Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

On 10/09/2024 07:13 AM, 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 one 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 some
useful rules.

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

 The rule of subset proves that every proper subset has fewer elements
than its superset. So there are more natural numbers than prime numbers,
|N| > |P|, and more complex numbers than real numbers, |C| > |R|.  Even
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.
So we have two sets of the same size, one of which is bigger than, the
other of which is smaller than another set.  Contradiction.
>
So your "rule of subset" is not coherent.
>

 The rule of construction yields the numbers of integers |Z| = 2|N| + 1
and the number of fractions |Q| = 2|N|^2 + 1 (there are fewer rational
numbers Q# ). Since all products of rational numbers with an irrational
number are irrational, there are many more irrational numbers than
rational numbers |X| > |Q#|.

>
There are many more irrational numbers than rational ones, but your
argument is not coherent.
>

 The rule of symmetry yields precisely the same number of real
geometric points  in every interval (n, n+1] and with at most a small
error same number of odd numbers and of even numbers in every finite
interval and in the whole real line.

>
I can't make out what you're trying to say, here.
>

The standard definition for infinite (or finite) sets being the same
size is the existence of a 1-1 correspondence between them.

>

You seem to be rejecting that definition.  What would you replace it by?
You have specified "bigger" for a special case.  What is your definition
for the general case?

>

Simply nonsense is the claim that there are as many algebraic numbers
as prime numbers.

>

It is not nonsense.  The prime numbers can be put into 1-1
correspondence with the algebraic numbers, therefore there are exactly
as many of each.

>

Nonsense. Only potential infinity is used. Never the main body is applied.

>
It's not nonsense.  Your last sentence doesn't make any sense.  Neither
does the middle one, without further context saying what "only"
potential infinity is used for.
>

For Cantor's enumeration of all fractions I have given a simple
disproof.

>

Your "proofs" tend to be nonsense.

>

It appears to you because you are unable to understand. Here is the
simplest:

>
No, it is clear to me BECAUSE I understand.  Being a graduate
mathematician, I can distinguish between maths and gobbledegook that
might look like maths.  Your "proofs" fall into the second category.
>

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.  What you do is revert to your fairy-story "dark
numbers" (which I've proven can't exist), and say all these alleged
infinitely many numbers are "dark".
>

Proof: If not, then there would be at least one endsegment with less
numbers.

>
... which is gobbledegook, not maths.
>

Note: The shrinking endsegments cannot acquire new numbers.

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

Regards, WM

>

How about Fred Katz' MIT Ph.D. Thesis "OUTPACING ...".
That at least makes generally clear a notion of nested sets
a size relation for them.
Or do your castigations of another sock-puppet troll
automatically apply to "Katz' MIT Ph.D. Thesis OUTPACING..."?
Don't get me wrong, it's sort of insulting everybody's intelligence,
especially the go-along-to-get-along crowd.