is zero.

So it appears to the naive mind. But there are infinitely many not indexed fractions initially

XOOO...
XOOO...
XOOO...
XOOO...
....

and provably their number does never change be exchanging X and O.

XXOO...
OOOO...
XOOO...
XOOO...
....

XXOO...
XOOO...
OOOO...
XOOO...
....

XXXO...
XOOO...
OOOO...
OOOO...
....

> > The number of fractions in all other columns
> > remains constant: infinite.

Regards, WM

William schrieb am Dienstag, 10. Mai 2022 um 22:44:07 UTC+2:
> On Tuesday, May 10, 2022 at 5:34:38 PM UTC-3, WM wrote:

> > What I meant is this: Divide the set 1, 2, 3, ... at any point you like. The lower set is finite.
> And the upper set is infinite, You can do this at any point you like, so you can do this at each of an infinite number of elements.

No. Always an infinite number must remain in the upper set. But it is impossible to divide 1, 2, 3, ... into two infinite sets.

> So there can be infinitely many elements each having infinitely many successors.

The lower set is never actually infinite (if the upper set is) because otherwise you would divide |N in two infinite sets.

Regards, WM

On 5/10/2022 3:50 PM, WM wrote:
> William schrieb am Dienstag, 10. Mai 2022 um 22:44:07 UTC+2:
>> On Tuesday, May 10, 2022 at 5:34:38 PM UTC-3, WM wrote:
>
>>> What I meant is this: Divide the set 1, 2, 3, ... at any point you like. The lower set is finite.
>> And the upper set is infinite, You can do this at any point you like, so you can do this at each of an infinite number of elements.
>
> No. Always an infinite number must remain in the upper set. But it is impossible to divide 1, 2, 3, ... into two infinite sets.
>
>> So there can be infinitely many elements each having infinitely many successors.
>
> The lower set is never actually infinite (if the upper set is) because otherwise you would divide |N in two infinite sets.
>
> Regards, WM

golly folks...

take the set of natural numbers N

1,2,3,...k-1,k,k+1,k+2,... (((Omega)))

and cut that baby in Half at k, thusly;

at k, NOW CHOP !!!

1,2,3,...k is called a finite FISON

k+1,k+2,... is called an infinite Endsegment

for any natural number k

they make spaghetti noodles that way, from the Mother Noodle in Italy,
the infinite noodle extruder machine, and they chop it at k, into convenient lengths, all you get is FISONs

Extra credit problem: where is the spaghetti noodle endsegment ?

On Tuesday, May 10, 2022 at 5:50:14 PM UTC-3, WM wrote:

>... it is impossible to divide 1, 2, 3, ... into two infinite sets.

Correct and totally irrelevant. What is possible is to divide {1,2,3,...} into a finite set followed by an infinite set. This can be done in an infinite number of different ways, a different way for each element of N_p. There are an infinite number of elements each with an infinite number of successors.

--
William Hughes