On 5/4/2022 2:14 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 4. Mai 2022 um 19:39:10 UTC+2:
>> On 5/4/2022 9:53 AM, WM wrote:
>>
>>> A difference in arithmetic is called a distance
>>> in geometry.
>> In geometry of the usual (Archimedean) variety,
>> for any two positive distances x and y,
>> there is a _finite ordinal_ n such that n*x > y.
>
> The usual is not Cantor's extension.
>
>> There _isn't_ room in a geometric line for omega-many
>> (meaning aleph_0) finite-length extensions.
>
> There is even much more room.
>
> Let the solutions of ω^x = x be the giants of the second number class
> Es fragt sich, in welchem Abstande von γ dieser Gigant δ liegt.
> (Abstand = Distance)
> 1886, 11. Oct. Cantor, letter to Goldscheider

misquoted.

>
>>> Before omega there is either nothing, i.e., a gap,
>>> or there is something. But this cannot be defined.
>> For each thing before omega,
>> there is room for at least one more before omega.
>> Therefore, nothing before omega is
>> immediately before omega.
>
> Then nothing is immediately before omega. That is called a gap.

to have a gap, you have to stop at k. from k to omega, your gap is full of natural numbers.

>
<snip>

>
> Regards, WM

FromTheRafters schrieb am Montag, 9. Mai 2022 um 17:59:21 UTC+2:
> WM laid this down on his screen :

> > That is not a valid argument because all j, p, q have more successors than
> > predecessors.
> Irrelevant. It doesn't make the argument invalid.

It excludes the second half.

> > Why don't you get that simple fact?
> Why worry yourself over such an irrelevant fact that infinite is larger
> than finite.

The infinite is larger than all finite initial segments. What is correct for them need not be correct for the infinite.

Regards, WM

William schrieb am Montag, 9. Mai 2022 um 21:18:25 UTC+2:
> On Monday, May 9, 2022 at 10:19:17 AM UTC-3, WM wrote
> > > Like any Peano set, N_p has cardinality aleph_0.
> > No, every element has finitely many predecessors but aleph_0 successors.
> Correct, And each of these successors is an element of N_p.

Impossible. Then the successors would be used up.
There canot be infinitely many elements havin infinitely many successors.

Regards, WM

On 5/9/2022 1:42 PM, WM wrote:
> William schrieb am Montag, 9. Mai 2022 um 21:18:25 UTC+2:
>> On Monday, May 9, 2022 at 10:19:17 AM UTC-3, WM wrote
>>>> Like any Peano set, N_p has cardinality aleph_0.
>>> No, every element has finitely many predecessors but aleph_0 successors.
>> Correct, And each of these successors is an element of N_p.
>
> Impossible. Then the successors would be used up.
> There canot be infinitely many elements havin infinitely many successors.

[0] = 0
[1] = 1/2
[2] = 1

The interval [0] and [1] has an infinite amount of numbers between them.

The interval [1] and [2] has an infinite amount of numbers between them.

Agreed?

On Monday, May 9, 2022 at 5:42:23 PM UTC-3, WM wrote:
> William schrieb am Montag, 9. Mai 2022 um 21:18:25 UTC+2:
> > On Monday, May 9, 2022 at 10:19:17 AM UTC-3, WM wrote
> > > > Like any Peano set, N_p has cardinality aleph_0.
> > > No, every element has finitely many predecessors but aleph_0 successors.
> > Correct, And each of these successors is an element of N_p.
> Impossible. Then the successors would be used up.
> 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. Given any n element of N_p, the set of elements that are used more than n times has cardinality aleph_0. There is no largest n, so the successors are not "used up",

--
William Hughes

On 5/9/2022 9:14 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 8. Mai 2022 um 19:04:48 UTC+2:
>> On 5/8/2022 8:42 AM, WM wrote:

>>> according to Cantor's function
>>> k = (m + n - 1)(m + n - 2)/2 + m
>>>
>>> then the fractions of the sequence
>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
>>> are indexed (or covered by X's)
>>> but the number of not indexed fractions (the O's)
>>> will remain the same forever.
>>
>> The number not-indexed is zero,
>
> It is infinite at the beginning

No.

The number of fractions m/n _not_ with
each non-empty sub-collection =< m or =< n
having a first and a last
_is zero_

....because
those are all the fractions in this discussion.

The number of fractions m/n _not_ with
each non-empty sub-collection =< (m+n-1)*(m+n-2)/2+m
having a first and a last
_is zero_

....because
the number of fractions m/n _not_ with
each non-empty sub-collection =< m+n
each non-empty sub-collection =< m+n-1
each non-empty sub-collection =< m+n-2
each non-empty sub-collection =< (m+n-1)*(m+n-2)
each non-empty sub-collection =< (m+n-1)*(m+n-2)/2
each non-empty sub-collection =< m+(m+n-1)*(m+n-2)/2
having a first and a last
_is zero_

Assign to fraction m/n the index m+(m+n-1)*(m+n-2)/2
and the number not-indexed is zero.

> and it will never decrease.
> The number of O's cannot decrease because
> there is purely exchanging of X and O.

Some collections can be matched to a proper subset.

_Up to_ some X or some O is NOT such a collection.
_All_ the X's and _all_ the O's are such a collection.

>> What you (WM) find obvious -- and which is wrong --
>> is that all collections can be ordered so that
>> each non-empty sub-collection
>> contains a first and a last.
>> Not all can be.
>
> All fractions can be ordered, according to set theory.

All fractions _cannot_ be ordered so that
each non-empty sub-collection
contains a first and a last.

>> You keep showing us proof that not all can be,
>> but you only ever interpret that as
>> _everyone else in the universe is wrong_
>
> Everyone in set theory is wrong.

Your "proof" that we're wrong is only
a proof that we're right, followed by
another declaration from you that,
since the proof gave the wrong answer,
the rest of us are wrong.
Multiplied by a million.

On 5/9/2022 3:42 PM, WM wrote:
> William schrieb am Montag, 9. Mai 2022 um 21:18:25 UTC+2:
>> On Monday, May 9, 2022 at 10:19:17 AM UTC-3, WM wrote
>>>> Like any Peano set, N_p has cardinality aleph_0.
>>> No, every element has finitely many predecessors but aleph_0 successors.
>> Correct, And each of these successors is an element of N_p.
>
> Impossible. Then the successors would be used up.

no, there are infinite number of successors for each and every n.

> There canot be infinitely many elements havin infinitely many successors.

It is so, accept that fact.

>
> Regards, WM