On Monday, 18 July 2022 at 18:42:57 UTC-3, WM wrote:
[...]

> But we know also that never an exchange of X and O will change the relative shares or delete any O.

The X's and the O's have the same cardinality (namely aleph_0) at the beginning and throughout. That you are too stupid, conceited and deluded to understand the significance of that makes *YOU* look like an idiot, and nobody else.

WM formulated the question :
> Fritz Feldhase schrieb am Montag, 18. Juli 2022 um 21:10:56 UTC+2:
>> On Monday, July 18, 2022 at 7:47:34 PM UTC+2, WM wrote:
>>
>>> Removing them leaves nothing.
>>> Removing them leaves infinitely many succesors.
>> See: https://en.wikipedia.org/wiki/Equivocation
>
>> 1. IN \ {n : n e IN} = {}
>> 2. An e IN: IN \ {1, ..., n} is infinite.
>>
>> Note that in mathematics the following "rule of derivation" is not valid: An
>> e IN: Phi({1, ..., n}) implies Phi(IN). Hence 1. and 2. do not "form" a
>> "contradiction" in mathematics.
>
> Note that removing all elements of a set is same as removing the set.

Do good sets go to set heaven when they are removed?

Why not speak of sets which you create rather than some notion of
change to an already existing set?

Gus Gassmann schrieb am Dienstag, 19. Juli 2022 um 01:53:25 UTC+2:
> On Monday, 18 July 2022 at 18:42:57 UTC-3, WM wrote:
> [...]
> > But we know also that never an exchange of X and O will change the relative shares or delete any O.
> The X's and the O's have the same cardinality

Cardinality is a fool's notion, as my proof shows.

Regards, WM

On 7/19/2022 6:04 AM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 19. Juli 2022 um 01:53:25 UTC+2:
>> On Monday, 18 July 2022 at 18:42:57 UTC-3, WM wrote:
>> [...]
>>> But we know also that never an exchange of X and O will change the relative shares or delete any O.
>> The X's and the O's have the same cardinality
>
> Cardinality is a fool's notion, as my proof shows.
>
> Regards, WM
>
>

your "proof" is a fool's notion, it is a spoof.

On Tuesday, 19 July 2022 at 08:04:24 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 19. Juli 2022 um 01:53:25 UTC+2:
> > On Monday, 18 July 2022 at 18:42:57 UTC-3, WM wrote:
> > [...]
> > > But we know also that never an exchange of X and O will change the relative shares or delete any O.
> > The X's and the O's have the same cardinality
> Cardinality is a fool's notion, as my proof shows.

HA! Good one, boss. You wouldn't get a mathematical concept if it came and bit you on the balls,

WM used his keyboard to write :
> Gus Gassmann schrieb am Dienstag, 19. Juli 2022 um 01:53:25 UTC+2:
>> On Monday, 18 July 2022 at 18:42:57 UTC-3, WM wrote:
>> [...]
>>> But we know also that never an exchange of X and O will change the relative
>>> shares or delete any O.
>> The X's and the O's have the same cardinality
>
> Cardinality is a fool's notion, as my proof shows.

Is the set of delusional numbers dark?

On Tuesday, July 19, 2022 at 1:04:24 PM UTC+2, WM wrote:

> ... as my proof shows.

Ha ha ha!!! Good one!

"Proof" --- looool!

On 7/19/2022 1:04 PM, Fritz Feldhase wrote:
> On Tuesday, July 19, 2022 at 1:04:24 PM UTC+2, WM wrote:
>
>> ... as my proof shows.
>
> Ha ha ha!!! Good one!
>
> "Proof" --- looool!

