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On 11/7/2024 12:59 PM, WM wrote:
On the contrary, every irrational (and every rational) is in contact with infinitely many intervals, because in the length of √2/5 there are infinitely many rationals and therefore infinitely many midpoints of intervals.When we cover the real axis by intervalsNo irrational is not in contact with
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J(n) = [n - √2/10, n + √2/10]
and shuffle them in a clever way,
then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.
the union of intervals.
The first is true because,If it is true, then the measure of the intervals of 3/10 of the real axis grows to infinitely many times the real axis.
for each irrational x,
each interval of which x is in its interior
holds rationals, and
rationals are points in the union of intervals.
There is an enumeration of ℚ⁺
The infinite sum of measures = 2³ᐟ²/9The intervals are shifted such that every rational number is the midpoint of an interval. Every point p on the real axis is covered by infinitely many intervals, namely by all intervals having midpoint rationals q with
d is _not in contact with_ each interval.
However,The intervals have length √2/5. There is no ε.
each interval of which d is in its interior
holds rationals, which are points in ⋃(ε.cover)
Thus, d is _in contact with_ ⋃(ε.cover)
but not with any of its intervals.
No, it violates mathematics and logic when by reordering the measure of a set of disjunct intervals grows.Do you believe this???Don't you believe this???
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