William schrieb am Dienstag, 25. Mai 2021 um 14:39:15 UTC+2:
> On Tuesday, May 25, 2021 at 8:33:16 AM UTC-3, WM wrote:
>
> > Every definable natnumber belongs to a FISON and has aleph_0 successors
> correct. Note that the set |N (the set you call the definable natural numbers ) only contains definable elements so all successors are definable elements

Then all successors could be defined and removed individually and nothing would remain. But as you know that is impossible.

> > which cannot all be definable because then there would be a last one having no successor.
> No, this does not follow and is not true.

If all are counted one by one such that none remains, then a last one is counted. If no last one is copunted, then not all are counted.
>
> I note that |N (the set that you call the definable natural numbers) only contains definable numbers.
As you note. |N is a Peano set

Yes, it does not contain aleph_0 elements because aleph_0 elements are following behind every defined n.

>and all Peano sets have cardinality aleph_0 .

Provably wrong. (0, 1/n) will never contain less than aleph_0 unit fractions. Therefore all definable unit fractions are less.

Regards, WM

William schrieb am Dienstag, 25. Mai 2021 um 14:57:17 UTC+2:
> On Tuesday, May 25, 2021 at 8:42:45 AM UTC-3, WM wrote:
> > every n has aleph_0 successors,
> trivial

They can be removed collectively but not individually.

> > aleph_0 of which cannot be represented by decimals because that would imply that none is remaining and that would imply a last represented one.
> No, there is no last decimal representation, so assuming that every element can be represented by a decimal does not mean there has to be a last element.

But removing all elements collectively shows that it is possible to leave no rest. Why is it not possible using the decimal representation?

> Note that in the set |N (the set that you call the definable natural numbers) every element has a decimal representation.

Of course. But removing every element that has a decimal representation will leave aleph_0 elements which cannot be removed individually but collectively.

Regards, WM

FromTheRafters schrieb am Dienstag, 25. Mai 2021 um 15:08:56 UTC+2:
> WM submitted this idea :

> > It does. There is no better definition of a natural numbers than by decimal
> > or binary representation
> Wrong, numbers are not defined by their various representations.

An individual number is best reprented by decimal or binary expansion.

> >> All we need to know is that 0 is definable and if n is definable then n+1 is
> >> definable.
> >
> > But, what is much, much more important, you forget that every n has aleph_0
> > successors,
> *Each* has one unique successor via the successor function.

And each has aleph_0 successors, aleph_0 of which are undefinable because they never are used up. They can be removed completely such that nothing remains, but not individually.

Regards, WM

On 5/25/2021 3:25 PM, WM wrote:
> William schrieb am Dienstag, 25. Mai 2021 um 14:39:15 UTC+2:
>> On Tuesday, May 25, 2021 at 8:33:16 AM UTC-3, WM wrote:
>>
>>> Every definable natnumber belongs to a FISON and has aleph_0 successors
>> correct. Note that the set |N (the set you call the definable natural numbers ) only contains definable elements so all successors are definable elements
>
> Then all successors could be defined and removed individually and nothing would remain. But as you know that is impossible.

only you say you "know" it, now prove it.

>
>>> which cannot all be definable because then there would be a last one having no successor.
>> No, this does not follow and is not true.
>
> If all are counted one by one such that none remains, then a last one is counted.

there is no last one in an infinite set, deal with it.

> If no last one is copunted, then not all are counted.
>>
>> I note that |N (the set that you call the definable natural numbers) only contains definable numbers.
> As you note. |N is a Peano set
>
> Yes, it does not contain aleph_0 elements because aleph_0 elements are following behind every defined n.

there is no such thing as a "defined" n

>
>> and all Peano sets have cardinality aleph_0 .
>
> Provably wrong. (0, 1/n) will never contain less than aleph_0 unit fractions. Therefore all definable unit fractions are less.
>
> Regards, WM
>
>
>

On 5/25/2021 3:29 PM, WM wrote:
> William schrieb am Dienstag, 25. Mai 2021 um 14:57:17 UTC+2:
>> On Tuesday, May 25, 2021 at 8:42:45 AM UTC-3, WM wrote:
>>> every n has aleph_0 successors,
>> trivial
>
> They can be removed collectively but not individually.

they are not carrots in a field, nor rocks on a beach.

>
>>> aleph_0 of which cannot be represented by decimals because that would imply that none is remaining and that would imply a last represented one.
>> No, there is no last decimal representation, so assuming that every element can be represented by a decimal does not mean there has to be a last element.
>
> But removing all elements collectively shows that it is possible to leave no rest. Why is it not possible using the decimal representation?

why is it not possible to print all natural numbers ?

