On Thursday, January 13, 2022 at 6:46:15 AM UTC+1, WM wrote:

> When all its elements have distance ℵo from omega, then the set has distance ℵo from omega.

Please define "distance from "omega" (a) for the elements in IN (i. e. natural numbers)and (b) for IN. Then PROVE YOUR CLAIM, you silly crank.

Hint: We might accept the following definition for elements in IN (i. e. natural numbers):

dist_omega(n) = card({m e IN : n < m < omega}). [n e IN]

Then we get:

An e IN: dist_omega(n) = aleph_0.

So we may claim that "all elements of IN have distance ℵo from omega" is true.

Now please define DIST_OMEGA(M) for subsets M of IN *in a reasonable way*. Then show that DIST_OMEGA(IN) = aleph_0.

Hint: Note that the definition

DIST_OMEGA(M) = aleph_0. [M c IN]

isn't a _reasonable_ definition [since the value of DIST_OMEGA(M) does not depend on M].

On the other and, the following might be a reasonable definition:

DIST_OMEGA(M) = card({n e IN : Am e M: m < n < omega}). [M c IN]

But in this case DIST_OMEGA(IN) = 0.

Hint: There IS NO ordinal number x such that x is bigger than all natural numbers, but smaller than omega:

~Ex e ORD: An e IN: n < x < omega.

There is nothing (no ordinal number) between "the/all elements of IN" and omega.

On Thursday, January 13, 2022 at 1:46:15 AM UTC-4, WM wrote:
> William schrieb am Mittwoch, 12. Januar 2022 um 23:43:31 UTC+1:
belong to the subset N_P.
> > However, it remains true that if you consider an element of N_P you are considering something that is not larger than all the elements that you are considering.
> You can find a larger element in every case.

You do not have to "find" it. All the elements you are considering exist.

> But this larger element has same distance from omega.

Correct, however it remains true that if you consider an element of N_P you are considering something that is not larger than all the elements that you are considering.

--
William Hughes

On Thursday, 13 January 2022 at 08:51:10 UTC-4, Fritz Feldhase wrote:
> On Thursday, January 13, 2022 at 6:46:15 AM UTC+1, WM wrote:
>
> > When all its elements have distance ℵo from omega, then the set has distance ℵo from omega.
> Please define "distance from "omega" (a) for the elements in IN (i. e. natural numbers)and (b) for IN. Then PROVE YOUR CLAIM, you silly crank.
>
> Hint: We might accept the following definition for elements in IN (i. e. natural numbers):
>
> dist_omega(n) = card({m e IN : n < m < omega}). [n e IN]
>
> Then we get:
>
> An e IN: dist_omega(n) = aleph_0.
>
> So we may claim that "all elements of IN have distance ℵo from omega" is true.
>
> Now please define DIST_OMEGA(M) for subsets M of IN *in a reasonable way*.. Then show that DIST_OMEGA(IN) = aleph_0.
>
> Hint: Note that the definition
>
> DIST_OMEGA(M) = aleph_0. [M c IN]
>
> isn't a _reasonable_ definition [since the value of DIST_OMEGA(M) does not depend on M].
>
> On the other and, the following might be a reasonable definition:
>
> DIST_OMEGA(M) = card({n e IN : Am e M: m < n < omega}). [M c IN]
>
> But in this case DIST_OMEGA(IN) = 0.
>
> Hint: There IS NO ordinal number x such that x is bigger than all natural numbers, but smaller than omega:
>
> ~Ex e ORD: An e IN: n < x < omega.
>
> There is nothing (no ordinal number) between "the/all elements of IN" and omega.

I can see several approaches to this:

1. DIST_OMEGA(M) = aleph_0 for every M c IN.
You explicitly call that not interesting, and I concur.

2. DIST_OMEGA(M) = inf{dist_omega(n): n in M}
This is not particularly interesting, either, because DIST_OMEGA(M) = aleph_0 for every subset of IN (and, of course, IN itself) --- and it is equivalent to 1. Not much you can do with it, and it runs counter to my intuition. But it is using dist_omega, and it depends on M...

3. DIST_OMEGA(M) = aleph_0 if M is finite, 0 otherwise
What is not particularly nice about this definition is that DIST_OMEGA for sets has no connection to dist_omega for the elements of the set. On the other hand, WM is clearly not going to like it, so it has that going for it....

4. Various ways equivalent to your card(.) approach above, which will all reduce to 3.

Fritz Feldhase schrieb am Donnerstag, 13. Januar 2022 um 11:33:31 UTC+1:
> On Thursday, January 13, 2022 at 6:36:37 AM UTC+1, WM wrote:
>
> > It is the subset ℕ_def.
> Hint: IN_def = IN.

