FromTheRafters schrieb am Freitag, 24. September 2021 um 16:01:20 UTC+2:
> After serious thinking WM wrote :

> > It is not difficult to prove you wrong. You claim that every natural number
> > has a decimal representation. Name the decimal representation of n. Fail.
> n.000...

Decimal means digits 0 to 9.

Regards, WM

zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:08:11 UTC+2:
> torsdag 23 september 2021 kl. 15:23:55 UTC+2 skrev WM:

> if I say n is a natural number, it is a natural number.

If you say you have proved the Riemann hypothesis, then you have proved it? If you say that set theory is mathematics, then it is mathematics? You need not say that you are a fool.

Regards, WM

Gus Gassmann schrieb am Donnerstag, 23. September 2021 um 22:17:43 UTC+2:
> On Thursday, 23 September 2021 at 16:59:32 UTC-3, WM wrote:
> > There is a reason why omega is called a limit ordinal, you know...

Yes, it is the confusion of potential and actual infinity. There is a reason why actual infinity is complete. There is no gap between natural numbers and omega.

> > There are definable natural numbers each of which is followed by infinitely many natural numbers. All natural numbers together are not followed by any natural number. Can you see a difference between "definable natural numbers" and "all natural numbers"? What causes this difference?

Of course no answer.

Regards, WM

FromTheRafters schrieb am Freitag, 24. September 2021 um 00:08:40 UTC+2:
> WM has brought this to us :
> > Gus Gassmann schrieb am Donnerstag, 23. September 2021 um 15:36:27 UTC+2:
> >> "Infinite" means that there is *no end* to the successor operation.
> >
> > Nevertheless there is an end to the natural numbers, namely omega. Like the
> > zero ends the sequence of unit fractions.
> Stop lying.

Start thinking.

> Each natural number is both definable and is "followed" by infinitely
> more definable natural nunbers.
> > All natural numbers together are not followed by any
> > natural number. Can you see a difference between "definable natural numbers"
> > and "all natural numbers"?

Consider all natural numbers which can be last of a FISON F(n). They are by definition the definable natnumbers.

∀n ∈ ℕ: ℕ \ F(n) =/= { }

When subtracting them from ℕ infinitely many numbers will remain.

Obviously collectively more can be subtracted, namely all:
ℕ \ ℕ = { }

What causes this difference?

> Your unfounded implication that the 'defined' ara a proper subset of
> the 'just plain old' naturals. they are not, you just have a half-baked
> notion that you want them to be.

I have the proof. What else causes the difference?

Regards, WM

zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:12:49 UTC+2:
>of course the set N is not followed by anything, it is a meaningless thing to ask because the successor function cannot take N in it as it is not defined for it!

It cannot take undefinable numbers. It can take every set of definable numbers.

> The natural numbers IN the set N however ALWAYS have successors,

Consider all natural numbers which can be last of a FISON F(n). They are by definition the definable natnumbers.

∀n ∈ ℕ: ℕ \ F(n) =/= { }

When subtracting them from ℕ infinitely many numbers will remain.

Obviously collectively more can be subtracted, namely all:
ℕ \ ℕ = { }

Regards, WM

Gus Gassmann schrieb am Donnerstag, 23. September 2021 um 22:07:38 UTC+2:
> On Thursday, 23 September 2021 at 16:49:25 UTC-3, WM wrote:
> > If two non-empty sets, A and B, have an empty intersection, then A has no element of B and B has no element of A. Then A and B are not inclusion monotonic.
> Perhaps it escaped your attention that A and B are but *TWO* sets.

That remains true for all sets. You seem to believe that as soon as a set of sets is infinite it may contain an orange orang utan.

> Two is a *FINITE* number, and the intersection of a finite number of inclusion ordered end segments was never at issue. It is equal to the smallest of these end segments, namely the one with the largest "base".

Every endsegment is infinite. That is large enough.