>
>> Note that in the set |N (the set that you call the definable natural numbers) every element has a decimal representation.
>
> Of course. But removing every element that has a decimal representation will leave aleph_0 elements which cannot be removed individually but collectively.

no. Try again.

>
> Regards, WM
>

On 5/25/2021 2:22 PM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 25. Mai 2021 um 16:18:47 UTC+2:
>> On Tuesday, 25 May 2021 at 08:42:45 UTC-3, WM wrote:
>> [...]
>>> It does. There is no better definition of a natural numbers than by decimal or binary representation
>> Hmmmh. The representation of a natural number is *equal* to the natural number???
>
> Unless you believe in platonic nonsense, there is nothing of a number but its set of representations.

go back to counting rock, start at 1, then 2, then 3, build up from
there, pretty soon you are at 17...

>
>> A natural number is *defined* by its representation???
>
> Yes, a definition says what the object is. This is best achieved by the representation.

1 there you go, no pesky decimals, no alien binary,

>
> Regards, WM
>
>

WM brought next idea :
> FromTheRafters schrieb am Dienstag, 25. Mai 2021 um 15:08:56 UTC+2:
>> WM submitted this idea :
>
>>> It does. There is no better definition of a natural numbers than by decimal
>>> or binary representation
>> Wrong, numbers are not defined by their various representations.
>
> An individual number is best reprented by decimal or binary expansion.
>
>>>> All we need to know is that 0 is definable and if n is definable then n+1
>>>> is definable.
>>>
>>> But, what is much, much more important, you forget that every n has aleph_0
>>> successors,
>> *Each* has one unique successor via the successor function.
>
> And each has aleph_0 successors,

Aleph-zero is not a number of elements, it is an infinite cardinality,
and each natural number has one unique successor.

FromTheRafters schrieb am Dienstag, 25. Mai 2021 um 22:53:53 UTC+2:
> WM brought next idea :
> > FromTheRafters schrieb am Dienstag, 25. Mai 2021 um 15:08:56 UTC+2:
> >> WM submitted this idea :
> >
> >>> It does. There is no better definition of a natural numbers than by decimal
> >>> or binary representation
> >> Wrong, numbers are not defined by their various representations.
> >
> > An individual number is best reprented by decimal or binary expansion.
> >
> >>>> All we need to know is that 0 is definable and if n is definable then n+1
> >>>> is definable.
> >>>
> >>> But, what is much, much more important, you forget that every n has aleph_0
> >>> successors,
> >> *Each* has one unique successor via the successor function.
> >
> > And each has aleph_0 successors,
> Aleph-zero is not a number of elements, it is an infinite cardinality,

It is defined as the cardinality of a set in bijection with |N. As I have shown there cannot be a bijection between infinite sets. Therefore I use aleph_0 as an abbreviation for actual infinity.

> and each natural number has one unique successor.

One unique immediate successor and aleph_0 successors of successors. This is abbreviated by aleph_0 successors.

Regards, WM

On Tuesday, May 25, 2021 at 5:25:57 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 25. Mai 2021 um 14:39:15 UTC+2:
> > On Tuesday, May 25, 2021 at 8:33:16 AM UTC-3, WM wrote:
> >
> > > Every definable natnumber belongs to a FISON and has aleph_0 successors
> > correct. Note that the set |N (the set you call the definable natural numbers ) only contains definable elements so all successors are definable elements
> Then all successors could be defined and removed individually and nothing would remain.

No, there is no reason to assume that defined numbers can be removed "individually" and nothing would remain.

> But as you know that is impossible.
> > > which cannot all be definable because then there would be a last one having no successor.
> > No, this does not follow and is not true.
> If all are counted one by one such that none remains, then a last one is counted. If no last one is copunted, then not all are counted.
> >
I note that |N (the set that you call the definable natural numbers) only contains only definable numbers. As you note. |N is a Peano set and all Peano sets have cardinality aleph_0 . (follows trivially from the fact that every element has a successor)/

==
William Hughes

On Tuesday, May 25, 2021 at 6:06:09 PM UTC-3, WM wrote:
> As I have shown there cannot be a bijection between infinite sets.

Piffle. There are lots (eg. the identity, f(x)=x for all x an element of S is a bijection between any set S and itself).