Hint: Since you are blind, you cannot see the difference.
>
> You know: An e IN: card({IN \ {1, 2, 3, ..., n}) = aleph_0.
>
> Hence IN_def = {n e IN : card({IN \ {1, 2, 3, ..., n}) = aleph_0} = IN. qed

If all n leave almost all elements of ℕ, what are these elements?

If ℕ is a fixed set, then all its elements could be removed. But all definable elements leave almost all elements of ℕ. Proof: You can't define more.

Regards, WM

FromTheRafters schrieb am Donnerstag, 13. Januar 2022 um 12:49:43 UTC+1:
> on 1/13/2022, WM supposed :

> > It is the subset ℕ_def.
> > In particular I am not changing the remaining ℵo elements of ℕ \ ℕ_def.
> You are not changing *anything*,

Adding or subtracting changes the set under consideration.

Regards, WM

William schrieb am Donnerstag, 13. Januar 2022 um 16:43:11 UTC+1:
> On Thursday, January 13, 2022 at 1:46:15 AM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 12. Januar 2022 um 23:43:31 UTC+1:
> belong to the subset N_P.
> > > However, it remains true that if you consider an element of N_P you are considering something that is not larger than all the elements that you are considering.
> > You can find a larger element in every case.
> You do not have to "find" it. All the elements you are considering exist.

But man elements cannot be considered although they exist.

> > But this larger element has same distance from omega.
> Correct, however it remains true that if you consider an element of N_P you are considering something that is not larger than all the elements that you are considering.

Yes. If I consider n then also n^n belongs to the set N_P. But for all we get
|ℕ \ {1, 2, 3, ..., n}| = ℵo
wheras
|ℕ \ {1, 2, 3, ...}| = 0 .

This is a difference.

Regards, WM

On 1/13/2022 12:41 PM, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 13. Januar 2022 um 11:33:31 UTC+1:
>> On Thursday, January 13, 2022 at 6:36:37 AM UTC+1, WM wrote:
>>
>>> It is the subset ℕ_def.
>> Hint: IN_def = IN.
>
> Hint: Since you are blind, you cannot see the difference.
>>
>> You know: An e IN: card({IN \ {1, 2, 3, ..., n}) = aleph_0.
>>
>> Hence IN_def = {n e IN : card({IN \ {1, 2, 3, ..., n}) = aleph_0} = IN. qed
>
> If all n leave almost all elements of ℕ, what are these elements?
>
> If ℕ is a fixed set, then all its elements could be removed. But all definable elements leave almost all elements of ℕ. Proof: You can't define more.
>
> Regards, WM

you fail to understand that sets are fixed.

If you add or delete an element, you have a different set, not the same set.

your rubber leaky sets are unworkable, and not math.

On 1/13/2022 12:44 PM, WM wrote:
> FromTheRafters schrieb am Donnerstag, 13. Januar 2022 um 12:49:43 UTC+1:
>> on 1/13/2022, WM supposed :
>
>>> It is the subset ℕ_def.
>>> In particular I am not changing the remaining ℵo elements of ℕ \ ℕ_def.
>> You are not changing *anything*,
>
> Adding or subtracting changes the set under consideration.
>

no. you have a different set.

you are trying to use sets as variables, fail.

you would know that if you understood proofs.

> Regards, WM

On Thursday, January 13, 2022 at 2:48:32 PM UTC-4, WM wrote:
> William schrieb am Donnerstag, 13. Januar 2022 um 16:43:11 UTC+1:
> > On Thursday, January 13, 2022 at 1:46:15 AM UTC-4, WM wrote:
> > > William schrieb am Mittwoch, 12. Januar 2022 um 23:43:31 UTC+1:

> > > > However, it remains true that if you consider an element of N_P you are considering something that is not larger than all the elements that you are considering.
> > > You can find a larger element in every case.
> > You do not have to "find" it. All the elements you are considering exist.
> But man[y] elements cannot be considered although they exist.

All elements of N_P can be considered. N_P is a Peano set. You can consider 1. If you can consider n you can consider n+1,
Therefore, by induction, you can consider any element of N_P

--
William Hughes

WM formulated on Thursday :
> Fritz Feldhase schrieb am Donnerstag, 13. Januar 2022 um 11:33:31 UTC+1:
>> On Thursday, January 13, 2022 at 6:36:37 AM UTC+1, WM wrote:
>>
>>> It is the subset ℕ_def.
>> Hint: IN_def = IN.
>
> Hint: Since you are blind, you cannot see the difference.
>>
>> You know: An e IN: card({IN \ {1, 2, 3, ..., n}) = aleph_0.
>>
>> Hence IN_def = {n e IN : card({IN \ {1, 2, 3, ..., n}) = aleph_0} = IN. qed
>
> If all n leave

Elements don't "leave" sets, but they can be used (considered) to make
new sets.

