fredag 13 januari 2023 kl. 13:26:13 UTC+1 skrev Eram semper recta:
> On Friday, 13 January 2023 at 13:34:41 UTC+2, zelos...@gmail.com wrote:
> > torsdag 12 januari 2023 kl. 14:30:14 UTC+1 skrev Eram semper recta:
> > > On Wednesday, 11 January 2023 at 12:33:16 UTC+2, zelos...@gmail.com wrote:
> > > > tisdag 10 januari 2023 kl. 08:08:28 UTC+1 skrev Eram semper recta:
> > > > > On Tuesday, 10 January 2023 at 08:08:42 UTC+2, zelos...@gmail.com wrote:
> > > > > > söndag 8 januari 2023 kl. 09:35:15 UTC+1 skrev WM:
> > > > > > > "It never happens that the (always finite number of) pathbunches split into 2^ℵo pathbunches. However, if we posit a time after an infinite number of steps, we can say that the split happened. This 'it never happens but it happened' is unique to processes with no last step. For a process with no last step, you cannot conclude something cannot have happened, because there is no step where it happens."
> > > > > > >
> > > > > > > "You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;"
> > > > > > >
> > > > > > > "At every level we have more lamettas than nodes in the binary tree. But through each of these nodes pass infinitely many paths. Maybe you can not distinguish them just by looking at the nodes but they do exist."
> > > > > > >
> > > > > > > "Unfortunately you have not provided a way of moving this from 'true for all finite' to 'true for infinite n', and induction won't take you from 'all finite n' to 'infinite n'."
> > > > > > >
> > > > > > > "If all nodes of a path have been deleted, that does not mean you ever deleted the path!!! You could delete all and only the finite paths from the tree, and that would delete all the nodes, yet you would never have deleted any infinite path, and they would all still exist in any case, regardless of what had been deleted!"
> > > > > > >
> > > > > > > Gruß, WM
> > > > > > your dark numbers do not exist as you cannot define them to be distinct from natural number.
> > > > > They don't exist anymore than your trailing "infinite 3s" do in 0..333...
> > > > >
> > > > > What a moron.
> > > >
> > > > False, those are valid and rigorously defined. His is not as he cannot even define it.
> > >
> > > No retard. They are identical. As for rigour, it passed away when morons like you accepted ZFC as formal beliefs.
> > They are not, One can be defined in First Order Logic, the second cannot.
> That's your two cents opinion which in reality is FALSE in every respect.

Nope, it is accurate. Cry me a river elsewhere :)

> >
> > ANd it is rigorous and ZFC is rigorous as well even if you are too retarded to understand it.
> If I were retarded like you, it is possible that I would have believed that I understand it. Chuckle. The fact that I am much smarter than you, tells me it is incomprehensible garbage.

You are not smarter than me, you are not even smarter than a potato!

You are so far below me in terms of intelligence and knowledge it is hilarious.

fredag 13 januari 2023 kl. 13:29:28 UTC+1 skrev Eram semper recta:
> On Friday, 13 January 2023 at 13:36:08 UTC+2, zelos...@gmail.com wrote:
> > fredag 13 januari 2023 kl. 11:19:58 UTC+1 skrev WM:
> > > Jim Burns schrieb am Donnerstag, 12. Januar 2023 um 20:33:01 UTC+1:
> > > > On 1/12/2023 3:58 AM, WM wrote:
> > >
> > > > > But the sequence (1/n) converges to 0 and
> > > > > does not cross it.
> > > > > Therefore it ends there or before.
> > > > The sequence ⟨1/n⟩ doesn't end.
> > > The sequence has ended befor 0.
> > > If you go from 0 to 1 in order to pick the first visible term you have already passed almost all its successors. Try to contradict me by finding a counter example, i.e., picking a term with less successors. But spare your silly proofs which have brain-damaged so many mathematicians.
> > >
> > > Regards, WM
> > THERE IS NO FUCKING END YOU RETARD!
> ZFC says there is. WM rejects ZFC and assumes it only to point out that you are wrong. Call yourself a retard because in reality you are!

