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

Am 09.10.2024 um 18:12 schrieb Alan Mackenzie:

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

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.

 End segments don't "lose" anything.  They are what they are, namely well
defined sets.  Note that your "True" in your last paragraph, agrees that
the intersection of all end segments is empty, which you immediately
contradict by asserting it is not empty.

You do not understand the least! The intersection of all endsegments is empty. The intersection of infinite endsegments is infinite.

 

[ .... ]

 

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}
{ } .

 

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

Obviously the above sets were meant.

 

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

 

You can't "complete" a set.

 

You can add elements.

 Then you get different sets, which weren't the ones you were trying to
reason about.

The completion of the above sets does not change the principle:
Non-empty inclusion-monotonic sets like infinite endsegments have a non-empty intersection. All endsegments have an empty intersection.
Regards, WM