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From: Alan Mackenzie <acm@muc.de>
Newsgroups: sci.math
Subject: Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)
Date: Wed, 9 Oct 2024 15:11:31 -0000 (UTC)
Organization: muc.de e.V.
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WM <wolfgang.mueckenheim@tha.de> wrote:
> On 09.10.2024 16:13, Alan Mackenzie wrote:
>> WM <wolfgang.mueckenheim@tha.de> wrote:
>>> On 08.10.2024 23:08, Alan Mackenzie wrote:
>>>> WM <wolfgang.mueckenheim@tha.de> wrote:
>>>>> On 07.10.2024 18:11, Alan Mackenzie wrote:
>>>>>> What I should have
>>>>>> written (WM please take note) is:

>>>>>> The idea of one countably infinite set being "bigger" than another
>>>>>> countably infinite set is simply nonsense.

>>>>> The idea is supported by the fact that set A as a superset of set B=
 is
>>>>> bigger than B.

>>>> What do you mean by "bigger" as applied to two infinite sets when on=
e of
>>>> them is not a subset of the other?

>>> That is not in every case defined. But here are some rules:
>>> Not all infinite sets can be compared by size, but we can establish s=
ome
>>> useful rules.

>> Possibly.  But these rules would require proof, which you haven't
>> supplied.

> These rules are self-evident.

They're anything but self-evident for infinite sets.  Presumably the
muddle you're in is much the same as what mathematicians were in a couple
of hundred years ago.  Things have advanced since then.

>>> =EF=83=A8 The rule of subset proves that every proper subset has fewe=
r elements
>>> than its superset. So there are more natural numbers than prime numbe=
rs,
>>> |N| > |P|, and more complex numbers than real numbers, |C| > |R|.  Ev=
en
>>> finitely many exceptions from the subset-relation are admitted for
>>> infinite subsets. Therefore there are more odd numbers than prime
>>> numbers |O| > |P|.

>> This breaks down in a contradiction, as shown by Richard D in another
>> post: To repeat his idea:

>> The set {0,    2,    4,    6, ...} is a subset of the natural numbers =
N,
>>          {0, 1, 2, 3, 4, 5, 6, ...}, thus is smaller than it.
>> We can replace the second set by one of the same "size" by multiplying
>> each of its members by 4.  We then get the set
>>          {0,          4,         8, ...}.
>> Now this third set is a subset of the first hence is smaller than it.

> No. When we *in actual infinity* multiply all |=E2=84=95|natural number=
s by 2,=20
> then we keep |=E2=84=95| numbers but only half of them are smaller than=
 =CF=89, i.e.,=20
> are natural numbers. The other half is larger than =CF=89.

Ha ha ha ha!  This is garbage.  If you think doubling some numbers gives
results which are "larger than =CF=89" you'd better be prepared to give a=
n
example of such a number.  But you're surely going to tell me that these
are "dark numbers" (which I've proved don't exist).

>>> Theorem: If every endsegment has infinitely many numbers, then
>>> infinitely many numbers are in all endsegments.

>> That is simply false.  You cannot specify a single number which is in
>> all endsegments.

> True. This proves dark numbers.

Dark numbers don't exist, or at least they're not natural numbers.  There
is no number in each and every end segment of N.

[ .... ]

>>> Note: The shrinking endsegments cannot acquire new numbers.

>> An end segment is what it is.  It doesn't change.

> But the terms of the sequence do. Here is a simple finite example:

> {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
> {2, 3, 4, 5, 6, 7, 8, 9, 10}
> {3, 4, 5, 6, 7, 8, 9, 10}
> {4, 5, 6, 7, 8, 9, 10}
> {5, 6, 7, 8, 9, 10}
> {6, 7, 8, 9, 10}
> {7, 8, 9, 10}
> {8, 9, 10}
> {9, 10}
> {10}
> { } .

Example of what?  The reasoning you might do on finite sets mostly isn't
applicable to infinite sets.

> Theorem: Every set that contains at least 3 numbers (call it TN-set)=20
> holds these numbers in common with all TN-sets.

Not even an ignorant schoolboy would maintain this.  The two TN-sets {0,
1, 2} and {3, 4, 5} have no numbers in common.

> Quantifier shift: There is a subset of three elements common to all
> TN-sets. Understood?

Yes, I understand completely.  What you've written is garbage.

> Now complete all sets by the natural numbers > 10 and complete the
> sequence.

You can't "complete" a set.  A set is what it is, defined by its well
defined members and is not subject to change.

> Then we get: All sets which have lost at most n elements have the=20
> remainder in common. Note: All sets which are infinite have lost at mos=
t=20
> a finite number of elements.

More gobbledegook which isn't maths.

> Regards, WM

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