thats a type-o, must be Goof or Spoof

On 7/18/2022 10:55 AM, Tom Bola wrote:
> sergi o schrieb:
>
>> On 7/18/2022 9:22 AM, WM wrote:
>>> Gus Gassmann schrieb am Montag, 18. Juli 2022 um 16:16:00 UTC+2:
>>>> On Saturday, 16 July 2022 at 11:20:26 UTC-3, William wrote:
>>>> [...]
>>>>>> XOOOO...
>>>>>> XOOOO...
>>>>>> XOOOO...
>>>>>> XOOOO...
>>>>>> XOOOO...
>>>>>> ...
>>>>> Starting with this matrix does not work. So it is completely irrelevant to the question of whether M exists.
>>>>
>>>> I would argue that one can make it work
>>>
>>> Nonsense. As JG has remarked completely correct, in every line there are far more fractions than integer fractions.
>>>
>>> Regarsd, WM
>>
>>
>>
>> It is a pile of steaming poo, WM. no matter how you swapparoo your piles, it will always stink.
>>
>> an endorsement by JG of your math, underscores your intent of misleading people with bad math, as that is what he does.
>>
>>
>>
>> borrowed from Dan's post;
>>
>> Even at his advanced age (60+?), John Gabriel is STILL struggling with basic, elementary-school arithmetic. As he has repeatedly posted here:
>>
>> "There are no points on a line."
>> --April 12, 2021
>>
>> "Pi is NOT a number of ANY kind!"
>> --July 10, 2020
>>
>> "1/2 not equal to 2/4"
>> --October 22, 2017
>>
>> “1/3 does NOT mean 1 divided by 3 and never has meant that”
>> -- February 8, 2015
>>
>> "3 =< 4 is nonsense.”
>> --October 28, 2017
>>
>> "Zero is not a number."
>> -- Dec. 2, 2019
>>
>> "0 is not required at all in mathematics, just like negative numbers."
>> -- Jan. 4, 2017
>>
>> “There is no such thing as an empty set.”
>> --Oct. 4, 2019
>>
>> “3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
>> --Oct. 22, 2019
>>
>> No math genius our JG, though he actually lists his job title as “mathematician” at Linkedin.com. Apparently, they do not verify your credentials.
>>
>> Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10,
>> 2020) at https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ
>
> Today:
>
> |IN| is an invariable constant
> |{1, 2, 3, ...}| = | IN |
> |{0, 1, 2, 3, ...}| = | IN | + 1
> |{2, 4, 6, ...}| = | IN |/2
>
>

On 7/5/2022 9:22 PM, sergi o wrote:
> On 7/5/2022 4:12 PM, WM wrote:
>> William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:
>>> On Tuesday, July 5, 2022 at 5:18:06 PM UTC-3, WM wrote:
>>>> William schrieb am Dienstag, 5. Juli 2022 um 17:50:30 UTC+2:
>>>> visible numbers.
>>>
>>> "Visible" One of your many many many many ways of saying "can be written down".
>>
>> Has a FISON.
>
> "Visible Ants" are already on the ANT LIST
>
>>
>>>>> There are methods by which you can obtain the state.
>>>> What methods?
>>> Induction for one. (1 is not in the intersection. If n is not in the intersection then n+1 is not in the intersection. Hence the intersection is empty.)
>>
>> The intersection is empty. But every definable endsegment (every endsegment that you can choose, every endsegment that closes a finite set of
>> endsegments) contains almost all natural numbers:
>>
>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀.
>
>
> ∩{E(1), E(2), ..., E(k)} = E(k)
>
> that is all the above says
>
>>
>> Do you accept that?
>>
>>> Or contradiction (Assume the intersection is not empty. Contradiction. Hence the intersection is empty)
>
> your MISTAKE is believing the intersection of all endsegments is an endsegment, it is not.
>
>>
>> The intersection is empty. But all endsegments indexed by visible numbers (= numbers having FISONs) contain almost all natural numbers. (ℵ₀ is the
>> complement of every |FISON|.)
>>
>>>>> These methods do not depend on the existence of "dark elements".
>>>>>
>>>> Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?
>>> Yes. But this triviality just means that a stepwise process will not work.
>>
>> No, it means that this stepwise process works until the empty set is reached. Otherwise you should not accept it.
>
> Wrong. You have a progression of endsegments there is no way of reaching the empty set, as it is not an endsegment.
>
> Fools parade.
>
>>
>> Regards, WM
>
>