Then his arguments are invalid, Retard

fredag 13 januari 2023 kl. 13:28:15 UTC+1 skrev Eram semper recta:
> On Friday, 13 January 2023 at 13:35:27 UTC+2, zelos...@gmail.com wrote:
> > torsdag 12 januari 2023 kl. 14:33:55 UTC+1 skrev Eram semper recta:
> > > On Tuesday, 10 January 2023 at 09:25:35 UTC+2, Eram semper recta wrote:
> > >
> > > Lots of dark 3s. 🤣🤣🤣🤣
> > >
> > > Proof:
> > > 1. There are ℵ dark numbers always as pointed out by WM.
> > > 2. For each 3 in the sequence 0.333..., there is an index by a natural number.
> > > 3. We cannot know the dark numbers that are indexes for ℵ threes, therefore those ℵ threes are dark threes.
> > > Q.E.D
> > >
> > >
> > > https://www.academia.edu/39981684/Proof_of_the_most_important_Number_theorem_that_mainstream_mathematics_academics_never_learned
> > Which is invalid because step 1 has not been proven but only assumed.
> Piffle. Step 1 is proved and established. Remember we are not assuming anything here except perhaps your bullshit ZFC "beliefs" which means that in the worst case, we are better off in every respect. 🤣🤣🤣🤣
>
> To deny that most of N cannot be known is to admit openly that you're a retard.
Nope, it is assumed, never once proven. He nor you can define what "dark" means without the set of "dark numbers" being empty.

On Monday, 16 January 2023 at 08:42:45 UTC+2, zelos...@gmail.com wrote:
> fredag 13 januari 2023 kl. 13:28:15 UTC+1 skrev Eram semper recta:
> > On Friday, 13 January 2023 at 13:35:27 UTC+2, zelos...@gmail.com wrote:
> > > torsdag 12 januari 2023 kl. 14:33:55 UTC+1 skrev Eram semper recta:
> > > > On Tuesday, 10 January 2023 at 09:25:35 UTC+2, Eram semper recta wrote:
> > > >
> > > > Lots of dark 3s. 🤣🤣🤣🤣
> > > >
> > > > Proof:
> > > > 1. There are ℵ dark numbers always as pointed out by WM.
> > > > 2. For each 3 in the sequence 0.333..., there is an index by a natural number.
> > > > 3. We cannot know the dark numbers that are indexes for ℵ threes, therefore those ℵ threes are dark threes.
> > > > Q.E.D
> > > >
> > > >
> > > > https://www.academia.edu/39981684/Proof_of_the_most_important_Number_theorem_that_mainstream_mathematics_academics_never_learned
> > > Which is invalid because step 1 has not been proven but only assumed.

🤣🤣🤣🤣 Oh, you mean the requirement of (d1d2d3...dn)/(10^n) which means that an equivalent fraction for any given p/q to be measurable in base 10?

Moron. That is proved in the statements that follow. Remember this?

https://www.academia.edu/48985883/The_Obelus_and_Division_in_Arithmetic

Your arse still hurting from that time when I last fucked your brains out, you retard.

> > Piffle. Step 1 is proved and established. Remember we are not assuming anything here except perhaps your bullshit ZFC "beliefs" which means that in the worst case, we are better off in every respect. 🤣🤣🤣🤣
> >
> > To deny that most of N cannot be known is to admit openly that you're a retard.
> Nope, it is assumed, never once proven.

Piffle. If anyone made the following claims about axiomatic beliefs:
1. Most of N cannot be known.
2. N contains all its elements.

The popular choice would be easy: Most of N cannot be known.

WM has proved it by showing a contradiction in set theory. The fact that you are a moron who cannot understand and/or refuses to understand (because even an idiot understands) does not mean WM is wrong in this regard. Chuckle..

Learn to clean your arse properly. Do you have bidets in Sweden? Next time I fuck you, I need your arse to be clean. 🤣🤣🤣🤣

Fritz Feldhase schrieb am Sonntag, 15. Januar 2023 um 15:51:43 UTC+1:
> On Sunday, January 15, 2023 at 12:38:01 PM UTC+1, WM wrote:
>
> > You can "prove" in set theory that every one can be picked.
>
> Please show us this proof.

Every set can be well-ordered. Every ne subset has a least element. To well-order a set allows to pick all elements which could act as least ones.

