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On 10/5/2024 3:33 PM, WM wrote: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.On 05.10.2024 20:12, Jim Burns wrote:On 10/5/2024 5:43 AM, WM wrote:Your argument,>[In] many cases it is correct.>
"Many cases" is insufficient when
the argument requires "all cases".
My argument requires only one case,
in order to _be an argument_
needs to _show_ its result is true.
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?My argument requires only one case,The intersection of all end.segments is
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 set of natural numbers in each end.segment.
Each natural number is not.in one or more end.segment.
That creates a problem (but for only you).Not at all.
Your attempt to fix the problem involves
a change of the definition of intersection, or
of natural number, or of something else.
Changing what you think "infinite" meansNo. Infinite means infinite. All infinite endsegments cotain more than any finite set of numbers.
from what we mean by "very large finite"
to what we mean by "infinite"
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