On 7/3/2022 12:43 PM, Jim Burns wrote:
> On 7/3/2022 10:23 AM, WM wrote:
>> Jim Burns schrieb
>> am Sonntag, 3. Juli 2022 um 00:10:53 UTC+2:
>>> On 7/2/2022 12:42 PM, WM wrote:
>
>>>> For all definable numbers we obtain
>>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>>>
>>> No, you don't obtain that.
>>>
>>> That definition is ill-formed.
>>
>> |∩{E(k) : k <10}| = ℵ₀
>> |∩{E(k) : k <100}| = ℵ₀
>> |∩{E(k) : k <10^10}| = ℵ₀
>> |∩{E(k) : k <10^10^10}| = ℵ₀
>> and so on for all definable numbers k.
>>
>> Can you find an exception?
>
> So, you want to say that ℕ_def is
> the collection of all m such that
> |∩{E(k) : k < m }| = ℵ₀
>
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> is not how one says that.
> How one says that (in one way) is
> ℕ_def =
> { m ∈ ℕ : |∩{E(k) : k < m }| = ℵ₀ }
>
>>> If we correct your definition to what you say,
>>> then you obtain |∩{E(k) : k ∈ ℕ_def}| = 0
>>
>> That is no correction but a misleading change,
>> because I use only such endsegments which satisfy
>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀.
>
> The intersection of all the end segments
> _which you intend to use_ is empty.
>
> {E(k) : k <10} does not have all the end segments.
> {E(k) : k <100} does not have all the end segments.
> {E(k) : k <10^10} does not have all the end segments.
> {E(k) : k <10^10^10} does not have all the end segments.
>
> The intersection of _all_ the end segments is empty.
> |⋂{ E(j) : | j ∈ {m∈ℕ:|∩{E(k):k<m}|=ℵ₀} )| = 0
>
>>> As a consequence,
>>> each element of ℕ can be reached by
>>> a terminating process of '+1' steps.
>>
>> Then try to reach elements with less than
>> ℵ₀ successors.
>
> Each element of ℕ has |ℕ|-many successors in ℕ.
>
> For each end segment,
> there are |ℕ|-many element in common with
> each _on the way_ to that end segment.
>
> Each element of ℕ has fewer than |ℕ|-many
> predecessors in ℕ.
>
> Each element is only in end segments _on the way_
> to its end segment,
> which are fewer than all the end segments,
> so not-in some of them,
> so not-in the intersection of all.
>
> Infinity is not
> a reallyreallyreallyreallyreallyreally
> large number.
>
>

On 7/12/2022 10:22 AM, sergi o wrote:
> On 7/12/2022 10:05 AM, WM wrote:
>> William schrieb am Montag, 11. Juli 2022 um 20:08:47 UTC+2:
>>> On Monday, July 11, 2022 at 2:52:12 PM UTC-3, WM wrote:
>>>
>>>> Never a decrease to less than infinitely many elements happens.
>>> And never a step that takes you "closer" to the final state happens.
>>
>> Then it is impossible to index all elements of an infinite set like the fractions.
>
> Wrong. Cantor already proved it.
>
>> Indexing is a process that can be followed step by step for all definable numbers.
>
> Indexing is a process that is commonly use without any "step by step".
>
>
>> It cannot be followed to completion.
>
> well, it is you that cannot follow to completion.  google  Limit
>
>
>> Either there is no completion or the completion happens in dark numbers.
>
> Your reliance on your "dark numbers" underscores you lack of Algebra.
>
> <snip crap>
>
>> Regards, WM
>
>

On 7/12/2022 11:02 PM, sergi o wrote:
> On 7/12/2022 6:52 PM, FromTheRafters wrote:
>> Fritz Feldhase laid this down on his screen :
>>> On Wednesday, July 13, 2022 at 1:02:02 AM UTC+2, FromTheRafters wrote:
>>>> Fritz Feldhase brought next idea :
>>>>> On Wednesday, July 13, 2022 at 12:23:59 AM UTC+2, FromTheRafters wrote:
>>>>>> What is increasing and/or decreasing if not size?
>>>>>
>>>>> *sigh* These are technical terms which are used in the context of (certain) _set sequences_.
>>>>> See: https://proofwiki.org/wiki/Definition:Monotone_Sequence_of_Sets
>>>>> Your question is ... well ... reather "misleading" (at best). A (mis)"leading  question". :-)
>>>>> "Why can't we walk through IR? After all it's a field!"
>>>>> "What is irrational in connection with irrational numbers?"
>>>>> etc.
>>>>>
>>>> You could have let him answer it. I often ask such questions to see what he's thinking.
>>>
>>> Sure. But in this case -believe it or not- he knows what he's talking about.
>>>
>>> He's has been using the phrases "decreasing sequence of set"/"increasing sequence of sets" for years now - correctly.
>>
>> And when someone asks what *he* means by it, he can't explain. We say "That doesn't help your case" and he doesn't know what that means. He insists
>> that that type of 'decreasing' will 'exhaust' the set only if not taken step by step and as such would require Dark Numbers.
>
> its like the gas gauge on your car, when driving, it keeps decreasing, and burned gasoline goes out the exhaust, dark smoke is dark numbers, and you
> need a tuneup.
>
> When you car runs out of gas, it is exhausted, and you have arrived at your travel limit,  closer to aleph-null, but not really.
>