Regards, WM

William schrieb am Sonntag, 15. Januar 2023 um 16:20:03 UTC+1:
> On Sunday, January 15, 2023 at 7:45:37 AM UTC-4, WM wrote:
> >Try it.
>
> "Pickable" is just one of your many many many ways of saying "can be written down"
>
> Each subset of a Peano set of cardinality 1 has the property that it can be written down (Note this does not mean that a set of cardinality aleph_0 that can be written down).

True, because there are not ℵo singletons in the Peano set. Otherwise ℵo singletons could be writte down.
>
> When I "try it" I find that I either pass no elements, or pass a set of l elements with cardinality aleph_0 Since a Peano set does not change each of these elements is part of the set, i.e. I do not add any element so the set. I cannot "stop at the first element that can be written down".

But every element where you stop can be written down. Infinitely many repetitions however will never show ℵo elements because ℵo elements are always passed without stopping.

Regards, WM

Eram semper recta schrieb am Sonntag, 15. Januar 2023 um 17:22:45 UTC+1:
> On Sunday, 15 January 2023 at 13:38:01 UTC+2, WM wrote:

> > > 1. If we say x+1 is dark, then it suggests that we have discerned (identified) x.
> > No. If you say 17, then you have discerned a number. If you say x, you have not.
> 17 is not dark. Read carefully!

Numbers having FISONs or digits are not dark.

Regards, WM

Jim Burns schrieb am Sonntag, 15. Januar 2023 um 17:43:06 UTC+1:

> I have looked at your proofs.
> Over and over and over,
> your essential step is that
> you (WM) _assert_ that there are dark numbers.

No, I assert that matematics and logic are true.

Therefore an infinite sequence of infinite and inclusion-monotonic elements cannot have an empty intersection.

Therefore McDuck cannot get bankrupt.

Therefore the matrix
XOOO...
XOOO...
XOOO...
XOOO...
....
cannot be filled with X by simply shuffling them.

All "proofs" refuting this are not worth to be considered in detail.

Regards, WM

On 1/16/2023 9:11 AM, WM wrote:
> William schrieb am Sonntag, 15. Januar 2023 um 16:20:03 UTC+1:
>> On Sunday, January 15, 2023 at 7:45:37 AM UTC-4, WM wrote:
>>> Try it.
>>
>> "Pickable" is just one of your many many many ways of saying "can be written down"
>>
>> Each subset of a Peano set of cardinality 1 has the property that it can be written down (Note this does not mean that a set of cardinality aleph_0 that can be written down).
>
> True, because there are not ℵo singletons in the Peano set. Otherwise ℵo singletons could be writte down.

Singleton Ants
Writte Ants
No ℵo Ants

>>
>> When I "try it" I find that I either pass no elements, or pass a set of l elements with cardinality aleph_0 Since a Peano set does not change each of these elements is part of the set, i.e. I do not add any element so the set. I cannot "stop at the first element that can be written down".
>
> But every element where you stop can be written down.

of course, else you dont know where you stopped.

> Infinitely many repetitions

there are no repetitions.

> however will never show ℵo elements because ℵo elements are always passed without stopping.

who is counting ?

>
> Regards, WM

On 1/16/2023 9:19 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 15. Januar 2023 um 17:43:06 UTC+1:
>
>> I have looked at your proofs.
>> Over and over and over,
>> your essential step is that
>> you (WM) _assert_ that there are dark numbers.
>

<snip crap>

>
> Regards, WM

On 1/16/2023 9:07 AM, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 15. Januar 2023 um 15:51:43 UTC+1:
>> On Sunday, January 15, 2023 at 12:38:01 PM UTC+1, WM wrote:
>>
>>> You can "prove" in set theory that every one can be picked.
>>
>> Please show us this proof.
>
> Every set can be well-ordered. Every ne subset has a least element. To well-order a set allows to pick all elements which could act as least ones.

that is not a proof.

you mean "plucked", not "picked". "pick" is chosen but not removed, "plucked" is chosen and removed

>
> Regards, WM

On 1/16/2023 9:14 AM, WM wrote:
> Eram semper recta schrieb am Sonntag, 15. Januar 2023 um 17:22:45 UTC+1:
>> On Sunday, 15 January 2023 at 13:38:01 UTC+2, WM wrote:
>
>>>> 1. If we say x+1 is dark, then it suggests that we have discerned (identified) x.
>>> No. If you say 17, then you have discerned a number. If you say x, you have not.
>> 17 is not dark. Read carefully!
>
> Numbers having FISONs or digits are not dark.
>
> Regards, WM

Which FISONs is 17 in ?

