Das Kalenderblatt 120530

29/05/2012 - 09:17 von WM | Report spam
All rational numbers of the unit interval [0, 1] can be covered by
countably many intervals, such that the n-th rational is covered by an
interval of measure 1/10^n. There remain countably many complementary
intervals of measure 8/9 in total.
Does each of the complementary intervals contain only one
irrational number? Then there would be only countably many which could
be covered by another set of countably many intervals of measure 1/9.
My question: Can this contradiction be formalized in ZFC?
[user31686, 15. 4. 2012]
http://math.stackexchange.com/quest...674#150674

It has been mentioned already that the irrationals x of the set X of
the remaining part of measure 8/9 (or more), that is not covered by
your intervals, form a totally disconnected space, so called "Cantor
dust". Every particle x in X is separated from every other particle x'
in X by at least one rational q_n, and, as every q_n is covered by an
interval I_n, it is separated by at least one interval I_n. Since the
end points a_n and b_n of the I_n are rational numbers too, also
being covered by their own intervals, the particles of Cantor dust can
only be limits of infinite sequences of endpoints of overlapping
intervals I_n. (If they don't overlap, then the limits must come
earlier, but in any case infinitely many endpoints are required to
form a limit.) Such an infinite set of overlapping intervals is called
a cluster. In principle, given a fixed enumeration of the rationals,
we can calculate every cluster C_k and the limits of its union. Since
two clusters are disjoint (by their limits), there are only countably
many clusters (disjoint subsets of the countable set of intervals
I_n). Therefore, every irrational x can be put in bijection with the
cluster lying right of it, say, between x and its next right
neighbour. So, by this bijection we prove that the set of uncovered
irrational numbers x in X is countable.
[Stentor Schicklgruber, StackExchange (2012)]
http://math.stackexchange.com/users...icklgruber

Let all rational numbers q_n of the interval (0,oo) be covered by
intervals I_n =[s_n, t_n ] of measure |I_n| = 2^−n, such that q_n is
the center of I_n. Then there remain uncountably many irrational
numbers as uncovered "Cantor-dust". Every uncovered irrational x_a
must be separated from every uncovered irrational x_b by at least one
rational, hence by at least one interval I_n covering that rational.
But as the end points s_n and t_n of the I_n also are rational numbers
and also are covered by their own intervals, the irrationals x_a can
only be limits of infinite sequences (s_μ) or (t_ν) of endpoints of
overlapping intervals I_n. In principle we can calculate the limit x_a
of every such sequence (s_μ) or (t_ν) of endpoints of overlapping
intervals. Therefore, every irrational x_a can be put in bijection
with the infinite set of intervals lying right of it, say, between x_a
and its right neighbour x_b. There are countably many disjoint sets
like {t | t in (t_ν)} of elements of the sequences (t_ν) converging to
one of the x_a. By this bijection we get a countable set of not
covered irrational numbers x_a. Where are the other irrational numbers
that are not covered by intervals I_n? Nowhere. Uncountability is
contradicted.
[Quidquid pro quo, MathOverflow (2012)]

Gruß, WM
 

Lesen sie die antworten

#1 Ralf Bader
29/05/2012 - 22:09 | Warnen spam
WM wrote:

All rational numbers of the unit interval [0, 1] can be covered by
countably many intervals, such that the n-th rational is covered by an
interval of measure 1/10^n. There remain countably many complementary
intervals of measure 8/9 in total.
Does each of the complementary intervals contain only one
irrational number? Then there would be only countably many which could
be covered by another set of countably many intervals of measure 1/9.
My question: Can this contradiction be formalized in ZFC?
[user31686, 15. 4. 2012]