WM has brought this to us :
> FromTheRafters schrieb am Donnerstag, 13. Januar 2022 um 12:49:43 UTC+1:
>> on 1/13/2022, WM supposed :
>
>>> It is the subset ℕ_def.
>>> In particular I am not changing the remaining ℵo elements of ℕ \ ℕ_def.
>> You are not changing *anything*,
>
> Adding or subtracting changes the set under consideration.

No, it defines the set you are building (considering).

onsdag 12 januari 2022 kl. 21:57:28 UTC+1 skrev WM:
> FromTheRafters schrieb am Mittwoch, 12. Januar 2022 um 10:06:42 UTC+1:
>
> > Even if 'improper' it were not easiliy misunderstood, WM would still
> > have trouble with it. He sees no difference between a set and its
> > elements.
> There is a difference. The elements are many, the set is a single entity. But I dare to consider all elements without treating them as a set.
> All elements which have ℵo elements between themselves and omega can be collected. By collecting them, this distance will not be extinguished. Matheologians claim that as soon as the elements are collected into a set, the distance has vanished. That is nonsense.
>
> Regards, WM

"distance" is only relevant in metric spaces, not sets

torsdag 13 januari 2022 kl. 19:41:41 UTC+1 skrev WM:
> Fritz Feldhase schrieb am Donnerstag, 13. Januar 2022 um 11:33:31 UTC+1:
> > On Thursday, January 13, 2022 at 6:36:37 AM UTC+1, WM wrote:
> >
> > > It is the subset ℕ_def.
> > Hint: IN_def = IN.
> Hint: Since you are blind, you cannot see the difference.
> >
> > You know: An e IN: card({IN \ {1, 2, 3, ..., n}) = aleph_0.
> >
> > Hence IN_def = {n e IN : card({IN \ {1, 2, 3, ..., n}) = aleph_0} = IN. qed
> If all n leave almost all elements of ℕ, what are these elements?
>
> If ℕ is a fixed set, then all its elements could be removed. But all definable elements leave almost all elements of ℕ. Proof: You can't define more.
>
> Regards, WM
your "definable" is meaningless

William schrieb am Donnerstag, 13. Januar 2022 um 21:12:11 UTC+1:
> On Thursday, January 13, 2022 at 2:48:32 PM UTC-4, WM wrote:
> > William schrieb am Donnerstag, 13. Januar 2022 um 16:43:11 UTC+1:
> > > On Thursday, January 13, 2022 at 1:46:15 AM UTC-4, WM wrote:
> > > > William schrieb am Mittwoch, 12. Januar 2022 um 23:43:31 UTC+1:
>
> > > > > However, it remains true that if you consider an element of N_P you are considering something that is not larger than all the elements that you are considering.
> > > > You can find a larger element in every case.
> > > You do not have to "find" it. All the elements you are considering exist.
> > But man[y] elements cannot be considered although they exist.
>
> All elements of N_P can be considered. N_P is a Peano set. You can consider 1. If you can consider n you can consider n+1,
> Therefore, by induction, you can consider any element of N_P

which, also by induction, has ℵo successors.

Regards, WM

On Friday, 14 January 2022 at 08:05:12 UTC-4, WM wrote:
> William schrieb am Donnerstag, 13. Januar 2022 um 21:12:11 UTC+1:
> > On Thursday, January 13, 2022 at 2:48:32 PM UTC-4, WM wrote:
> > > William schrieb am Donnerstag, 13. Januar 2022 um 16:43:11 UTC+1:
> > > > On Thursday, January 13, 2022 at 1:46:15 AM UTC-4, WM wrote:
> > > > > William schrieb am Mittwoch, 12. Januar 2022 um 23:43:31 UTC+1:
> >
> > > > > > However, it remains true that if you consider an element of N_P you are considering something that is not larger than all the elements that you are considering.
> > > > > You can find a larger element in every case.
> > > > You do not have to "find" it. All the elements you are considering exist.
> > > But man[y] elements cannot be considered although they exist.
> >
> > All elements of N_P can be considered. N_P is a Peano set. You can consider 1. If you can consider n you can consider n+1,
> > Therefore, by induction, you can consider any element of N_P
> which, also by induction, has ℵo successors.

Each of which, yet again by induction, can be considered.

On Friday, January 14, 2022 at 8:05:12 AM UTC-4, WM wrote:
> William schrieb am Donnerstag, 13. Januar 2022 um 21:12:11 UTC+1:
> > On Thursday, January 13, 2022 at 2:48:32 PM UTC-4, WM wrote:

> > > But man[y] elements cannot be considered although they exist.
> >
> > All elements of N_P can be considered. N_P is a Peano set. You can consider 1. If you can consider n you can consider n+1,
> > Therefore, by induction, you can consider any element of N_P
> which, also by induction, has ℵo successors.

