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.
Best Answer
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 $(x-\varepsilon,x+\varepsilon) \cap X = \{x\}$.