WM submitted this idea :
> FromTheRafters schrieb am Mittwoch, 13. Juli 2022 um 01:02:02 UTC+2:
>
>> You could have let him answer it. I often ask such questions to see
>> what he's thinking.
>
> Try to comprehend it if you can:
> The sequence
> {1, 2, 3, ...}
> {2, 3, 4, ...}
> {3, 4, 5, ...}
> ...
> {n, n+1, n+2, ...}
> ...
> has an infinite intersection, as long as every term is infinite.
>
> It has a finite intersection, if only the last natnumbers are remaining.
> Since they are dark, they have no order. Therefore I denote the last numbers
> 100 by D(100).
>
> {1, 2, 3, ..., D(100)}
> {2, 3, 4, ..., D(100)}
> {3, 4, 5, ..., D(100)}
> ...
> {n, n+1, n+2, ..., D(100)}
> ...
> {D(100)}
>
> The intersection is D(100) now. But even this cannot be calculated because
> infinitely many natnumbers are dark. We cannot discern the last 100.
> Nevertheless they are either present, or there are no endsegments at all.
>
> Regards, WM

The question was, what is *decreasing* about your proposed sequence?

On 8/8/2022 10:45 AM, FromTheRafters wrote:
> WM submitted this idea :
>> FromTheRafters schrieb am Mittwoch, 13. Juli 2022 um 01:02:02 UTC+2:
>>
>>> You could have let him answer it. I often ask such questions to see what he's thinking.
>>
>> Try to comprehend it if you can:
>> The sequence
>> {1, 2, 3, ...}
>> {2, 3, 4, ...}
>> {3, 4, 5, ...}
>> ...
>> {n, n+1, n+2, ...}
>> ...
>> has an infinite intersection, as long as every term is infinite.
>>
>> It has a finite intersection, if only the last natnumbers are remaining. Since they are dark, they have no order. Therefore I denote the last numbers
>> 100 by D(100).
>>
>> {1, 2, 3, ..., D(100)}
>> {2, 3, 4, ..., D(100)}
>> {3, 4, 5, ..., D(100)}
>> ...
>> {n, n+1, n+2, ..., D(100)}
>> ...
>> {D(100)}
>>
>> The intersection is D(100) now. But even this cannot be calculated because infinitely many natnumbers are dark. We cannot discern the last 100.
>> Nevertheless they are either present, or there are no endsegments at all.
>>
>> Regards, WM
>
> The question was, what is *decreasing* about your proposed sequence?

I cant figure out what the D(100) is for, is WM applying a LABEL to a Dark Number ?

Sergio formulated the question :
> On 8/8/2022 10:45 AM, FromTheRafters wrote:
>> WM submitted this idea :
>>> FromTheRafters schrieb am Mittwoch, 13. Juli 2022 um 01:02:02 UTC+2:
>>>
>>>> You could have let him answer it. I often ask such questions to see what
>>>> he's thinking.
>>>
>>> Try to comprehend it if you can:
>>> The sequence
>>> {1, 2, 3, ...}
>>> {2, 3, 4, ...}
>>> {3, 4, 5, ...}
>>> ...
>>> {n, n+1, n+2, ...}
>>> ...
>>> has an infinite intersection, as long as every term is infinite.
>>>
>>> It has a finite intersection, if only the last natnumbers are remaining.
>>> Since they are dark, they have no order. Therefore I denote the last
>>> numbers 100 by D(100).
>>>
>>> {1, 2, 3, ..., D(100)}
>>> {2, 3, 4, ..., D(100)}
>>> {3, 4, 5, ..., D(100)}
>>> ...
>>> {n, n+1, n+2, ..., D(100)}
>>> ...
>>> {D(100)}
>>>
>>> The intersection is D(100) now. But even this cannot be calculated because
>>> infinitely many natnumbers are dark. We cannot discern the last 100.
>>> Nevertheless they are either present, or there are no endsegments at all.
>>>
>>> Regards, WM
>>
>> The question was, what is *decreasing* about your proposed sequence?
>
> I cant figure out what the D(100) is for, is WM applying a LABEL to a Dark
> Number ?

I would like him to answer the question instead of deflecting.

Yes, he wants infinity to be mistaken for a very large
(largest/last/greatest) natural number so he can have the intersection
be non-empty.

He probably doesn't realize that we define things so that they work as
they do *including* infinity.