Regards, WM

On Friday, September 24, 2021 at 6:39:51 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Freitag, 24. September 2021 um 05:27:09 UTC+2:
> > On Thursday, September 23, 2021 at 10:08:15 PM UTC+2, WM wrote:
> >
> > Let n e IN \ {1}.
> >
> > > These things 1,2,3,4,5,... are just those which have [a] unique prime [factorization].
> >
> > Certainly 1 is not one of "these things". Btw. these things are the natural numbers > 1. n is one of them.
> >
> They have prime factorizations that can be known. n has not.

n does not need to have a KNOWN prime factorization.

Hint: The Schnirelmann constant, which certainly is a natural number, doesn't have a KNOWN prime factorization either, you silly idiot!

Jim Burns schrieb am Donnerstag, 23. September 2021 um 22:23:58 UTC+2:
> On 9/23/2021 3:38 PM, WM wrote:
>
> > Simplified it is this:
> > There are definable natural numbers
> > each of which is followed by infinitely many natural numbers.
> forall j, exists k, k > j
> > All natural numbers are not followed by any natural number.
> not exists j, forall k, k =< j

These are the correct facts. Why are they so?

> > Can you see a difference between "definable natural numbers"
> > and "all natural numbers"?

Of course: A definable natural number is the last element of a FISON and has infinitely many successors. All natural numbers including the successors of all definable numbers have no successors (because these are included).

> That's not a difference.

That's a huge difference!

Regards, WM

FromTheRafters schrieb am Donnerstag, 23. September 2021 um 23:45:14 UTC+2:
> WM wrote on 9/23/2021 :

> > Then A and B are not inclusion monotonic.
> I prefer my disjoint sets to be so.

Disjoint is correct. Inclusion monotonic sets are not disjoint.
Therefore they share at least one element unless at least one of them is empty.

Regards, WM

On Friday, September 24, 2021 at 6:43:02 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:08:11 UTC+2:
> > torsdag 23 september 2021 kl. 15:23:55 UTC+2 skrev WM:
> >
> > If

Zelos defines /n/ to be

> > a[n arbitrary] natural number, it is a natural number

in that [i.e. his] context.

Don't you get that you silly crank?

Serg io schrieb am Freitag, 24. September 2021 um 01:48:30 UTC+2:
>
> There is no element common to all endsegments.

But all have infinitely many elements in common with E(1). So they must be disjoint. Do you claim that?

Regards, WM

Greg Cunt schrieb am Freitag, 24. September 2021 um 05:14:15 UTC+2:
> On Thursday, September 23, 2021 at 2:57:01 PM UTC+2, WM wrote:

Proof: Let E be an infinite endsegment. It has infinitely many elements in common with all infinite endsegments by inclusion monotony. Otherwise there was a first endsegment having less elements in common with all. QED

> Proof: Let be E is an endsegment. Then there is a unique index i e IN such that E = E(i). Let n e IN be this index. Hence E = E(n), i.e. E = {m e IN : m >= n}. Since min({m e IN : m >= n}) = n we get that min(E) = n. Hence min(E) is E's index and E contains all natural numbers >= min(E) (and nothing else). Since E was an arbitrary endsegment, this holds for each and every endsegment, i.e. for ALL endsegments. qed

Contradiction!

Regards, WM

Greg Cunt schrieb am Freitag, 24. September 2021 um 05:16:58 UTC+2:
> On Thursday, September 23, 2021 at 9:49:25 PM UTC+2, WM wrote:
>
> > By definition: E(n) = (n, n+1, n+2, ...)
> Nope. E(n) = {n, n+1, n+2, ...}
>
> That's a BIG difference in the context set theory

But not in the context of mathematics.

Regards, WM

zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:10:55 UTC+2:
> all natural numbers are followed by infinitely many more natural numbers.

Then the former would not be all. More than all is possible in matheology?

Regards, WM

zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:10:55 UTC+2:

> No because the "proeprty" you state that "definable" means apply to all natural numbers.

Not to those which are emptying the set |N by subtracting them.

Regards, WM

On 9/24/2021 11:54 AM, WM wrote:
> Jim Burns schrieb am Donnerstag, 23. September 2021 um 22:23:58 UTC+2:
>> On 9/23/2021 3:38 PM, WM wrote:
>>
>>> Simplified it is this:
>>> There are definable natural numbers
>>> each of which is followed by infinitely many natural numbers.
>> forall j, exists k, k > j
>>> All natural numbers are not followed by any natural number.
>> not exists j, forall k, k =< j
>
> These are the correct facts. Why are they so?
>
>>> Can you see a difference between "definable natural numbers"
>>> and "all natural numbers"?
>
> Of course: A definable natural number is the last element of a FISON

any natural number can be the last element of a FISON.

