JVR schrieb am Sonntag, 16. Januar 2022 um 11:22:38 UTC+1:
> On Sunday, January 16, 2022 at 9:46:10 AM UTC+1, WM wrote:
> > Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> > > On 1/15/2022 5:38 PM, WM wrote:
> > > > Jim Burns schrieb
> > > > am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
> > > There is no first smaller.
> > > For each smaller there are infinitely-many between
> > > -- which is why there is no first smaller.
> > But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
> > That is potential infinity.
> >
>
> \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.

How can you reconcile this with the fact that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!

Note that the so-called Matheologial Explosion is not acceptable in rational mathematics.

> Now shut up until you have understood this fact.

This fact is easily understood and reconciled with the constancy of the numbers of O's by recognizing dark numbers. Do you have a better idea?

Regards, WM

On 1/16/2022 2:46 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
>> On 1/15/2022 5:38 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
>
>> There is no first smaller.
>> For each smaller there are infinitely-many between
>> -- which is why there is no first smaller.
>
> But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
> That is potential infinity.
>
> Regards, WM

being 'chosen' is an 'external event'. Each 'chosen one' is the last one YOU 'chose'.

this has nothing to do with sets or numbers at all.

On Sunday, January 16, 2022 at 4:28:49 PM UTC+1, WM wrote:
> JVR schrieb am Sonntag, 16. Januar 2022 um 11:22:38 UTC+1:
> > On Sunday, January 16, 2022 at 9:46:10 AM UTC+1, WM wrote:
> > > Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> > > > On 1/15/2022 5:38 PM, WM wrote:
> > > > > Jim Burns schrieb
> > > > > am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
> > > > There is no first smaller.
> > > > For each smaller there are infinitely-many between
> > > > -- which is why there is no first smaller.
> > > But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
> > > That is potential infinity.
> > >
> >
> > \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.
> How can you reconcile this with the fact that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!
>
> Note that the so-called Matheologial Explosion is not acceptable in rational mathematics.
> > Now shut up until you have understood this fact.
> This fact is easily understood and reconciled with the constancy of the numbers of O's by recognizing dark numbers. Do you have a better idea?
>
> Regards, WM

Yes, I have a better idea. Learn to understand the obvious facts such as
\pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.
Then find the errors in your simplistic argument yourself.

On 1/16/2022 2:45 AM, WM wrote:
> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>
>> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
>
> Construct the largest number you can. It is the the largest one, but only as long as you do not construct a larger one. And after that one, you can construct an even larger one. That is potential infinity.
>
> Regards, WM

following your instructions;

Ok, I construct the number 17.

it is the largest one, I have not constructed a larger one.

after that, I can construct an even larger one, Ok how about 19.

that is 'potential infinity' (??)

no, that is silly. I have a card with 19 on it.

that is called "observer dependent Math", the math changes depending upon the observer.

you are missing some lines to get from picking an number, to a statement about (non) infinity.
that is where Math comes in, use math to be clear.

JVR schrieb am Sonntag, 16. Januar 2022 um 16:36:20 UTC+1:
> On Sunday, January 16, 2022 at 4:28:49 PM UTC+1, WM wrote:
> > JVR schrieb am Sonntag, 16. Januar 2022 um 11:22:38 UTC+1:
> > > On Sunday, January 16, 2022 at 9:46:10 AM UTC+1, WM wrote:
> > > > Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> > > > > On 1/15/2022 5:38 PM, WM wrote:
> > > > > > Jim Burns schrieb
> > > > > > am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
> > > > > There is no first smaller.
> > > > > For each smaller there are infinitely-many between
> > > > > -- which is why there is no first smaller.
> > > > But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
> > > > That is potential infinity.
> > > >
> > >
> > > \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.
> > How can you reconcile this with the fact that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!
> >
> > Note that the so-called Matheologial Explosion is not acceptable in rational mathematics.
> > > Now shut up until you have understood this fact.
> > This fact is easily understood and reconciled with the constancy of the numbers of O's by recognizing dark numbers. Do you have a better idea?
> >
> Yes, I have a better idea. Learn to understand the obvious facts such as
> \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.

