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On 08.10.2024 19:28, Jim Burns wrote:On 10/8/2024 6:18 AM, WM wrote:No. Very large is not more than any finite set.All infinite endsegments contain more than any finite set of numbers....still true if 'infinite' means "very large".
With each, but not with all at once (cf. quantifier shift).An end segment E is infinite because each natural number has aTherefore every infinite endsegment has infinitely many elements with
successor,
so no element of E is max.E so not all nonempty subsets are two.ended
so E is not finite.
each predecessor in common. This is valid for all infinite endsegments.
Intersection with what?There are no other end segments, none are finite.They all have an infinite intersection.
Consider end segment E(k+1)No, but by definition there are infinitely many numbers. They are dark.
k+1 is in E(k+1)
Is k+1 in each end segment? Is k+1 in E(k+2)?
Is E(k+1) the set of natural numbers in each end.segment?
not only one but infinitely manyWhy else should they be infinite?Because each natural number is followed by a natural number,
Invalid quantifier shift.because 'infinite' DOES NOT mean 'very large'.Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.
Proof: If not, then there would be at least one endsegment with lessNo. Why do you think that?
numbers.
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