William schrieb am Dienstag, 17. August 2021 um 19:41:33 UTC+2:
> On Tuesday, August 17, 2021 at 8:06:37 AM UTC-4, WM wrote:
> > William schrieb am Montag, 16. August 2021 um 23:26:53 UTC+2:
>
> > > Correct, there is no *element of the set of FISONs* that is not finite, i.e. an *element of the set of FISONs* contains only enough elements to distinguish a finite number of FISONs. So what?
> > > The *set of FISONs* contains an infinite number of elements.
> > No. The set is a single object. Of course it has other properties than its elements. But it cannot increase their properties. All [elements of the set of FISONs] are finite by the pigeon hole principle.
>
> So what? No *element of the set of FISONs* is infinite (duh). Knowing that every *element of S* had property P does not mean *the set S* has property P.

But it means that all elements have that property.

> *The set of FISONs" is infinite.

Correct for potential infinity.
Easy to contradict for actual infinity.

Actual infinity meass larger than every natural number or larger than every FISON. The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.

Proof:
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Regards, WM

Jim Burns schrieb am Dienstag, 17. August 2021 um 20:11:24 UTC+2:
> On 8/17/2021 8:36 AM, WM wrote:

> >>> "No interval has all of its rationals indexed before
> >>> any other interval has all of its rationals indexed."
> >>> Jim Burns

> To review, I am answering your claim
> <WM>
> > Since in the natural order of |N, used by Cantor, never two
> > consecutive sets of cardinality ℵo can exist, this means that
> > up to this step only finitely many natural numbers can have
> > been applied. - For ℵo intervals! Therefore it is impossible
> > that Cantor's enumeration of the positive rationals is complete.
> </WM>
>
> There are no _consecutive_ sets of aleph_0 cardinality
> in the enumeration.

If in every interval at least one fractions is indexed, then aleph_0 indexes must have been supplied.
>
> _Up to_ any finite index, no more than finitely-many positive
> rationals in no more than finitely-many integer-ended intervals
> are indexed.

That is true. Therefore the rest cannot be distinguished by counting.
>
> Including _all_ the finite indexes and _all_ the positive
> rationals, infinitely-many positive rationals in infinitely-many
> integer-ended intervals are indexed (that is, all of them).

But not by a finite number of definable indexes?!

> > Would be nice whether you agreed too; or do you believe that
> > the aleph_0 unit intervals are completed by indexes which
> > can be distinguished?
> No integer-ended interval has a last-indexed rational
> in it.

Correct. Most are and remain without index.

> That is apparently what you mean by "completed".

Not at all. I explicitly excluded a last indexed rational. I only used the term completed by what you understand by completed.

> All of the rationals in all of the intervals are indexed.

That's just what I mean.
> These are different claims.

No, I mean exactly that. In every interval there is a first rational indexed. This requires at least aleph_0 rationals for aleph_0 intervals. Further all intervals are completed: "All of the rationals in all of the intervals are indexed." This requires another aleph_0 indexes. But we know that no interval is indexed before any other. We cannot count the indexes any longer. We cannot find out their order as we could with the first index per interval. They are dark.

Regards, WM

Sergio schrieb am Dienstag, 17. August 2021 um 20:23:24 UTC+2:
> On 8/17/2021 7:16 AM, WM wrote:

> > "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns
> each interval has an infinity of rationals in it, there is no "before", so his statement is true.

For each interval we can determine the first index issued to this interval. This requires at least as many indexes as there are intervals. Every interval is finished, i.e., completely indexed. But we cannot determine the order since no interval has all of its rationals indexed before any other interval has all of its rationals indexed.

This is a plain contrast. First we can count, later we cannot count.

Regards, WM

William schrieb am Dienstag, 17. August 2021 um 19:47:58 UTC+2:
> On Tuesday, August 17, 2021 at 8:31:42 AM UTC-4, WM wrote:

> Correct, and as no endsegment is not ordered, there is no last endsegment..
>
> because most are dark,
>
> Nope, no element of |N_F is dark (dark to you means more than just "cannot be written down")..