==
William Hughes

WM submitted this idea :
> FromTheRafters schrieb am Dienstag, 25. Mai 2021 um 22:53:53 UTC+2:
>> WM brought next idea :
>>> FromTheRafters schrieb am Dienstag, 25. Mai 2021 um 15:08:56 UTC+2:
>>>> WM submitted this idea :
>>>
>>>>> It does. There is no better definition of a natural numbers than by
>>>>> decimal or binary representation
>>>> Wrong, numbers are not defined by their various representations.
>>>
>>> An individual number is best reprented by decimal or binary expansion.
>>>
>>>>>> All we need to know is that 0 is definable and if n is definable then
>>>>>> n+1 is definable.
>>>>>
>>>>> But, what is much, much more important, you forget that every n has
>>>>> aleph_0 successors,
>>>> *Each* has one unique successor via the successor function.
>>>
>>> And each has aleph_0 successors,
>> Aleph-zero is not a number of elements, it is an infinite cardinality,
>
> It is defined as the cardinality of a set in bijection with |N. As I have
> shown

Failed to show you mean.

> there cannot be a bijection between infinite sets. Therefore I use
> aleph_0 as an abbreviation for actual infinity.

Actual infinity isn't a number either.

>> and each natural number has one unique successor.
>
> One unique immediate successor and aleph_0 successors of successors. This is
> abbreviated by aleph_0 successors.

Okay. Can you have successors when you have no elements?

On Tuesday, May 25, 2021 at 5:25:57 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 25. Mai 2021 um 14:39:15 UTC+2:
> > On Tuesday, May 25, 2021 at 8:33:16 AM UTC-3, WM wrote:
> >
> > > Every definable natnumber belongs to a FISON and has aleph_0 successors
> > correct. Note that the set |N (the set you call the definable natural numbers ) only contains definable elements so all successors are definable elements
> Then all successors could be defined

Nope. There is an infinite set of *definable* numbers. You can only *define* a finite set.

Simple counterexample. |N (the set that you call the definable natural numbers) only contains definable numbers.
As you note. |N is a Peano set and all Peano sets have cardinality aleph_0. (This follows trivially from the fact that every element has a successor.)

--
William Hughes

On Tuesday, 25 May 2021 at 16:22:16 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 25. Mai 2021 um 16:18:47 UTC+2:
> > On Tuesday, 25 May 2021 at 08:42:45 UTC-3, WM wrote:
> > [...]
> > > It does. There is no better definition of a natural numbers than by decimal or binary representation
> > Hmmmh. The representation of a natural number is *equal* to the natural number???
> Unless you believe in platonic nonsense, there is nothing of a number but its set of representations.
> > A natural number is *defined* by its representation???
> Yes, a definition says what the object is. This is best achieved by the representation.

And yet there are competing representations: 1, I, α, ١, א ,.... At most one of them can be the definition of the concept '1'. You are full of shit, as usual.

On 5/26/2021 6:57 AM, Gus Gassmann wrote:
> On Tuesday, 25 May 2021 at 16:22:16 UTC-3, WM wrote:
>> Gus Gassmann schrieb am Dienstag, 25. Mai 2021 um 16:18:47 UTC+2:
>>> On Tuesday, 25 May 2021 at 08:42:45 UTC-3, WM wrote:
>>> [...]
>>>> It does. There is no better definition of a natural numbers than by decimal or binary representation
>>> Hmmmh. The representation of a natural number is *equal* to the natural number???
>> Unless you believe in platonic nonsense, there is nothing of a number but its set of representations.
>>> A natural number is *defined* by its representation???
>> Yes, a definition says what the object is. This is best achieved by the representation.
>
> And yet there are competing representations: 1, I, α, ١, א ,... At most one of them can be the definition of the concept '1'. You are full of shit, as usual.
>

WM makes rocks exist by counting them (in linear order)

William schrieb am Dienstag, 25. Mai 2021 um 23:27:08 UTC+2:
> On Tuesday, May 25, 2021 at 6:06:09 PM UTC-3, WM wrote:
> > As I have shown there cannot be a bijection between infinite sets.
> Piffle. There are lots (eg. the identity, f(x)=x for all x an element of S is a bijection between any set S and itself).

It seems so to the uncritical mind but it is not. Let's use natnumbers because rationals or even reals would stretch you too far. Every natural n from (1/n, 1] can be put in bijections with itself. This can be proved by looking at n. But almost all naturals from (0, 1/n) cannot be looked at. Well you can say by definition n is mapped on n also in dark cases. But you can only define it for the typical element n which is a variable but not a natural number. You cannot verify it for most numbers because you have no access to them. And if more complicated mappings are under discussion then dark numbers will fail with certainty.

Regards, WM