That is not an idea how to reconcile complete indexing with the constancy of the numbers of O's.

> Then find the errors in your simplistic argument yourself.

Unfortunately there are no errors. The O's could only decrease during transpositions at finite steps. But they don't. Any idea?

Regards, WM

On 1/16/2022 9:41 AM, WM wrote:
> JVR schrieb am Sonntag, 16. Januar 2022 um 16:36:20 UTC+1:
>> On Sunday, January 16, 2022 at 4:28:49 PM UTC+1, WM wrote:
>>> JVR schrieb am Sonntag, 16. Januar 2022 um 11:22:38 UTC+1:
>>>> On Sunday, January 16, 2022 at 9:46:10 AM UTC+1, WM wrote:
>>>>> Jim Burns schrieb am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
>>>>>> On 1/15/2022 5:38 PM, WM wrote:
>>>>>>> Jim Burns schrieb
>>>>>>> am Samstag, 15. Januar 2022 um 21:15:35 UTC+1:
>>>>>> There is no first smaller.
>>>>>> For each smaller there are infinitely-many between
>>>>>> -- which is why there is no first smaller.
>>>>> But not all can be chosen. Each chosen one is the last one only as long as you do not construct a larger one.
>>>>> That is potential infinity.
>>>>>
>>>>
>>>> \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.
>>> How can you reconcile this with the fact that the number of indexes X does never change, i.e., in every finite step (where it only could change), it does not. The number of O's remains constant and much larger than the number of X's. After every finite step!
>>>
>>> Note that the so-called Matheologial Explosion is not acceptable in rational mathematics.
>>>> Now shut up until you have understood this fact.
>>> This fact is easily understood and reconciled with the constancy of the numbers of O's by recognizing dark numbers. Do you have a better idea?
>>>
>> Yes, I have a better idea. Learn to understand the obvious facts such as
>> \pi(x,y) = \frac{1}{2} (x + y) (x + y + 1) + y is bijective.
>
> That is not an idea how to reconcile complete indexing with the constancy of the numbers of O's.
>
>> Then find the errors in your simplistic argument yourself.
>
> Unfortunately there are no errors. The O's could only decrease during transpositions at finite steps. But they don't. Any idea?
>
> Regards, WM

wrong, you made several intentional mistakes, you *disappeared them* in your construct.

Quote your snippage, which shows your 2 errors;

"Now every index is moved to the place which Cantor has prescribed: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... . At this place
there appears an X and where the *index has been taken* from there appears an O. After eight steps for instance we see the following picture: 1/1, 1/2,
2/1, 1/3, 3/1, 1/4, 2/3, 3/2 are indexed whereas 4/1, 5/1, 6/1, 7/1, 8/1 *have lost their indexes*: "

On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>
> > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> Construct the largest number you can.

It is an element of N_P. And it was an element of N_P before you "constructed" it. N_P does not change.

> It is the the largest one, but only as long as you do not construct a larger one.

It may be the largest element that you have "constucted" but it is not the largest element of N_P.
N_P does not change. N_P is not something you have to "construct".

--
William Hughes

On 1/16/2022 3:46 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:

>> There is no first smaller.
>> For each smaller there are infinitely-many between
>> -- which is why there is no first smaller.
>
> But not all can be chosen.

All k which end some two-ended-counting-ordered
{0,...,k} beginning at 0
end some two-ended-counting-ordered
{0,...,k} beginning at 0

All k which end some two-ended-counting-ordered
{0,...,k} beginning at 0
are before omega.

No k which ends some two-ended-counting-ordered
{0,...,k} beginning at 0
is first smaller than omega.

> Each chosen one is the last one only as long as
> you do not construct a larger one.
> That is potential infinity.

Then potential infinity has nothing to say about
k which end some two-ended-counting-ordered
{0,...,k} beginning at 0

William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> >
> > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > Construct the largest number you can.
> It is an element of N_P. And it was an element of N_P before you "constructed" it.

Who told you, if you are the first to construct it? Nobody. Therefore you cannot be sure. N_P may be potentially infinite.

> N_P does not change.

If you are sure and right, then there are many, many elements of N_P which never will be used. That means they cannot not have been used at the end of times. That means they are dark.

