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

On 09.10.2024 17:11, Alan Mackenzie wrote:

WM <wolfgang.mueckenheim@tha.de> wrote:

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

{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
Should all places ω+2, ω+4, ω+6, ... remain empty?
Should the even numbers in spite of doubling remain below ω?
Then they must occupy places not existing before. That means the original set had not contained all natural numbers. That mans no actual or complete infinity.

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.

True. But those endsegments which have lost only finitely many numbers and yet contain infinitely many, have an infinite intersection.

 [ .... ]
 

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.

Why not? The essence is that only finitely many numbers have been lost and the rest is remaining.

 

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.

These sets do not belong to the above example. They are not TN-sets.

 

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

 Yes, I understand completely.

No.

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

 You can't "complete" a set.

You can add elements.
Regards, WM