On Wednesday, August 10, 2022 at 1:10:31 AM UTC+2, Sergio wrote:
> On 8/8/2022 10:45 AM, FromTheRafters wrote:
> > WM submitted this idea :
> >> FromTheRafters schrieb am Mittwoch, 13. Juli 2022 um 01:02:02 UTC+2:
> >>
> >>> You could have let him answer it. I often ask such questions to see what he's thinking.
> >>
> >> Try to comprehend it if you can:
> >> The sequence
> >> {1, 2, 3, ...}
> >> {2, 3, 4, ...}
> >> {3, 4, 5, ...}
> >> ...
> >> {n, n+1, n+2, ...}
> >> ...
> >> has an infinite intersection, as long as every term is infinite.
> >>
> > > It has a finite intersection, if only the last natnumbers are remaining. Since they are dark, they have no order. Therefore I denote the last 100 numbers
> > > by D(100).
> >>
> >> {1, 2, 3, ..., D(100)}
> >> {2, 3, 4, ..., D(100)}
> >> {3, 4, 5, ..., D(100)}
> >> ...
> >> {n, n+1, n+2, ..., D(100)}
> >> ...
> >> {D(100)}
> >>
> >> The intersection is D(100) now. But even this cannot be calculated because infinitely many natnumbers are dark. We cannot discern the last 100.
> >> Nevertheless they are either present, or there are no endsegments at all.
> >>
> >> Regards, WM

> I cant figure out what the D(100) is for, is WM applying a LABEL to a Dark Number ?

You did't pay attention. He wrote:

"I denote the last 100 numbers by D(100)." [WM - word order corrected]

Since this last 100 (natural) numbers are dark, he wrote "D" (for dark).

Note, that this is a rather delicate matter: "We cannot discern the last 100. Nevertheless they are either present, or there are no endsegments at all.." [WM]

Remember: "[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

On 8/10/2022 7:40 AM, Fritz Feldhase wrote:
> On Wednesday, August 10, 2022 at 1:10:31 AM UTC+2, Sergio wrote:
>> On 8/8/2022 10:45 AM, FromTheRafters wrote:
>>> WM submitted this idea :
>>>> FromTheRafters schrieb am Mittwoch, 13. Juli 2022 um 01:02:02 UTC+2:
>>>>
>>>>> You could have let him answer it. I often ask such questions to see what he's thinking.
>>>>
>>>> Try to comprehend it if you can:
>>>> The sequence
>>>> {1, 2, 3, ...}
>>>> {2, 3, 4, ...}
>>>> {3, 4, 5, ...}
>>>> ...
>>>> {n, n+1, n+2, ...}
>>>> ...
>>>> has an infinite intersection, as long as every term is infinite.
>>>>
>>>> It has a finite intersection, if only the last natnumbers are remaining. Since they are dark, they have no order. Therefore I denote the last 100 numbers
>>>> by D(100).
>>>>
>>>> {1, 2, 3, ..., D(100)}
>>>> {2, 3, 4, ..., D(100)}
>>>> {3, 4, 5, ..., D(100)}
>>>> ...
>>>> {n, n+1, n+2, ..., D(100)}
>>>> ...
>>>> {D(100)}
>>>>
>>>> The intersection is D(100) now. But even this cannot be calculated because infinitely many natnumbers are dark. We cannot discern the last 100.
>>>> Nevertheless they are either present, or there are no endsegments at all.
>>>>
>>>> Regards, WM
>
>> I cant figure out what the D(100) is for, is WM applying a LABEL to a Dark Number ?
>
> You did't pay attention. He wrote:
>
> "I denote the last 100 numbers by D(100)." [WM - word order corrected]

yes, the D(100) Ant Tribe, nasty little critters, can't tell them apart, and they aren't there anyway.

>
> Since this last 100 (natural) numbers are dark, he wrote "D" (for dark).
>
> Note, that this is a rather delicate matter: "We cannot discern the last 100. Nevertheless they are either present, or there are no endsegments at all." [WM]

wonder what D(1) is ? Oh, its dark, unless someone instantates, or defines it. I mean it is very important! it is in ALL endsegments, and they cannot
exist without D(1), and since we know D(1) is included in D(2), we can use inclusion monotony to extend on down to some k, like 17 or so. IAW WMs
conclusions.

remap N using alpeh_0 - n = D(n) (?)
but none of this is proof, rather a goof an spoof..