Whatever you choose, the alternatives are potential infinity or dark numbers.

> N_P does not change. N_P is not something you have to "construct".

Let's assume you are right. Then N_P is something most elements of which cannot be used. They are dark.

Regards, WM

Jim Burns schrieb am Sonntag, 16. Januar 2022 um 19:18:56 UTC+1:
> On 1/16/2022 3:46 AM, WM wrote:
> > Jim Burns schrieb
> > am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
> >> There is no first smaller.
> >> For each smaller there are infinitely-many between
> >> -- which is why there is no first smaller.
> >
> > But not all can be chosen.
>
> All k which end some two-ended-counting-ordered
> {0,...,k} beginning at 0
> are before omega.

But not all can be chosen and used. Almost all will not have been used at the end of times. That means they cannot have been used. That means they are dark.

Regards, WM

On 1/16/2022 3:29 PM, WM wrote:
> Jim Burns schrieb am Sonntag, 16. Januar 2022 um 19:18:56 UTC+1:
>> On 1/16/2022 3:46 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:
>>>> There is no first smaller.
>>>> For each smaller there are infinitely-many between
>>>> -- which is why there is no first smaller.
>>>
>>> But not all can be chosen.
>>
>> All k which end some two-ended-counting-ordered
>> {0,...,k} beginning at 0
>> are before omega.
>
> But not all can be chosen and used. Almost all will not have been used at the end of times. That means they cannot have been used. That means they are dark.
>
> Regards, WM

No, you are using "Observer Dependent" Math, which is totally unworkable.

Your chooser is some external person, you require time for your math operations, two more flaws.

On 1/16/2022 3:24 PM, WM wrote:
> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
>> On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
>>> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>>>
>>>> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
>>> Construct the largest number you can.
>> It is an element of N_P. And it was an element of N_P before you "constructed" it.
>
> Who told you, if you are the first to construct it? Nobody. Therefore you cannot be sure. N_P may be potentially infinite.

Observer Dependent math again. Fail.

>
>> N_P does not change.
>
> If you are sure and right, then there are many, many elements of N_P which never will be used. That means they cannot not have been used at the end of times. That means they are dark.

Used Numbers ? Yuck! (same with used Ants) There is no end of time.

>
> Whatever you choose, the alternatives are potential infinity or dark numbers.

false choice.

>
>> N_P does not change. N_P is not something you have to "construct".
>
> Let's assume you are right. Then N_P is something most elements of which cannot be used. They are dark.

your assumption, not fact.

>
> Regards, WM

On 1/16/2022 4:29 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 16. Januar 2022 um 19:18:56 UTC+1:
>> On 1/16/2022 3:46 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Sonntag, 16. Januar 2022 um 02:16:18 UTC+1:

>>>> There is no first smaller.
>>>> For each smaller there are infinitely-many between
>>>> -- which is why there is no first smaller.
>>>
>>> But not all can be chosen.
>>
>> All k which end some two-ended-counting-ordered
>> {0,...,k} beginning at 0
>> are before omega.
>
> But not all can be chosen and used.
> Almost all will not have been used at the end of times.
> That means they cannot have been used.
> That means they are dark.

We have a technique for reasoning about infinitely-many
individuals. Start with some claim we know is true of
each individual of the infinitely-many. Take only
reliable statement-steps from there.

Which infinitely-many? Which starting-claims?
The answers to these two questions are tightly linked.

We can make a claim about one of infinitely-many
right triangles that one of its angles is 90 degrees,
without knowing which right triangle our claim is
about.

We _can'_t make that claim about one of infinitely-many
_triangles_ not without knowing more. Some have a
90 degree angle, but some do not.

You tell us that "k has been chosen and used" is not
a claim that can be made for each one of the k which
end some two-ended-counting-ordered {0,...,k} which
begins at 0.

Okay, then. "k has been chosen and used" is not the
sort of starting-claim which we should use to reason
about one of the k which end some two-ended-counting-
-ordered {0,...,k} which begins at 0.

Since we wouldn't do that anyway, it's very easy
to follow that recommendation.

What else is there to say about this?
You got what you wanted didn't you?