http://math.stackexchange.com/quest...674#150674

It has been mentioned already that the irrationals x of the set X of
the remaining part of measure 8/9 (or more), that is not covered by
your intervals, form a totally disconnected space, so called "Cantor
dust". Every particle x in X is separated from every other particle x'
in X by at least one rational q_n, and, as every q_n is covered by an
interval I_n, it is separated by at least one interval I_n. Since the
end points a_n and b_n of the I_n are rational numbers too, also
being covered by their own intervals, the particles of Cantor dust can
only be limits of infinite sequences of endpoints of overlapping
intervals I_n. (If they don't overlap, then the limits must come
earlier, but in any case infinitely many endpoints are required to
form a limit.) Such an infinite set of overlapping intervals is called
a cluster. In principle, given a fixed enumeration of the rationals,
we can calculate every cluster C_k and the limits of its union. Since
two clusters are disjoint (by their limits), there are only countably
many clusters (disjoint subsets of the countable set of intervals
I_n). Therefore, every irrational x can be put in bijection with the
cluster lying right of it, say, between x and its next right
neighbour. So, by this bijection we prove that the set of uncovered
irrational numbers x in X is countable.
[Stentor Schicklgruber, StackExchange (2012)]
http://math.stackexchange.com/users...icklgruber



Wie schon bemerkt wurde,
http://chat.stackexchange.com/trans...36?mG45038
ist "Stentor Schicklgruber" wohl wieder eine Ihrer Sockenpuppen; und Godwin
kennt den Namen "Schicklgruber". Der über die sich Ihrer
studentenverblödenden Mitwirkung berühmen könnende Halbbildungsanstalt
pràsidierende Esomedizinfan sollte sich allmàhlich wirklich einmal Gedanken
über Ihren Beitrag zum Ruf des Ladens machen.

Let all rational numbers q_n of the interval (0,oo) be covered by
intervals I_n =[s_n, t_n ] of measure |I_n| = 2^−n, such that q_n is
the center of I_n. Then there remain uncountably many irrational
numbers as uncovered "Cantor-dust". Every uncovered irrational x_a
must be separated from every uncovered irrational x_b by at least one
rational, hence by at least one interval I_n covering that rational.
But as the end points s_n and t_n of the I_n also are rational numbers
and also are covered by their own intervals, the irrationals x_a can
only be limits of infinite sequences (s_μ) or (t_ν) of endpoints of
overlapping intervals I_n. In principle we can calculate the limit x_a
of every such sequence (s_μ) or (t_ν) of endpoints of overlapping
intervals. Therefore, every irrational x_a can be put in bijection
with the infinite set of intervals lying right of it, say, between x_a
and its right neighbour x_b. There are countably many disjoint sets
like {t | t in (t_ν)} of elements of the sequences (t_ν) converging to
one of the x_a. By this bijection we get a countable set of not
covered irrational numbers x_a. Where are the other irrational numbers
that are not covered by intervals I_n? Nowhere. Uncountability is
contradicted.
[Quidquid pro quo, MathOverflow (2012)]



Und diese Sockenpuppe ist wohl mal wieder rausgeflogen.

Ihr Geschwàtz, Mückenheim, über diese Intervalle beweist übrigens in der Tat
etwas: Daß Sie nàmlich keinsterlei Ahnung haben, wie Grenzwerte
funktionieren. Diesmal besteht Ihr Fehler im Einschmuggeln der blödsinnigen
Annahme, der Abschluß des Grenzwerts einer konvergenten Folge von
Teilmengen des Einheitsintervalls stimme mit dem Grenzwert der Folge der
Abschlüsse überein. Nee, Sie brauchen das nicht zu beschwafeln, da kommt eh
nur Krampf dabei heraus. Aber diese Annahme, und mithin Ihr
Intervallgeschwàtz, ist in jeder Variante von Analysis, ob klassisch,
nichtstandard, konstruktiv, pràdikativ, intuitionistisch oder sonstwas,
falsch. Sie wird nàmlich falsch, sobald es eine einzige nichttriviale
konvergente Folge und ihren Grenzwert gibt.

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