On Monday, January 16, 2023 at 11:11:05 AM UTC-4, WM wrote:
> William schrieb am Sonntag, 15. Januar 2023 um 16:20:03 UTC+1:

>ℵo elements are always passed without stopping.

So any "try" is doomed to fail.

William schrieb am Montag, 16. Januar 2023 um 16:54:53 UTC+1:
> On Monday, January 16, 2023 at 11:11:05 AM UTC-4, WM wrote:
> > William schrieb am Sonntag, 15. Januar 2023 um 16:20:03 UTC+1:
>
>
> >ℵo elements are always passed without stopping.
> So any "try" is doomed to fail.

Yes, but it proves the existence of dark numbers - if actual infinity is assumed (otherwise there would be nothing) and if Cantor's axiom is assumed that every number has a point on the real line.

Regards, WM

On 1/16/2023 10:27 AM, WM wrote:
> William schrieb am Montag, 16. Januar 2023 um 16:54:53 UTC+1:
>> On Monday, January 16, 2023 at 11:11:05 AM UTC-4, WM wrote:
>>> William schrieb am Sonntag, 15. Januar 2023 um 16:20:03 UTC+1:
>>
>>
>>> ℵo elements are always passed without stopping.
>> So any "try" is doomed to fail.
>
> Yes, but it proves the existence of dark numbers - if actual infinity is assumed (otherwise there would be nothing) and if Cantor's axiom is assumed that every number has a point on the real line.
>
> Regards, WM

diversion, you switch from natural numbers to real numbers.

On Monday, January 16, 2023 at 12:27:05 PM UTC-4, WM wrote:
> William schrieb am Montag, 16. Januar 2023 um 16:54:53 UTC+1:
> > On Monday, January 16, 2023 at 11:11:05 AM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 15. Januar 2023 um 16:20:03 UTC+1:
> >
> >
> > >ℵo elements are always passed without stopping.
> > So any "try" is doomed to fail.
> Yes

WM: It is fact. Try it.

"trying" if failed, a Peano set is "potentially infinite" is not a fact.

A Peano set does not change so a Peano set is not "potentially infinite".

On Monday, January 16, 2023 at 11:19:31 AM UTC-4, WM wrote:
> an infinite sequence of infinite and inclusion-monotonic elements cannot have an empty intersection.

The fact that an infinite sequence of infinitie and inclution-monotonic elements can have an empty intersection is not a contradiction, it is a result you do not like (you find it repugnant to the nature of logic).

On 1/16/2023 10:19 AM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 15. Januar 2023 um 17:43:06 UTC+1:

>> I have looked at your proofs.
>> Over and over and over,
>> your essential step is that
>> you (WM) _assert_ that there are dark numbers.
>
> No, I assert that matematics and logic are true.

The results of mathematics and logic
can be seen to be in sequences of claims in which
each claim is not first possibly-false.

For any sequence,
if any claim is possibly-false
then some claim is first possibly-false,
thus
if each claim is not first possibly-false,
then each claim is not possibly-false.

Some claims can be seen to be
not first possibly-false.
For example,
if we can can see that
𝓅 and 𝓅⇒𝓆 precede 𝓆
then we can see that
𝓆 is either not first or not possibly-false
and 𝓆 is not-first-possibly-false.

The results of mathematics and logic,
in all-not-first-possibly-false sequences,
can be seen to be not possibly-false.

> Therefore an infinite sequence of
> infinite and inclusion-monotonic [sets]
> cannot have an empty intersection.

𝓐 = ⟨ 𝐴₁ 𝐴₂ 𝐴₃ ... ⟩
is an infinite sequence 𝓐 of infinite and
inclusion-monotonic sets.
Each 𝐴ⱼ is infinite.
Following sets are proper subsets.

