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On 09.09.2024 13:49, joes wrote:The wrong quantifier order is: „There is a fixed infinite set of unitAm Mon, 09 Sep 2024 12:27:47 +0200 schrieb WM:The quantifier order related to the problem is this: NUF(x) = ℵo means:On 08.09.2024 22:11, Richard Damon wrote:More precisely: every positive x has infinitely many smaller unitOn 9/8/24 3:48 PM, WM wrote:>ℵo unit fractions are claimed to be smaller than every x > 0. If thatSelect any gap between one of the first ℵo unit fractions and itsBut that is changing the value of x in the middle of the problem
neighbour. Call its size x. Then ℵo unit fractions cannot fit into
the interval (0, x), independent of the actual size.
which isn't allowed.
Given that new x, we can choose a new set of Aleph_0 unit fractions
below that x.
is true then I can choose as the x one of the ℵo intervals between two
of them.
fractions (mind the quantifier order).
There exist ℵo unit fractions smaller than any x > 0. If this is not
true, then there are fewer. How many unit fractions are smaller than any
x > 0. THAT is the question. None. But all are differente. Hence there
must be a first one smaller than all other unit fractions.
But still infinitely many, since only finitely many are missing.A number is not an interval.An interval has a length that can be expressed by a real number:
1/n - 1/(n+1) = x .
Then the interval (0, x) contains not all unit fractions, for instance
not 1/n.
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