Liste des Groupes | Revenir à s math |
On 12/6/2024 3:19 AM, WM wrote:
Show two endsegments which do not hold common content.But no common.to.all finite.cardinals.⎜ With {} NOT as an end.segment,>
all endsegments hold content.
More than finitely many are finitely many, unless they are actually infinitely many. Therefore they are no enough.More.than.finitely.many are enough to⎜ there STILL are>
⎜ more.than.finite.many end.segments,
Not actually infinitely many however.
break the rules we devise for finitely.many.
For each finite.cardinal,All the fite cardinals are actually infinitely many. That is impossible as long as an upper bound rests in the contentents of endsegments.
up.to.that.cardinal are finitely.many.
A rule for finitely.many holds.
All.the.finite.cardinals are more.than.finitely.many.
A rule for more.than.finitely.many holds.
By an unfortunate definition (made by myself) there is always one cardinal content and index: E(2) = {2, 3, 4, ...}. But that is not really a problem.If all endsegments have content,That seems to be based on the idea that
then not all natnumbers are indices,
no finite.cardinal is both index and content.
Elsewhere, considering one set, that's true."All at once" is the seductive attempt of tricksters. All that happens in a sequence can be investigated at every desired step.
No element is both
index(minimum) and content(non.minimum).
However,
here, we're considering all the end segments.
Each content is index in a later set.
Each non.zero index is content in an earlier set.
Each content is index in a later set.Only if all content is lost. That is not possible for visible endsegments. They all are infinite and therefore are finitely many.
Yes, until about six years ago.Wasn't there a time when you (WM)⎜ And therefore,>
⎜ the intersection of all
⎝ STILL holds no finite cardinal.
No definable finite cardinal.
thought 'undefinable finite.cardinal'
was contradictory?
Round up the usual suspects∀k ∈ ℕ : E(k+1) = E(k) \ {k} cannot come down to the empty set in definable numbers. No other way however is accessible.
and label them 'definable'.
The intersection of all non.empty.end.segmentsA clear selfcontradiction because of inclusion monotony.
of the definable finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.
Generalizing,It is violating mathematics and logic. Like Bob.
the intersection of all non.empty end.segments is empty.
It is an argument considering finites,
of which there are more.than.finitely.many.
Les messages affichés proviennent d'usenet.