FISON is diversion, red herring.

>and has infinitely many successors.

not in the set!

> All natural numbers including the successors of all definable numbers have no successors (because these are included).

All the Ants in the universe including their successors, have no successors.

>
>> That's not a difference.
>
> That's a huge difference!
>
> Regards, WM
>

On 9/24/2021 12:03 PM, Transfinity wrote:
> zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:10:55 UTC+2:
>> all natural numbers are followed by infinitely many more natural numbers.
>
> Then the former would not be all. More than all is possible in matheology?
>
> Regards, WM
>

More > All

On 9/24/2021 12:54 PM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 23. September 2021 um 22:23:58 UTC+2:
>> On 9/23/2021 3:38 PM, WM wrote:

>>> Simplified it is this:
>>> There are definable natural numbers
>>> each of which is followed by infinitely many natural numbers.
>>
>> forall j, exists k, k > j
>>
>>> All natural numbers are not followed by any natural number.
>>
>> not exists j, forall k, k =< j
>
> These are the correct facts. Why are they so?

1a and 1b contradict each other.

1a. j > k
1b. j =< k

2a and 2b contradict each other.

2a. for all j, [1b]
2b. exists j, not [1b]

3a and 3b contradict each other.

3a. for all k, [2b']
3b. exists k, not [2b']

When these facts are combined, we see that
3a' and 3b' contradict each other.

3a'. for all k, exists j, j > k
3b'. exists k, for all j, j =< k

3a'. (all members have successors)
3b'. (all these members together have a successor).

>>> Can you see a difference between "definable natural numbers"
>>> and "all natural numbers"?
>
> Of course:
> A definable natural number is the last element of a FISON

A natural number which cannot be counted to even in principle
is not a natural number.
Throw it in the recycle bin with the four-cornered triangles.

> and has infinitely many successors.

Each FISON is followed by infinitely-many FISONs.

> All natural numbers
> including the successors of all definable numbers
> have no successors (because these are included).

No, that's not the reason.
Here's the reason:

(j > k) xor (j =< k)

(for all j, [1b]) xor (exists j, not [1b])

(for all k, [2b']) xor (exists k, not [2b'])

When these facts are combined, we see that
3a' and 3b' contradict each other.

3a'. for all k, exists j, j > k
3b'. exists k, for all j, j =< k

3a'. (all members have successors)
3b'. (all these members together have a successor).

>> That's not a difference.
>
> That's a huge difference!

| for all k, exists j, j > k
and
| not exists k, for all j, j =< k

are not different. Even given an exclamation point.

WM formulated the question :
> FromTheRafters schrieb am Freitag, 24. September 2021 um 16:01:20 UTC+2:
>> After serious thinking WM wrote :
>
>>> It is not difficult to prove you wrong. You claim that every natural number
>>> has a decimal representation. Name the decimal representation of n. Fail.
>> n.000...
>
> Decimal means digits 0 to 9.

Yes, and the fractional part of the expansion uses zeros.

How about using nines:

(n-1).999...

for *another* CDE representation of the same exact number for n > zero.

WM was thinking very hard :
> zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:08:11 UTC+2:
>> torsdag 23 september 2021 kl. 15:23:55 UTC+2 skrev WM:
>
>> if I say n is a natural number, it is a natural number.
>
> If you say you have proved the Riemann hypothesis, then you have proved it?
> If you say that set theory is mathematics, then it is mathematics? You need
> not say that you are a fool.

You are unaware of what a statement is?

You are the fool.