Correct and trivial. Every element of N_P is follow by an infinite number of elements of N_P (which means, of course, that N_P is infinite). Does not change the facts that any element of N_P can be considered and if you are considering an element of N_P you are considering something that is not larger than all the elements you are considering.

--
William Hughes

William schrieb am Freitag, 14. Januar 2022 um 16:46:01 UTC+1:
> On Friday, January 14, 2022 at 8:05:12 AM UTC-4, WM wrote:

> > > Therefore, by induction, you can consider any element of N_P
> > which, also by induction, has ℵo successors.
> Correct and trivial. Every element of N_P is follow by an infinite number of elements of N_P (which means, of course, that N_P is infinite). Does not change the facts that any element of N_P can be considered

No. But you cannot understand. You might understand that not every fraction can be indexed and from that, that dark numbers must exist. I see no point to continue. Please switch to the new thread if you like: The Methological Explosion.

Regards, WM

On 1/15/2022 8:50 AM, WM wrote:
> William schrieb am Freitag, 14. Januar 2022 um 16:46:01 UTC+1:
>> On Friday, January 14, 2022 at 8:05:12 AM UTC-4, WM wrote:
>
>>>> Therefore, by induction, you can consider any element of N_P
>>> which, also by induction, has ℵo successors.
>> Correct and trivial. Every element of N_P is follow by an infinite number of elements of N_P (which means, of course, that N_P is infinite). Does not change the facts that any element of N_P can be considered
>
> No. But you cannot understand. You might understand that not every fraction can be indexed

no, all fractions are indexed, which fraction cannot be indexed ?

> and from that, that dark numbers must exist. I see no point to continue.

ditto, and spread your falsehoods elsewhere. post in alt.fake.math

> Please switch to the new thread if you like: The Methological Explosion.
>
> Regards, WM

On Saturday, January 15, 2022 at 10:50:43 AM UTC-4, WM wrote:
> William schrieb am Freitag, 14. Januar 2022 um 16:46:01 UTC+1:
> > On Friday, January 14, 2022 at 8:05:12 AM UTC-4, WM wrote:
>
> > > > Therefore, by induction, you can consider any element of N_P
> > > which, also by induction, has ℵo successors.
> > Correct and trivial. Every element of N_P is follow by an infinite number of elements of N_P (which means, of course, that N_P is infinite). Does not change the facts that any element of N_P can be considered
> No.

Yes. The fact that every natural number has an infinite number of successors (which we know by the induction we have done are elements that can be considered does not change the fact that when you consider an element of N_P you are considering an element that is not larger than any of the elements you are considering..

--
William Hughes

On 1/15/2022 9:50 AM, WM wrote:

> You might understand that
> not every fraction can be indexed

Every fraction can be indexed.

A fraction n/d has numerator, denominator n,d
which are positive integers.

Because we know n,d are positive integers,
we know i = (n+d-1)*(n+d-2)/2 + n is a positive integer.

For positive integers n, d, where
| i = (n+d-1)*(n+d-2)/2 + n
we know i is unique.

For positive integer i, where
| s = floor( 3/2 + sqrt( 2*i - 3/4 ) )
| t = (s-1)*(s-2)/2
| n = i - t
| d = s - n
we know n,d are unique.

Eppur si muove.

> and from that, that dark numbers must exist.

Ex falso quodlibet.

Jim Burns has brought this to us :
> On 1/15/2022 9:50 AM, WM wrote:
>
>> You might understand that
>> not every fraction can be indexed
>
> Every fraction can be indexed.
>
> A fraction n/d has numerator, denominator n,d
> which are positive integers.

For positive rational fractions.

On 1/15/2022 3:00 PM, FromTheRafters wrote:
> Jim Burns has brought this to us :
>> On 1/15/2022 9:50 AM, WM wrote:

>>> You might understand that
>>> not every fraction can be indexed
>>
>> Every fraction can be indexed.
>>
>> A fraction n/d has numerator, denominator n,d
>> which are positive integers.
>
> For positive rational fractions.

You are right of course.

lördag 15 januari 2022 kl. 15:50:43 UTC+1 skrev WM:
> William schrieb am Freitag, 14. Januar 2022 um 16:46:01 UTC+1:
> > On Friday, January 14, 2022 at 8:05:12 AM UTC-4, WM wrote:
>
> > > > Therefore, by induction, you can consider any element of N_P
> > > which, also by induction, has ℵo successors.
> > Correct and trivial. Every element of N_P is follow by an infinite number of elements of N_P (which means, of course, that N_P is infinite). Does not change the facts that any element of N_P can be considered
> No. But you cannot understand. You might understand that not every fraction can be indexed and from that, that dark numbers must exist. I see no point to continue. Please switch to the new thread if you like: The Methological Explosion.
>
> Regards, WM
except we have a bijection between N and Q so you are wrong.