On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > >
> > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > Construct the largest number you can.
> > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> Who told you

It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,

--
William Hughes

söndag 16 januari 2022 kl. 22:24:29 UTC+1 skrev WM:
> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > >
> > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > Construct the largest number you can.
> > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> Who told you, if you are the first to construct it? Nobody. Therefore you cannot be sure. N_P may be potentially infinite.
>
> > N_P does not change.
>
> If you are sure and right, then there are many, many elements of N_P which never will be used. That means they cannot not have been used at the end of times. That means they are dark.
>
> Whatever you choose, the alternatives are potential infinity or dark numbers.
> > N_P does not change. N_P is not something you have to "construct".
> Let's assume you are right. Then N_P is something most elements of which cannot be used. They are dark.
>
> Regards, WM
your "dark" is irrelevant here and serves no purpose.

Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
> On 1/16/2022 4:29 PM, WM wrote:

> > But not all can be chosen and used.
> > Almost all will not have been used at the end of times.
> > That means they cannot have been used.
> > That means they are dark.

> We have a technique for reasoning about infinitely-many
> individuals.

Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.

Regards, WM

William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
> On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> > William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > > >
> > > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > > Construct the largest number you can.
> > > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> > Who told you
> It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,

By induction we can show that almost all elements are dark and will never have been used:
∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
That is tantamount to they cannot be used. They can only be used collectively:
|ℕ \ {1, 2, 3, ...}| = 0 .

Regards, WM

On Monday, 17 January 2022 at 06:56:35 UTC-4, WM wrote:
> Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
> > We have a technique for reasoning about infinitely-many
> > individuals.
> Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.

Logic really is not your strong suit. There is no way to say of any particular number that it will never be used in the future. Any one of them can be reasoned about, any one of them is linkable by a chain of successor operations back to 0, and any one of them may turn out not only to be used, but to be very useful indeed. So stop your inane whining.

horand....@gmail.com schrieb am Montag, 17. Januar 2022 um 13:15:43 UTC+1:
> On Monday, 17 January 2022 at 06:56:35 UTC-4, WM wrote:
> > Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
> > > We have a technique for reasoning about infinitely-many
> > > individuals.
> > Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.
> There is no way to say of any particular number that it will never be used in the future.

Of course not. Thar one would have been used.

> Any one of them can be reasoned about,

Almost all cannot. Proof: Each used one has ℵo succesors, ℵo of which will be successors of every used number.

Regards, WM

On Monday, January 17, 2022 at 2:31:38 PM UTC+1, WM wrote:
> horand....@gmail.com schrieb am Montag, 17. Januar 2022 um 13:15:43 UTC+1:
> >
> > Any one of them can be reasoned about
> >
> Almost all cannot.

Which one has been left in this case:

An e IN: n + n = 2*n.

> Proof: Each [...] one has ℵo succesors,

Yeah: An e IN: card({m e IN : m > n}) = aleph_0.

> ℵo of which will be successors of every [...] number.

Mückenheim's rule?

An e IN: EM c IN (card(M) = aleph_0): [Am e M: n < m]

implies

EM c IN (card(M) = aleph_0): An e IN: [Am e M: n < m]

?

Fascinaring! Hint: This rule boils down to the following "principle":

An e IN Em e M: n < m

implies

Em e M: An e IN: n < m.

A truism in mückenmath.

On Monday, January 17, 2022 at 2:31:38 PM UTC+1, WM wrote:
> horand....@gmail.com schrieb am Montag, 17. Januar 2022 um 13:15:43 UTC+1:
> >
> > Any one of them can be reasoned about
> >
> Almost all cannot.

Which one has been left out in this case:

An e IN: n + n = 2*n

?

> Proof: Each [...] one has ℵo succesors,

Yeah: An e IN: card({m e IN : m > n}) = aleph_0.

> ℵo of which will be successors of every [...] number.

Mückenheim's rule?

An e IN: EM c IN (card(M) = aleph_0): [Am e M: n < m]

implies

EM c IN (card(M) = aleph_0): An e IN: [Am e M: n < m]

?

Fascinating! Hint: This rule boils down to the following "principle":

An e IN: Em e M: n < m

implies

Em e M: An e IN: n < m.