On Friday, 24 September 2021 at 13:45:07 UTC-3, WM wrote:
> Gus Gassmann schrieb am Donnerstag, 23. September 2021 um 22:17:43 UTC+2:
> > On Thursday, 23 September 2021 at 16:59:32 UTC-3, WM wrote:
> > > There is a reason why omega is called a limit ordinal, you know...
> Yes, it is the confusion of potential and actual infinity. There is a reason why actual infinity is complete. There is no gap between natural numbers and omega.
> > > There are definable natural numbers each of which is followed by infinitely many natural numbers. All natural numbers together are not followed by any natural number. Can you see a difference between "definable natural numbers" and "all natural numbers"? What causes this difference?
> Of course no answer.

Cute! You snip what I write, and then you complain about not getting an answer. I repeat: Your statement above is gobbledygook, and until you clarify what the hell you mean, there is not going to come another answer from me. I will not try to second-guess your ambiguous use of natural language any longer. PFO.

On Friday, 24 September 2021 at 13:52:02 UTC-3, WM wrote:
> Gus Gassmann schrieb am Donnerstag, 23. September 2021 um 22:07:38 UTC+2:
> > On Thursday, 23 September 2021 at 16:49:25 UTC-3, WM wrote:
> > > If two non-empty sets, A and B, have an empty intersection, then A has no element of B and B has no element of A. Then A and B are not inclusion monotonic.
> > Perhaps it escaped your attention that A and B are but *TWO* sets.

> That remains true for all sets. You seem to believe that as soon as a set of sets is infinite it may contain an orange orang utan.

You do not even have a single clue of the issues. *You* seem to think that as long as an infinite collection of sets is monotone inclusive, it has to contain an "orange orang utan", as you so quaintly put it. To you there has to be something there that is *NOT* a definable integer.

> > Two is a *FINITE* number, and the intersection of a finite number of inclusion ordered end segments was never at issue. It is equal to the smallest of these end segments, namely the one with the largest "base".
> Every endsegment is infinite. That is large enough.

Nope, not even close. The intersection over infinitely many end segments is empty. Unless, of course, it contains an orang utan or some other weird creature springing from your demented hallucinations.

WM used his keyboard to write :
> FromTheRafters schrieb am Freitag, 24. September 2021 um 00:08:40 UTC+2:
>> WM has brought this to us :
>>> Gus Gassmann schrieb am Donnerstag, 23. September 2021 um 15:36:27 UTC+2:
>>>> "Infinite" means that there is *no end* to the successor operation.
>>>
>>> Nevertheless there is an end to the natural numbers, namely omega. Like the
>>> zero ends the sequence of unit fractions.
>> Stop lying.
>
> Start thinking.

Stop lying.

Zero is not in segment (0,1/n] with a one as numerator no natural
number as denominator will suffice.

[...]

> Consider all natural numbers which can be last of a FISON F(n). They are by
> definition the definable natnumbers.

Since each and every natural number has that property, it is the set of
natural numbers wearing party hats.

Snipped more lies.

Greg Cunt schrieb am Freitag, 24. September 2021 um 18:53:27 UTC+2:
> On Friday, September 24, 2021 at 6:39:51 PM UTC+2, WM wrote:
> > Greg Cunt schrieb am Freitag, 24. September 2021 um 05:27:09 UTC+2:
> > > On Thursday, September 23, 2021 at 10:08:15 PM UTC+2, WM wrote:
> > >
> > > Let n e IN \ {1}.
> > >
> > > > These things 1,2,3,4,5,... are just those which have [a] unique prime [factorization].
> > >
> > > Certainly 1 is not one of "these things". Btw. these things are the natural numbers > 1. n is one of them.
> > >
> > They have prime factorizations that can be known. n has not.
> n does not need to have a KNOWN prime factorization.
>
> Hint: The Schnirelmann constant, which certainly is a natural number,

Contrary to n which is no number with certainty.

Regards, WM

WM used his keyboard to write :
> FromTheRafters schrieb am Freitag, 24. September 2021 um 00:13:58 UTC+2:
>> After serious thinking WM wrote :
>
>>> I did not change. These things 1,2,3,4,5,... are just those which have
>>> unique prime decompositions.
>> Four is the only composite number listed explicitly, how do you
>> decompose that which is not composite?
>
> How do you raise a number to the power of 1?

Non sequitur. I'm just saying that natural numbers have a unique prime
factorization, but only "composite numbers" can have "decompositions".