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

On 10/8/24 6:18 AM, WM wrote:

On 07.10.2024 20:05, Jim Burns wrote:

On 10/5/2024 3:33 PM, WM wrote:

On 05.10.2024 20:12, Jim Burns wrote:

On 10/5/2024 5:43 AM, WM wrote:

>

[In] many cases it is correct.

>
"Many cases" is insufficient when
the argument requires "all cases".

>
My argument requires only one case,

>
Your argument,
in order to _be an argument_
needs to _show_ its result is true.

 It is easy: Consider a function of shrinking sets which remain infinite. Then there is an infinite subset or core common to all of them.

Right, which shows that with infinte sets, subsets and their superset can be of the "same" size.

 

My argument requires only one case,
best demonstrated with endsegments E(n).
The intersection of all *infinite* endsegments
is infinite, because
they all contain the same natural numbers which
have not yet been eliminated by
the process E(n+1) = E(n) \ {n}.

>
The intersection of all end.segments is
the set of natural numbers in each end.segment.
Each natural number is not.in one or more end.segment.

 That is true but wrong in case of infinite endsegments which are infinite because they have not lost all natural numbers. Why else should they be infinite?

But that just shows that each of your FISONs is not an infinite set.
There may be an infinite set of FISONs, but none of them are infinite in and of themselves.
WHen talking of natural number segments, there is no such thing as a infinite endpoint, just that you have segments that don't HAVE and (upper) endpoint.

>
That creates a problem (but for only you).
Your attempt to fix the problem involves
a change of the definition of intersection, or
of natural number, or of something else.

 Not at all.

>
Changing what you think "infinite" means
from what we mean by "very large finite"
to what we mean by "infinite"

 No. Infinite means infinite. All infinite endsegments cotain more than any finite set of numbers.
 Regards, WM