>
> Remember: "[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

On 8/10/2022 8:54 AM, Sergio wrote:
> On 8/10/2022 7:40 AM, Fritz Feldhase wrote:
>> On Wednesday, August 10, 2022 at 1:10:31 AM UTC+2, Sergio wrote:
>>> On 8/8/2022 10:45 AM, FromTheRafters wrote:
>>>> WM submitted this idea :
>>>>> FromTheRafters schrieb am Mittwoch, 13. Juli 2022 um 01:02:02 UTC+2:
>>>>>
>>>>>> You could have let him answer it. I often ask such questions to see what he's thinking.
>>>>>
>>>>> Try to comprehend it if you can:
>>>>> The sequence
>>>>> {1, 2, 3, ...}
>>>>> {2, 3, 4, ...}
>>>>> {3, 4, 5, ...}
>>>>> ...
>>>>> {n, n+1, n+2, ...}
>>>>> ...
>>>>> has an infinite intersection, as long as every term is infinite.
>>>>>
>>>>> It has a finite intersection, if only the last natnumbers are remaining. Since they are dark, they have no order. Therefore I denote the last 100
>>>>> numbers
>>>>> by D(100).
>>>>>
>>>>> {1, 2, 3, ..., D(100)}
>>>>> {2, 3, 4, ..., D(100)}
>>>>> {3, 4, 5, ..., D(100)}
>>>>> ...
>>>>> {n, n+1, n+2, ..., D(100)}
>>>>> ...
>>>>> {D(100)}
>>>>>
>>>>> The intersection is D(100) now. But even this cannot be calculated because infinitely many natnumbers are dark. We cannot discern the last 100.
>>>>> Nevertheless they are either present, or there are no endsegments at all.
>>>>>
>>>>> Regards, WM
>>
>>> I cant figure out what the D(100) is for, is WM applying a LABEL to a Dark Number ?
>>
>> You did't pay attention. He wrote:
>>
>> "I denote the last 100 numbers by D(100)." [WM - word order corrected]
>
> yes, the D(100) Ant Tribe, nasty little critters, can't tell them apart, and they aren't there anyway.
>
>>
>> Since this last 100 (natural) numbers are dark, he wrote "D" (for dark).
>>
>> Note, that this is a rather delicate matter: "We cannot discern the last 100. Nevertheless they are either present, or there are no endsegments at
>> all." [WM]
>
> wonder what D(1) is ?   Oh, its dark, unless someone instantates, or defines it. I mean it is very important! it is in ALL endsegments, and they cannot
> exist without D(1), and since we know D(1) is included in D(2), we can use inclusion monotony to extend on down to some k, like 17 or so. IAW WMs
> conclusions.
>
> remap N using  alpeh_0 - n = D(n) (?)
> but none of this is proof, rather a goof an spoof..
>
>>
>> Remember: "[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental
>> misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."
>
>

On 7/14/2022 5:46 AM, Britt Robustelli wrote:
> zelos...@gmail.com wrote:
>
>>>> Not relevant to modern mathematics
>>> "modern mathematics" doesn't even exists, possibly, idiot. This
>>> imbecile doesn't know what mathematics is. You may say, for instance,
>>> "engineering", but not "modern mathematics". What an idiot.
>> Modern mathematics does exist because we do mathematics today in the
>> modern era.
>
> you just broke my bullshitmeter. Another one.
>

On 7/18/2022 8:13 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 17. Juli 2022 um 20:55:11 UTC+2:
>
>> Hence,
>> the same individuals have finitely-many predecessors
>> and infinitely-many successors.
>
> If there is a bijection, say natural and rationals, are there for every paired natnumber infinitely many not paired successors? Or are also all successors paired such that all paired natnumbers have no successors? Note that every paired natnumber can be named.
> If you define pi by an infinite digit sequence, are there for every digit infinitely many successors such that the value pi is never realized by digits?
>>
>> The step from
>> | each individual has infinitely-many (different) after
>> to
>> | infinitely many (same) are after each individual
>> is invalid.
>
> But the step from "every exchange of X and O in
> XOOOO...
> XOOOO...
> XOOOO...
> XOOOO...
> XOOOO...
> ...
> does not change the relative shares" to "all O's have disappeared" is valid?
>
> Regards, WM
?

On 7/18/2022 8:13 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 17. Juli 2022 um 20:55:11 UTC+2:
>
>> Hence,
>> the same individuals have finitely-many predecessors
>> and infinitely-many successors.
>
> If there is a bijection, say natural and rationals, are there for every paired natnumber infinitely many not paired successors? Or are also all successors paired such that all paired natnumbers have no successors? Note that every paired natnumber can be named.
> If you define pi by an infinite digit sequence, are there for every digit infinitely many successors such that the value pi is never realized by digits?
>>
>> The step from
>> | each individual has infinitely-many (different) after
>> to
>> | infinitely many (same) are after each individual
>> is invalid.
>
> But the step from "every exchange of X and O in
> XOOOO...
> XOOOO...
> XOOOO...
> XOOOO...
> XOOOO...
> ...
> does not change the relative shares" to "all O's have disappeared" is valid?
>
> Regards, WM
>

where are you putting the Os when your done with them ?