This is what I mean:

When indexing all positive rationals, then for each interval (k, k+1] we can determine the first index issued to this interval. This requires at least as many indexes as there are intervals: aleph_0. According to set theory every interval is finished, i.e., completely indexed. But we cannot determine the order how this happens since no interval has all of its rationals indexed before any other interval has all of its rationals indexed.

This is a plain contrast. First we can count, later we cannot count. That is what I call dark. It is indeed more than "cannot be written down".

Regards, WM

On Tuesday, August 17, 2021 at 3:50:50 PM UTC-4, WM wrote:

> The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.

Given any elementof the set of FISONs x, I can find an element of the set of FISONS k(x) such that k(x) > x. Thus for every element x of the set of FISONs the cardinality of the *set of FISONs* is greater than the cardinality of x.. So what if I cannot find a single element of the *set of FISONs* that works for every *element of the set of FISONs*?

--
William Hughes

On 8/17/2021 3:13 PM, WM wrote:
> William schrieb am Dienstag, 17. August 2021 um 19:47:58 UTC+2:
>> On Tuesday, August 17, 2021 at 8:31:42 AM UTC-4, WM wrote:
>
>> Correct, and as no endsegment is not ordered, there is no last endsegment.
>>
>> because most are dark,
>>
>> Nope, no element of |N_F is dark (dark to you means more than just "cannot be written down")..
>
> This is what I mean:
>
> When indexing all positive rationals,

the positive rationals are already indexed. you can find it online, there are a few different orderings.

> then for each interval (k, k+1] we can determine the first index issued to this interval.

so what ?

> This requires at least as many indexes as there are intervals: aleph_0.

wrong. You did not specifically state that k is infinite.

> According to set theory every interval is finished, i.e., completely indexed.

so you accept the fact that positive rationals are already indexed

> But we cannot determine the order

wrong. The order is known, as the rationals are indexed, go look it up online.

> how this happens since no interval has all of its rationals indexed before any other interval has all of its rationals indexed.

so what ? each interval has space for infinite number of rationals

>
> This is a plain contrast.

to What ?

> First we can count, later we cannot count.

Wrong. positive rationals are already counted, google for it.

> That is what I call dark.

I call your dark, Quack.

> It is indeed more than "cannot be written down".

nothing here at all.

>
> Regards, WM
>
>

On 8/17/2021 3:08 PM, WM wrote:
> Sergio schrieb am Dienstag, 17. August 2021 um 20:23:24 UTC+2:
>> On 8/17/2021 7:16 AM, WM wrote:
>
>>> "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns
>> each interval has an infinity of rationals in it, there is no "before", so his statement is true.
>
> For each interval we can determine the first index issued to this interval. This requires at least as many indexes as there are intervals. Every interval is finished, i.e., completely indexed. But we cannot determine the order since no interval has all of its rationals indexed before any other interval has all of its rationals indexed.
>
> This is a plain contrast. First we can count, later we cannot count.
>
> Regards, WM
>

Errors

1 "...we cannot determine the order...", wrong => the rationals are already indexed, it is online.
2. "First we can count, later we cannot count" wrong => you stop at k
3. Conclusion is wrong, based on wrong assumptions "This is a plain contrast"

> When indexing all positive rationals, then for each interval (k, k+1] we can determine the first index issued to this interval. This requires at least as many indexes as there are intervals: aleph_0. According to set theory every interval is finished, i.e., completely indexed. But we cannot determine the order how this happens

because the term "order" has no meaning here. Either no intervals are finished or all intervals are finished. There is no order in which the intervals are completed. Note that for an element of the set |N_F the term order does have meaning. This is true even for elements of |N_F that you cannot write down.

--
William Hughes

On Tuesday, 17 August 2021 at 16:50:50 UTC-3, WM wrote:
[...]
> Actual infinity meass larger than every natural number or larger than every FISON. The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.

Every day you are losing more of your cognitive faculties. Do you really not get that?

The set of FISONs has the same cardinality as the set of natural numbers, i..e., IN. This cardinality is aleph_0. Every natural number is finite, and still the *CARDINALITY* of the *SET* of natural numbers is aleph_0. Same with FISONs. Every Fison is a finite set, and yet the set of FISONs, namely {FISON(1), FISON(2), FISON(3), ...} has cardinality aleph_0.

