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I was thinking: if we have any countable infinite set of real numbers (because it is a set it must have infinite different values) let's say for example:

$X = \{\frac1n , \forall n \geq 1, n \in \mathbb N\}$

can we build a sequence $a_n$ only with the elements of this set (and that covers the whole set), such that $a_n$ is monotone?

My first tought was yes, but them I remembered that the rational numbers fit in this definition. And I can't really answer this question.

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    $\begingroup$ It seems to me that there is plenty of strictly monotone sequences $a_\bullet:\Bbb N\to\Bbb Q$. For instance $a_n=n$. $\endgroup$
    – user239203
    May 13, 2020 at 8:51
  • $\begingroup$ My guess is that you meant to write “a countable infinite set of real numbers” instead of “a set of countable infinite real numbers”. $\endgroup$ May 13, 2020 at 8:51
  • $\begingroup$ It took your comments and edited the question, thank you both! $\endgroup$ May 13, 2020 at 9:32

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The questions in the title and in the message are not equivalent. You clearly can build a monotone sequence $a_n$ of only rational numbers.

When it comes to only and all rational numbers, you can't do that for at least two reasons. First of all, the set is unbounded from below, so you can't have $a_0$. Second of all, it is dense, meaning that between any $p,q$ there is $r: p<r<q$. So even if were able to find $a_0$, there's no immediate succesor, meaning that you aren't able to find $a_1$.

So creating a monotone sequence of only and all numbers of a set $X \subseteq \mathbb R$ is only possible when $X$ has a minimum, and when every of its elements is isolated -- ie. if for every $x \in X$ there is $\varepsilon >0$ such that $(x-\varepsilon,x+\varepsilon) \cap X = \{x\}$.

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  • $\begingroup$ is there a proof for your statement? "If X is bounded and it's elements are isolated we can order it." $\endgroup$ May 13, 2020 at 8:55
  • $\begingroup$ I'm a bit out of shape, but let's try. Well, first of all you have to see that a set with isolated elements is closed. So if it has an infimum, it has a minimum. So our set has a minimum, $a_0$. The set $X\setminus \{a_0\}$ retains the property of isolated elements, so it too has a minimu, $a_1$. Then consider the set $X \setminus \{a_0,a_1\}$ and so on. $\endgroup$ May 13, 2020 at 9:07
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    $\begingroup$ @MichałZapała This does define a monotone sequence in $X$ starting from its smallest value but one must still show that all elements of $X$ are reached. A priori, $X$ could have order type $\omega + 1$, whence this procedure has image $\omega$. $\endgroup$ May 13, 2020 at 9:13
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    $\begingroup$ @paulblartmathcop Real numbers ar Archimedean, so no set with order type greater than $\omega$ can be isolated. Of course that's something I neglected in my message. $\endgroup$ May 13, 2020 at 9:15
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    $\begingroup$ A set with isolated elements is not necessarily closed. $\endgroup$
    – Paramanand Singh
    May 13, 2020 at 10:14
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Your sequence you mentioned is already monotone (decreasing). If you mean a arbitrary set (countable) to be ordered the answer is yes.

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