On 7/18/2022 8:13 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 17. Juli 2022 um 20:55:11 UTC+2:
>
>> Hence,
>> the same individuals have finitely-many predecessors
>> and infinitely-many successors.
>
> If there is a bijection, say natural and rationals, are there for every paired natnumber infinitely many not paired successors? Or are also all successors paired such that all paired natnumbers have no successors? Note that every paired natnumber can be named.
> If you define pi by an infinite digit sequence, are there for every digit infinitely many successors such that the value pi is never realized by digits?
>>
>> The step from
>> | each individual has infinitely-many (different) after
>> to
>> | infinitely many (same) are after each individual
>> is invalid.
>
> But the step from "every exchange of X and O in
> XOOOO...
> XOOOO...
> XOOOO...
> XOOOO...
> XOOOO...
> ...
> does not change the relative shares" to "all O's have disappeared" is valid?
>
> Regards, WM
>

WM & Sergi_o ██۞██ Moscow electric blackouts██۞██ Ben Bacarisse are the scopes Oval or Ellipse??

W. Mueckenheim 15 year long nonstop spam of sci.math too dumb to even admit slant cut of cone is Oval, not ellipse.
Sergio 137 "A new approach"
William & Ben Bacarisse 502 "Questions on Hilbert's..."

Chris Thomasson 486 "Questions on Hilbert's Hotel" was that hotel not blown up by Iran drones????
> 
> > Ben Bacarisse 292, "Three proofs of dark numbers..."
> > Can Ben Bacarisse, Fritz Feldhase & WM & Gottingen Uni build drones better than they can do math, for clearly they failed math.
> >
> > ██۞██ Moscow electric blackouts██۞██
> > Kibo Parry M. is he using bola or javelin short circuit weapons???
> > Kibo, why not just use drones, I was reading that France makes the worlds best drones.
> > Kibo, does bow and arrow with aluminum foil streamers tied to arrow, work better than javelin or bola???
> >
> > Fritz Feldhase 477 "Question on Hilbert's Hotel"
> >
> > Kibo was the Hotel hit by Iranian drone????
> > Prof. Cynthia A. Volkert,Metin Tolin,BenBacarisse,Anja Karliczek,Sabine Doering-Manteuffel,Prof. Dr. Angela Rizzi, are they forever going to be mindless idiots of math especially geometry. And is Ben B and WM keep spamming sci.math as a Gay pick up bar, is that what they want for Germany, Gottingen Univ, no math but gay pick up bar????
> > > > > > And here you can plainly see the side EC is smaller than CF, while a ellipse requires them to be equal
> > > > > > ......./\E
> > > > > > ...../ c.\
> > > > > > F / .......\
> > > > > >
> > > > > > entrance E of planar cut, and "c" the center axis, and F the exit of planar cut, cF is far larger than Ec.
> >
> > Is WM and his spam gang trying to turn sci.math into a gay pick up bar, using math to pick up??????
> >
> > Chris Thomasson 166 "Question on..."
> > Chris Thomasson 238 " Two similar...."
> > Ben Bacarisse 292, "Three proofs of dark numbers..."
> > WM 39 "Die 7 Erklarungen..."
> >
> > Ben Bacarisse 287 "Three proofs of dark numbers..."
> > > > > FromTheRafters 160"Two similar..." spamming jackarses who cannot admit slant cut of cone is Oval, never ellipse
> > > > > Sergi_o 150 "Two similar..."
> > > > > Ben Bacarisse 285 "Three proofs of dark numbers..."
> > > > > Ben Bacarisse 135, "Two similar properties..."
> > > > > >
> > > > > >
> > > > > >
> > > > > > WM turning Gottingen Germany into geometry failures.
> > > > > > Kibo chasing corpses in WM gay pick up bar, for Kibo sure does not know the difference between Oval and ellipse as seen in his latest stupidity-- a revolving axis as Kibo the moron and BWR describes it--
> > > > > >
> > > > > > Let's try again with your little diagram, fixed.
> > > > > > > > >>> .......A
> > > > > > > > >>> ....../.\E
> > > > > > > > >>> ...../.C.\
> > > > > > > > >>> ..../B....\
> > > > > > > > >>> .../.......\
> > > > > > > > >>> .F/....G....\
> > > > > > > >
> > ██۞██ Moscow electric blackouts██۞██
> > Kibo Parry M. is he using bola or javelin short circuit weapons???
> > Kibo, why not just use drones, I was reading that France makes the worlds best drones.
> > Kibo, does bow and arrow with aluminum foil streamers tied to arrow, work better than javelin or bola???
> > Kibo Parry CIA undercover