How can you function with your brain fogged in like that???

On Tuesday, August 17, 2021 at 9:50:50 PM UTC+2, WM wrote:
> William schrieb am Dienstag, 17. August 2021 um 19:41:33 UTC+2:
> >
> > *The set of FISONs" is infinite.
> >
> Easy to contradict for actual infinity.

Lock, you blithering idiot, if the natural numbers are defined due to von Neuman (which is the standard approach in ZFC these days) the set of FISONs is IDENTICAL with IN.

Now IN -as you certainly know- is the "prototype" of an "actually infinite" set in ZFC. So, no, it's not "easy to contradict for actual infinity", you silly crank.

On Tuesday, August 17, 2021 at 9:50:50 PM UTC+2, WM wrote:

> The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.

It's getting worse with you, Mückenheim.

(a) The set of FISONs is larger than each and every FISON, just because each and every FISION is FINITE, while the set of FISONs is INFINITE. Morevoer we have AX e set_of_FISONs: X c set_of_FISONs [assuming that the natural numbers are defined due to von Neumann]. So the set of FISONs it's actually c-larger than each and every FISON, since each and every FISON is a proper subset of the set of FISONs.

(b) Indeed: "The set of FISONs does not contain anything larger than every FISON." That's another fascinating insight, Mückenheim! In other words, the set IN does not contain anything [which is] larger than every natural number. -- Incredible!

I mean, in symbols your claim is: ~Ey e IN: Ax e IN: x < y. You see: Ey e IN: Ax e IN: x < y could only hold if there were a natural number n such that n < n.

So your "argument" has the form: 1 + 2 = 4 because 0 =/= 1.

In short: It's nonsense.

tisdag 17 augusti 2021 kl. 14:18:12 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 17. August 2021 um 09:20:48 UTC+2:
> > >Of course. Every definable element is the last one of a finite initial segment. How could it be else?
> > Because it is so.
> >
> > If you define |∩{E(k) : k ∈ ℕ_def}| = ℵo by this property, then your N_def = N because N has the exact same property.
> No, only the definable endsegments have this property. Most however are undefinable. Like the natural numbers: "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns
>
> Regards, WM

You assume that N_def is finite and this is a fault assumption, all "definable", whatever that means, endsegments is still N

William schrieb am Dienstag, 17. August 2021 um 22:44:16 UTC+2:
> On Tuesday, August 17, 2021 at 3:50:50 PM UTC-4, WM wrote:
>
> > The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.
> Given any elementof the set of FISONs x, I can find an element of the set of FISONS k(x) such that k(x) > x.

Yes.

> Thus for every element x of the set of FISONs the cardinality of the *set of FISONs* is greater than the cardinality of x.

There is no cardinality of potentially infinite sets. Cardinality is a property of actually infinite sets, i.e. sets which are complete. You cannot find a natural number outside of |N. Otherwise nobody could claim a bijection between natural numbers and fractions. But you can always find a larger natural number than n.

> So what if I cannot find a single element of the *set of FISONs* that works for every *element of the set of FISONs*?

It does not exist. |N is the limit of the sequence of FISONs. But you will never reach it. Most of the distance will be dark forever.

Regards, WM

William schrieb am Dienstag, 17. August 2021 um 22:54:58 UTC+2:
> > When indexing all positive rationals, then for each interval (k, k+1] we can determine the first index issued to this interval. This requires at least as many indexes as there are intervals: aleph_0. According to set theory every interval is finished, i.e., completely indexed. But we cannot determine the order how this happens
> because the term "order" has no meaning here.

Why not? For every definable index it has meaning. If it has no meaning later, then there are no definable indices.

> Either no intervals are finished or all intervals are finished.

This cannot happen by one index. At least aleph_0 indices are required. But between finishing one interval and all intervals no order and no indices can be distinguished.

> There is no order in which the intervals are completed.

If they were completed by definable indices, then there would be an order. But there is none.

> Note that for an element of the set |N_F the term order does have meaning..

Of course!

> This is true even for elements of |N_F that you cannot write down.

