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

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 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.

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.

 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.

No. When we *in actual infinity* multiply all |ℕ|natural numbers by 2,
then we keep |ℕ| numbers but only half of them are smaller than ω, i.e.,
are natural numbers. The other half is larger than ω.

Ha ha ha ha!  This is garbage.  If you think doubling some numbers gives
results which are "larger than ω" you'd better be prepared to give an
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)
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
remainder in common. Note: All sets which are infinite have lost at most
a finite number of elements.

More gobbledegook which isn't maths.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).