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Let $\mathbb{Q}$ be the set of rational numbers. Is it possible, relabeling if needed, to list $\mathbb{Q}$ such that the result set to be a monotonic sequence? If not, why? If it is true, where is $\frac{r_1+r_2}{2}?$

This question occurs to me when I am reading Apostol's book, Mathematical Analysis, Page 68, Exercise 3.42. That exercise is : Consider the metric space ℚ of rational numbers with the Euclidean metric of ℝ. Let S consist of all rational numbers in the open interval (a,b), where a and b are irrational. Then S is is a closed and bounded subset of ℚ which is not compact.

When was ready to prove that $S$ is not compact, I wanted to construct an open covering of $S$ whose any finite subcovering can not cover $S$. Then I asked myself the above question. I feel that the answer is negative, but I can not find any explanation. Can anyone help me?

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    $\begingroup$ Good question, but not appropriate for mathoverflow, where we do research. $\endgroup$ Commented Apr 10, 2014 at 0:40
  • $\begingroup$ @GerryMyerson, I am very sorry for my improper question. Actually, this question occurs to me when I was doing an exercise in Apostol's book, Mathematical Analysis, Page 68, Exercise 3.42. That exercise is : Consider the metric space $\mathbb{Q}$ of rational numbers with the Euclidean metric of $\mathbb{R}.$ Let $S$ consist of all rational numbers in the open interval $(a,b),$ where $a$ and $b$ are irrational. Then $S$ is is a closed and bounded subset of $\mathbb{Q}$ which is not compact. Maybe my question is not a research. $\endgroup$
    – nuage
    Commented Apr 10, 2014 at 1:00

3 Answers 3

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Assume such a listing exists. Consider two successive terms $a_n$ and $a_{n+1}$. Then the number $\frac12(a_n+ a_{n+1})$ whether it appears after the $(n+1)$th term or before the $n$th term it will violate monotonicity. So, it is not possible.

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Regarding the finite subcover question, you don't necessarily need such an ordering of the rationals if you just choose the rational endpoints to your open cover appropriately. For example, take $U_{n}=(a_{n},b_{n})\cap\mathbb{Q}$ for all $n\in\mathbb{N}$, where $$a_{n}\in (a+\frac{1}{n+1},a+\frac{1}{n})\cap\mathbb{Q}$$ and $$b_{n}\in (b-\frac{1}{n},b-\frac{1}{n+1})\cap\mathbb{Q}.$$ If $a$ and $b$ are close to each other you can start labelling these sets from a large $n\in\mathbb{N}$ to ensure you get subsets of $S$. Note that $\mathbb{Q}$ is dense in $\mathbb{R}$ so each $a_{n}$ and $b_{n}$ exists, and by definition of the subspace topology each $U_{n}$ is open in $\mathbb{Q}$. Now $(a_{n})_{n=1}^{\infty}$ is a strictly decreasing sequence of rational numbers with $$\lim_{n\to\infty}a_{n}=a$$ and $(b_{n})_{n=1}^{\infty}$ is a strictly increasing sequence of rational numbers with $$\lim_{n\to\infty}b_{n}=b.$$ You can use the sandwich theorem for example to conclude that. Hence $S=\bigcup_{n=1}^{\infty}U_{n}$, and $(U_{n})_{n=1}^{\infty}$ is a strictly increasing sequence, i.e. $U_{n}\subset U_{n+1}$ for all $n\in\mathbb{N}$. What happens if you take a finite subcover? There will exist $N\in\mathbb{N}$ for instance so that this subcover fails to cover all $q\in (b-\frac{1}{N},b)\cap\mathbb{Q}\subseteq S$.

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Assume it is possible. Restrict to an interval I . Then, e.g., by Bolzano-Weirstass' thm., there is a limit point 'a' of the sequence of Rationals in I, and all-but-finitely-many Rationals in the interval are in $(a-r,a+r)$, and then there are subintervals $I' \subset I$ with no Rationals in them.

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