That's why "dark" means more than not writable. If all rationals can be enumerated then the linearity of the set |N forces the completion of one interval first. This can only be veiled by dark indices (and rationals).

Regards, WM

Gus Gassmann schrieb am Dienstag, 17. August 2021 um 22:59:55 UTC+2:
> On Tuesday, 17 August 2021 at 16:50:50 UTC-3, WM wrote:
> [...]
> > Actual infinity meass larger than every natural number or larger than every FISON. The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.

> The set of FISONs has the same cardinality as the set of natural numbers, i.e., IN.

No.

> This cardinality is aleph_0. Every natural number is finite, and still the *CARDINALITY* of the *SET* of natural numbers is aleph_0. Same with FISONs. Every Fison is a finite set, and yet the set of FISONs, namely {FISON(1), FISON(2), FISON(3), ...} has cardinality aleph_0.

No.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 18. August 2021 um 07:36:05 UTC+2:
> tisdag 17 augusti 2021 kl. 14:18:12 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Dienstag, 17. August 2021 um 09:20:48 UTC+2:
> > > >Of course. Every definable element is the last one of a finite initial segment. How could it be else?
> > > Because it is so.
> > >
> > > If you define |∩{E(k) : k ∈ ℕ_def}| = ℵo by this property, then your N_def = N because N has the exact same property.
> > No, only the definable endsegments have this property. Most however are undefinable. Like the natural numbers: "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns

> You assume that N_def is finite and this is a fault assumption, all "definable", whatever that means, endsegments is still N

The definable natural numbers have a discernible well-ordering. Would they complete the i ndexing of the rationals, then we could find out which interval is completed first.

Regards, WM

Greg Cunt schrieb am Mittwoch, 18. August 2021 um 02:08:49 UTC+2:
> if the natural numbers are defined due to von Neuman (which is the standard approach in ZFC these days) the set of FISONs is IDENTICAL with IN.

ZFC is a big blunder. It is hard to understand that a hyper-intelligent person like on Neumann fell prey to this nonsense. Obviously some kind of herd instinct.
>
> Now IN -as you certainly know- is the "prototype" of an "actually infinite" set in ZFC. So, no, it's not "easy to contradict for actual infinity",

But here it has been achieved: The definable natural numbers have a discernible well-ordering. Would they complete the indexing of the rationals, then we could find out which interval was completed first.

Regards, WM

Greg Cunt schrieb am Mittwoch, 18. August 2021 um 02:27:05 UTC+2:
> On Tuesday, August 17, 2021 at 9:50:50 PM UTC+2, WM wrote:
> > The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.

> (a) The set of FISONs is larger than each and every FISON, just because each and every FISION is FINITE, while the set of FISONs is INFINITE.

No. You claimed once upon a time that the union of FISONs is larger than all FISONs. That is impossible. The set of FISONs is not a set but a potentially infinite collection. It has no fixed cardinality.

> (b) Indeed: "The set of FISONs does not contain anything larger than every FISON." That's another fascinating insight, Mückenheim! In other words, the set IN does not contain anything [which is] larger than every natural number. -- Incredible!

But it contains more than any definable natural number, i.e. a number which is last element of a FISON.

> So your "argument" has the form: 1 + 2 = 4 because 0 =/= 1.
>
> In short: It's nonsense.

On the contrary: Your argument is: "all natural numbers are definable". In fact almost all natural numbers are undefinable.

Regards, WM

On Wednesday, August 18, 2021 at 1:45:19 PM UTC+2, WM wrote:
> Gus Gassmann schrieb am Dienstag, 17. August 2021 um 22:59:55 UTC+2:
> >
> > The set of FISONs has the same cardinality as the set of natural numbers, i.e., IN.
> >
> No.

Yes.

Proof: n |-> {1, ..., n} (n e IN) is a bijection, and card(A) = card(B) iff A ~ B for any two sets A, B. qed

> > This cardinality is aleph_0.

Right.

