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

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