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From: Alan Mackenzie <acm@muc.de>
Newsgroups: sci.math
Subject: Re: The non-existence of "dark numbers"
Date: Thu, 13 Mar 2025 17:53:30 -0000 (UTC)
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WM <wolfgang.mueckenheim@tha.de> wrote:
> On 13.03.2025 13:59, Alan Mackenzie wrote:
>> WM <wolfgang.mueckenheim@tha.de> wrote:

>>> I know that it is self-contradictory because it cannot distinguish
>>> potential and actual infinity.

>> It can, but doesn't need to.  Potential and actual infinity are needle=
ss
>> concepts which only serve to confuse and obfuscate.  If you disagree,
>> feel free to cite a standard result in standard mathematics which depe=
nds
>> on these notions.

>>> When |=E2=84=95| \ |{1, 2, 3, ..., n}| =3D =E2=84=B5o, ....

>> The difference operator \ applies to sets, not to cardinal numbers.

> I know, but erroneously I had used the sets. I corrected that but=20
> without correcting the sign

It would aid communication enormously if you would use standard
mathematical symbols and words in the same way they are used by
mathematicians.

>>> .... then |=E2=84=95| \ |{1, 2, 3, ..., n+1}| =3D =E2=84=B5o. This ho=
lds for all elements
>>> of the inductive set, i.e., all FISONs F(n) or numbers n which have
>>> more successors than predecessors.

>> I.e. all natural numbers.

> No. All numbers can be subtracted from =E2=84=95 such that none remains=
:
> =E2=84=95 \ {1, 2, 3, ...} =3D { }, let alone =E2=84=B5o.

Yes.  {1, 2, 3, ...} is N, and trivially N \ N is the empty set.  What
are you trying to say with "let along aleph0"?

>>> Only those contribute to the inductive set!

>> The inductive set is all natural numbers.  Why must you make such a so=
ng
>> and dance about it?

> Because when only definable numbers are subtracted from =E2=84=95, ....

"Definable number" has not been defined by you, except in a sociological
sense.

> .... then =E2=88=80n =E2=88=88 =E2=84=95_def: |=E2=84=95 \ {1, 2, 3, ..=
.., n}| =3D =E2=84=B5o infinitely many
> numbers remain. That is the difference between dark and defiable
> numbers.

Rubbish!  It's just that the set difference between an infinite set and a
one of its finite subsets remains infinite.  That doesn't shed any light
on "dark" or "defi[n]able" numbers.

>>> Modern mathematics must claim that contrary to the definition =E2=84=B5=
o
>>> vanishes to 0 because =E2=84=95 \ {1, 2, 3, ...} =3D { }.  That is bl=
atantly
>>> wrong and shows that modern mathematicians believe in miracles.
>>> Matheology.

>> Modern mathematics need not and does not claim such a ridiculous thing=
..

> =E2=84=95 \ {1, 2, 3, ...} =3D { } is wrong?

Don't be obtuse.  It's the assertion you made in your previous paragraph
that is ridiculous.  The assertion that "aleph0 vanishes to 0".

>>>> You didn't point out any mistake in 3.  I doubt you can.

>>> I told you that potential infinity has no last element, therefore the=
re
>>> is no first dark number.

>> The second part of your sentence does not follow clearly from the firs=
t,
>> therefore the sentence is false.  And even if it were not false, it ha=
s
>> no bearing on my item 3.

> Try to think better. =E2=84=95_def is a subset of =E2=84=95. If =E2=84=95=
_def had a last=20
> element, the successor would be the first dark number.

If, if, if, ....  "N_def" remains undefined, so it is not sensible to
make assertions about it.  Whether or not it has a last element awaits
its definition.

>> But I can agree with you that there is no first "dark number".  That i=
s
>> what I have proven.  There is a theorem that every non-empty subset of
>> the natural numbers has a least member.

> That theorem is wrong in case of dark numbers.

That's a very bold claim.  Without further evidence, I think it's fair to
say you are simply mistaken here.

>>>>>>> Try to remove all numbers individually from the harmonic series s=
uch
>>>>>>> that none remains. If you can't, find the first one which resists=
..

>>>>>> Why should I want to do that?

>>>>> In order to experience that dark numbers exist and can't be manipul=
ated.

>>>> Dark numbers don't exist, as Jim and I have proven.

>>> When |=E2=84=95 \ {1, 2, 3, ..., n}| =3D =E2=84=B5o, then |=E2=84=95 =
\ {1, 2, 3, ..., n+1}| =3D
>>> =E2=84=B5o. How do the =E2=84=B5o dark numbers get visible?

There are no such things as "dark numbers", so talking about their
visibility is not sensible.

>> There is no such thing as a "dark number".  It's a figment of your
>> imagination and faulty intuition.

> Above we have an inductive definition of all elements which have=20
> infinitely many dark successors.

"Dark number" remains undefined, except in a sociological sense.  "Dark
successor" is likewise undefined.

>>>>> Induction cannot cover all natural numbers but only less than remai=
n
>>>>> uncovered.

>>>> The second part of that sentence is gibberish.  Nobody has been talk=
ing
>>>> about "uncovering" numbers, whatever that might mean.  Induction
>>>> encompasses all natural numbers.  Anything it doesn't cover is not a
>>>> natural number, by definition.

>>> Every defined number leaves =E2=84=B5o undefined numbers. Try to find=
 a
>>> counterexample. Fail.

>> What the heck are you talking about?  What does it even mean for a num=
ber
>> to "leave" a set of numbers?

> The set =E2=84=95_def defined by induction does not include =E2=84=B5o =
undefined numbers.

The set N doesn't include ANY undefined numbers.  Such talk is idiotic.

>>  Quite aside from the fact that there is no
>> mathematical definition of a "defined" number.  The "definition" you g=
ave
>> a few posts back was sociological (talking about how people interacted
>> with eachother) not mathematical.

> Mathematics is social, even when talking to oneself. Things which canno=
t=20
> be represented in any mind cannot be treated.

Natural numbers can be "represented in a mind", in fact in any
mathematician's mind.  It would appear certain such things can't be
represented in your mind.  That is not our problem.

> Regards, WM

--=20
Alan Mackenzie (Nuremberg, Germany).