> > On Tuesday, December 20, 2022 at 10:26:03 PM UTC-6, Michael Moroney wrote:
> >
> > Kibo Parry Moroney in 1997 blows his CIA cover-- to the entire world, mind you---
> > Re: Archimedes Vanadium, America's most beloved poster
> > >> In article <5nefan$i06$9...@news.thecia.net> kibo greps <ki...@shell..thecia.net> writes:
> > > >
> >
> > > Moscow electric Blackouts██۞██
> >
> > James Parry dunce in sci.math & sci,physics
> > On Tuesday, December 20, 2022 at 11:49:40 AM UTC-6, Michael Moroney wrote:
> > > Pluto, so you finally tracked down James Parry in Moscow?
> > >
> > > But once again, that sentence no verb.
> > >
> > > Kibo WAS short circuiting Moscow electricity?
> > > Kibo WILL BE short circuiting Moscow electricity?
> > > Kibo PLANS ON short circuiting Moscow electricity?
> > > Kibo THINKS ABOUT short circuiting Moscow electricity?
> > >
> > > ..?
> > >
> > > Now that you found Kibo, will you stop pretending that I am Kibo?
> > > Hopefully he won't be using your aluminum pole idea, otherwise there
> > > will be an ex-Kibo. As in:
> > >
> > > Kibo WAS VAPORIZED WHILE short circuiting Moscow electricity.
> >
> > Kibo, is Xi aware of the fact that Outer Manchuria is 28 times bigger than Taiwan????
> > >
> > >
> > >
> > > > And as the Baby Xi grew up from the rice paddies and reeds of Outer Manchuria, stolen by the Naxi and Zani Dictator Putin in Moscow, Xi learned in school in chemical engineering that Taiwan was 1/28 the size of Outer Manchuria, as Putin bombs Ukraine. And the nascent Xi orders 1,000 divisions to the Outer Manchuria border to regain back the stolen Old China.
> >
> > > > > > > > Kibo in a javelin throw to knock out Moscow high voltage lines
> > > > > > > >> ........///// /////////////////////////////////////////////////////////////////////////|
> > > > > > > >> ......///// /////////////////////////////////////////////////////////////////////////..|
> > > > > > > >> ....///// /////////////////////////////////////////////////////////////////////////....| Moscow
> > > > > > > >> ..///// /////////////////////////////////////////////////////////////////////////......| Electric
> > > > > > > >> ///// /////////////////////////////////////////////////////////////////////////........| Tower
> > > > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|......../
> > > > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|....../
> > > > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|..../
> > > > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|../
> > > > > > > >> |__| |__|__|__| |__|__|__|__|__| |__|__|__|__|__| |__|__|__|__|__|/
> >
> >
> > > Vitaly a freedom fighter in Russia
> > ..............................^
> > ..............................||
> > ..............................||
> > .....................//////// ||
> > .....................|:---[.] ||
> > ......................|( _J ||
> > ....................| ^ ( _| ||
> > ................../ \_____)|| Kibo Parry with fake wig
> > ................. / _____ \|| Long Aluminum **javelin** to throw to short circuit
> > ..................... | / \ |...||
> > ..................... | | | |...||
> > ................. \ /\ /\__ | ||
> > ..................... | | \/--- ||
> > ...................... \ \ )....||
> > ................... | >____||/_____)
> > ................... \______||____/
> > ......................... / \
> > ...................... | | ~~
> > ..................... \ \\
> > .................... \ |\
> > ...................., \ | \
> > .................... \ | |
> > .................... | | |
> > .................... | | |
> > ..................... | / |
> > .................... |___/____|
> > .................... (____)____)
> > ==============================
> > freedom fighter from the tiresome dictator
> > Yea, the dictator has no power and no hot water, no heat, no electricity for his house, offices.

Gus Gassmann,Thorsten Theobald,Yury Person, Wolfgang Mueckenheim ever admit slant cut of cone is oval, never the ellipse, ever do a geometry proof of FTC, no all failures of mathematics, and idling away.

Germany's Muck the Puke taking up oxygen out of sci.math with his endless and mindless dark numbers, ellipse a conic when that is an oval, and the failure of all of Germany-- never a geometry proof of Fundamental Theorem of Calculus