Proof: n |-> {1, ..., n} (n e IN) is a bijection, card(A) = card(B) iff A ~ B for any two sets A, B, and card(IN) = aleph_0. qed

> > Every natural number is finite, and still the *CARDINALITY* of the *SET* of natural numbers is aleph_0. Same with FISONs. > > Every Fison is a finite set, and yet the set of FISONs, namely {FISON(1), FISON(2), FISON(3), ...} has cardinality aleph_0.
> >
> No.

Yes.

Look, you silly crank: There's a PROOF for this fact. (See proof above.)

On Wednesday, August 18, 2021 at 1:47:28 PM UTC+2, WM wrote:

> The [...] natural numbers [are] well-order[ed by <=]. Would they complete the indexing of the rationals, then we could find out which interval is completed first.

No, we "would't", since we can't.

Actually, ***no*** intervall "is completed first", you silly crank.

HOLY SHIT!!!

On Wednesday, August 18, 2021 at 1:59:25 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Mittwoch, 18. August 2021 um 02:27:05 UTC+2:
> >
> > (a) The set of FISONs is larger than each and every FISON, just because each and every FISION is FINITE, while the set of FISONs is INFINITE.
> >
> No.

Yes.

Once more:

1.) AF e Set_of_FISONs: card(F) < aleph_0
2.) card(Set_of_FISONs) = aleph_0

Hence: AF e Set_of_FISONs: card(F) < card(Set_of_FISONs).

> The set of FISONs is not a set

Errrr? Time to quite, Mückenheim. Really.

On Wednesday, 18 August 2021 at 08:45:19 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 17. August 2021 um 22:59:55 UTC+2:
> > On Tuesday, 17 August 2021 at 16:50:50 UTC-3, WM wrote:
> > [...]
> > > Actual infinity meass larger than every natural number or larger than every FISON. The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.
> > The set of FISONs has the same cardinality as the set of natural numbers, i.e., IN.
> No.

Yes. Your delusions are getting worse, and they are blinding you!

> > This cardinality is aleph_0. Every natural number is finite, and still the *CARDINALITY* of the *SET* of natural numbers is aleph_0. Same with FISONs. Every Fison is a finite set, and yet the set of FISONs, namely {FISON(1), FISON(2), FISON(3), ...} has cardinality aleph_0.
> No.

Yes. (Again.)

Greg Cunt schrieb am Mittwoch, 18. August 2021 um 15:03:01 UTC+2:
> On Wednesday, August 18, 2021 at 1:47:28 PM UTC+2, WM wrote:
>
> > The [...] natural numbers [are] well-order[ed by <=]. Would they complete the indexing of the rationals, then we could find out which interval is completed first.
>
> No, we "would't", since we can't.
>
> Actually, ***no*** intervall "is completed first",

If all intervals are completed, then one of them must be completed first because there is a linear sequence of natnumbers which are used one after the other, and one natnumber cannot complete more than one interval - if completion is possible at all.

Regards, WM

WM formulated the question :
> William schrieb am Dienstag, 17. August 2021 um 22:44:16 UTC+2:
>> On Tuesday, August 17, 2021 at 3:50:50 PM UTC-4, WM wrote:
>>
>>> The set of FISONs is not larger than every FISON because it does not
>>> contain anything larger than every FISON.
>> Given any elementof the set of FISONs x, I can find an element of the set of
>> FISONS k(x) such that k(x) > x.
>
> Yes.
>
>> Thus for every element x of the set of FISONs the cardinality of the *set of
>> FISONs* is greater than the cardinality of x.
>
> There is no cardinality of potentially infinite sets. Cardinality is a
> property of actually infinite sets, i.e. sets which are complete. You cannot
> find a natural number outside of |N. Otherwise nobody could claim a bijection
> between natural numbers and fractions. But you can always find a larger
> natural number than n.
>
>> So what if I cannot find a single element of the *set of FISONs* that works
>> for every *element of the set of FISONs*?
>
> It does not exist. |N is the limit of the sequence of FISONs.

The sequence of FISONs has a limit?

On Wednesday, August 18, 2021 at 4:12:35 PM UTC+2, WM wrote:

> If all intervals are completed, then <etc.>

Well, right, it's your claim that they "are completed" (whatever that means).

Fact: There's a bijection between IN and IN x IN.

See: https://en.wikipedia.org/wiki/Pairing_function