Define the infinite sequence 𝓓 of
set differences, 𝐷ₖ = 𝐴ₖ\𝐴ₖ₊₁
𝓓 = ⟨ 𝐷₁ 𝐷₂ 𝐷₃ ... ⟩

𝓓 is all-non-empty and disjoint.

Define the infinite sequence 𝓔 of
end unions of differences, 𝐸ᵢ = ⋃[i ≤ k] 𝐷ₖ
𝓔 = ⟨ 𝐸₁ 𝐸₂ 𝐸₃ ... ⟩

Each 𝐸ᵢ is infinite, since it is
the union of infinitely-many disjoint 𝐷ₖ
Following sets are proper subsets.

Their intersection ⋂𝓔 is empty.

| Assume otherwise.
| Assume x is in ⋂𝓔
| | x is in each 𝐸ᵢ in 𝓔
| | However,
| x is in some set difference 𝐷ⱼ
|| Otherwise
|| x is in no end union of differences.
|| and not in all of them, in ⋂𝓔
| The 𝐷ₖ are disjoint
| x is not in ⋃[j+1 ≤ k] 𝐷ₖ = 𝐸ⱼ₊₁
| x is not in each 𝐸ᵢ in 𝓔
| Contradiction.

Therefore,
Their intersection ⋂𝓔 is empty.

> Therefore an infinite sequence of
> infinite and inclusion-monotonic [sets]
> cannot have an empty intersection.

No.
𝓔 = ⟨ 𝐸₁ 𝐸₂ 𝐸₃ ... ⟩
is an infinite sequence of
infinite and inclusion-monotonic sets.
𝓔 has an empty intersection.

𝓔 is a _counter-example_ to your claim.

> Therefore McDuck cannot get bankrupt.
>
> Therefore the matrix
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
> cannot be filled with X by simply shuffling them.
>
> All "proofs" refuting this are not worth to be
> considered in detail.

It is by the details that we know these results.

No claim in the sequence is first possibly-false.
If the details show us that's so,
we know the result is necessarily-true,
whatever anyone's feelings on the matter might be.

Please re-consider your position on "proofs".

It happens that WM formulated :
> William schrieb am Sonntag, 15. Januar 2023 um 16:20:03 UTC+1:
>> On Sunday, January 15, 2023 at 7:45:37 AM UTC-4, WM wrote:
>>> Try it.
>>
>> "Pickable" is just one of your many many many ways of saying "can be written
>> down"
>>
>> Each subset of a Peano set of cardinality 1 has the property that it can be
>> written down (Note this does not mean that a set of cardinality aleph_0 that
>> can be written down).
>
> True, because there are not ℵo singletons in the Peano set

How do you know, did you look at them each individually to check?

WM was thinking very hard :
> Eram semper recta schrieb am Sonntag, 15. Januar 2023 um 17:22:45 UTC+1:
>> On Sunday, 15 January 2023 at 13:38:01 UTC+2, WM wrote:
>
>>>> 1. If we say x+1 is dark, then it suggests that we have discerned
>>>> (identified) x.
>>> No. If you say 17, then you have discerned a number. If you say x, you have
>>> not.
>> 17 is not dark. Read carefully!
>
> Numbers having FISONs or digits are not dark.

So you've changed your mind about dark numbers then?

On 1/16/2023 1:27 PM, FromTheRafters wrote:
> It happens that WM formulated :
>> William schrieb am Sonntag, 15. Januar 2023 um 16:20:03 UTC+1:
>>> On Sunday, January 15, 2023 at 7:45:37 AM UTC-4, WM wrote:
>>>> Try it.
>>>
>>> "Pickable" is just one of your many many many ways of saying "can be written down"
>>> Each subset of a Peano set of cardinality 1 has the property that it can be written down (Note this does not mean that a set of cardinality aleph_0
>>> that can be written down).
>>
>> True, because there are not ℵo singletons in the Peano set
>
> How do you know, did you look at them each individually to check?

its the one with the O stickie still on it, someone forgot to take it off

On 1/16/2023 10:14 AM, WM wrote:

> Numbers having FISONs or digits are not dark.