A truism in mückenmath.

On 1/17/2022 4:56 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
>> On 1/16/2022 4:29 PM, WM wrote:
>
>>> But not all can be chosen and used.
>>> Almost all will not have been used at the end of times.
>>> That means they cannot have been used.
>>> That means they are dark.
>
>> We have a technique for reasoning about infinitely-many
>> individuals.
>
> Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.
>
> Regards, WM

wrong. That is observer dependent math, Fail.

Any number may or can be used, which directly conflicts with your personal decision *they cannot be used*

On 1/17/2022 7:31 AM, WM wrote:
> horand....@gmail.com schrieb am Montag, 17. Januar 2022 um 13:15:43 UTC+1:
>> On Monday, 17 January 2022 at 06:56:35 UTC-4, WM wrote:
>>> Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
>>>> We have a technique for reasoning about infinitely-many
>>>> individuals.
>>> Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.
>> There is no way to say of any particular number that it will never be used in the future.
>
> Of course not. Thar one would have been used.
>
>> Any one of them can be reasoned about,
>
> Almost all cannot. Proof: Each used one has ℵo succesors, ℵo of which will be successors of every used number.

that is spoof, not a proof.

the number of successors has no bearing at all on weather a number is "used" (by an external observer)

You idea that numbers cannot be used, and therefore you can discount, remove, or turn them dark, is failure.

Numbers are not an extension of a persons personality

>
> Regards, WM
>
>

On 1/17/2022 4:59 AM, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
>> On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
>>> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
>>>> On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
>>>>> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>>>>>
>>>>>> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
>>>>> Construct the largest number you can.
>>>> It is an element of N_P. And it was an element of N_P before you "constructed" it.
>>> Who told you
>> It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
>
> By induction we can show that almost all elements are dark and will never have been used:
> ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

wrong. all that says is if you remove a finite set from an infinite set, you have an infinite set. Fail.

> That is tantamount to they cannot be used. They can only be used collectively:
> |ℕ \ {1, 2, 3, ...}| = 0 .

wrong. all that says is if you remove all the elements of a set there are no elements left. Fail.

>
> Regards, WM

On 1/17/2022 5:56 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
>> On 1/16/2022 4:29 PM, WM wrote:

>>> But not all can be chosen and used.
>>> Almost all will not have been used at the end of times.
>>> That means they cannot have been used.
>>> That means they are dark.
>
>> We have a technique for reasoning about infinitely-many
>> individuals.
>
> Yes, and we can reason that most of them will never have
> been used. That is tantamount to they cannot be used.

I continue...
Start with some claim we know is true of
each individual of the infinitely-many. Take only
reliable statement-steps from there.

However, the step you make from
| For each j, there are infinitely-many k > j
to
| There are infinitely-many k, for each j, k > j
is not a reliable statement-step.

By "reliable", I mean that, in any case in which
the first claim(s) is (are) true, the second
claim is true. (AKA "valid")

There are cases in which the second claim
(AKA "conclusion") is provably true. But the
quantifier shift is not used in the proof in those
cases.

There are cases in which the second claim is
provably false. It is because of these cases that
we cannot use a quantifier shift in the proof
even in case in which the second claim is true.
A reliable step cannot be otherwise: that's what
justifies the second claim.

An example of what looks like a quantifier shift
but isn't:
From:
| For each inductive set, there is an inductive set
| set which is contained in each of its inductive subsets.
To:
| There is an inductive set which, for each inductive
| set, is contained in each of its inductive subsets.

That's true, that's provable, but not by using a quantifier
shift. We use the fact that the intersection of inductive
sets is an inductive set.

An example showing why a quantifier shift is not reliable:
From:
| For each natural j, there is a natural k, k > j
To:
| There is NOT a natural j, for each natural k, k =< j

THIS step is reliable, because k > j xor k =< j

Also, this is why we can't justify a claim by
quantifier shift.

Also also, the "To" contradicts
| There are infinitely-many k, for each j, k > j

That's NOT a case of proofs contradicting each other.
There is one proof, ours, and one not-proof using a
quantifier shift, yours.