Dark numbers defined to be
in ℕ_Cantor but not in ℕ_Peano

ℕ_Peano is the smallest set such that
for each FISON 𝐹
if k is in 𝐹 then k is in ℕ_Peano

ℕ_Cantor is the largest set such that,
for each inductive set 𝑀
if m is not in 𝑀 then m is not in ℕ_Cantor

Define ℕ_Peano and ℕ_Cantor so that,
for each FISON 𝐹 and each inductive 𝑀
𝐹 ⊆ ℕ_Peano ⊆ ℕ_Cantor ⊆ 𝑀

----
∀k ∈ ℕ_Peano: k⁺⁺ ∈ ℕ_Peano

| Assume k ∈ ℕ_Peano
| | k ∈ 𝐹 for some FISON 𝐹
|| otherwise, ℕ_Peano\{k} is smaller
| | Let 𝐹⌝ be the end of FISON 𝐹
| Let 𝐹⌝⁺⁺ be the successor of 𝐹⌝
| Let 𝐹⊕⟨𝐹⌝⁺⁺⟩ be F⌝⁺⁺ appended to 𝐹
| || For each split of 𝐹⊕⟨𝐹⌝⁺⁺⟩
|| some i is last-before
|| and i++ is first-after
| | 𝐹⊕⟨𝐹⌝⁺⁺⟩ is a FISON
| 𝐹⊕⟨𝐹⌝⁺⁺⟩ ⊆ N_Peano
| k⁺⁺ ∈ 𝐹⊕⟨𝐹⌝⁺⁺⟩
| | k⁺⁺ ∈ N_Peano

Therefore,
∀k ∈ ℕ_Peano: k⁺⁺ ∈ ℕ_Peano

ℕ_Peano is inductive.
ℕ_Peano ⊆ ℕ_Cantor ⊆ ℕ_Peano
ℕ_Peano = ℕ_Cantor = ℕ_Peano

By definition, dark numbers are
in ℕ_Cantor but not in ℕ_Peano
But ℕ_Peano = ℕ_Cantor

Therefore,
dark numbers do not exist

because,
for each split of 𝐹⊕⟨𝐹⌝⁺⁺⟩
some i is last-before
and i++ is first-after

On Monday, January 16, 2023 at 10:07:39 PM UTC+1, Jim Burns wrote:

> By definition, dark numbers are
> in ℕ_Cantor but not in ℕ_Peano

Well, if they are defined that way...

> But ℕ_Peano = ℕ_Cantor

Agree. :-)

(Sometimes - outside of WM's psychotic kingdom - just called "IN".)

> Therefore,
> dark numbers do not exist

Indeed! Didn't Mückenheim claim to adhere to logic?

On 1/16/2023 4:28 PM, Fritz Feldhase wrote:
> On Monday, January 16, 2023
> at 10:07:39 PM UTC+1, Jim Burns wrote:

>> Therefore,
>> dark numbers do not exist
>
> Indeed!
> Didn't Mückenheim claim to adhere to logic?

My theory is that
WM believes that
we are bluffing about the existence of
all-not-first-possibly-false sequences of claims
== proofs,
and so he thinks his bluffing is "in return".

His way of ignoring proofs makes it much easier
for him to keep on believing that we're bluffing.

I wouldn't recommend posting the way I post
to anyone bothered by wasting their time.

On Monday, January 16, 2023 at 10:45:10 PM UTC+1, Jim Burns wrote:

> My theory is that
> WM believes that
> we are bluffing about the existence of
> all-not-first-possibly-false sequences of claims
> == proofs,
> and so he thinks his bluffing is "in return".
>
> His way of ignoring proofs makes it much easier
> for him to keep on believing that we're bluffing.

I see. But imho (I really believe this) he's suffering from a psychiatric condition.

Set theory **is [must be] wrong**. Period. No matter what you or anyone else says.

So you are trying to "discuss" with a lunatic who MUST believe that set theory "is wrong", For him that's an unshakeable truth. (The idea that it could be otherwise means pure horror to him. Hence he avoids that idea _at all_ costs.)

> I wouldn't recommend posting the way I post
> to anyone bothered by wasting their time.

Agree. :-P