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In this link, I am not getting why is $A_n$ is finite? Prove that function is continuous in all irrational points

Why is the number of such positive integers at most $[\frac {1} {\epsilon}]$ , the integral part of ${1\over\epsilon}$? And why is the set of all those rationals numbers $\frac pq \in (0,1)$ with $\frac 1q \geq \epsilon $ is $ \{\frac pq : 1\le q \lt [\frac 1\epsilon] ,1 \le p \lt q\}$? And most important of all they call this (or denote this) set afterwards by , $\{r_k= {{p_k}\over{q_k}} : 1 \le k \le N\}$.

I got the idea of the proof, but not getting the actual proof; why are all these happening?

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  • $\begingroup$ Finiteness of $A_n$ follows from the fact that $1/n\to 0$ $\endgroup$
    – Naive
    Jul 31, 2017 at 13:05
  • $\begingroup$ No but it has to be related with $\epsilon$ in $\epsilon \delta $ definition $\endgroup$ Jul 31, 2017 at 13:07
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    $\begingroup$ Given $\epsilon>0\;\; \exists N\in \Bbb N\; $ such that $|1/n-0|<\epsilon\;\;\forall n \ge N \implies 1/n<\epsilon \;\;\forall n\ge N\implies1/n\ge \epsilon $ for $n=1,...N-1$ $\endgroup$
    – Naive
    Jul 31, 2017 at 13:12
  • $\begingroup$ Ok so this is N. What about $p_k and q_k$ appeared suddenly? There are other links like this math.stackexchange.com/questions/530097/… but none of them contains this type of arguments. $\endgroup$ Jul 31, 2017 at 13:14
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    $\begingroup$ Once you know that a (non-empty) set $S$ is finite, say with size $N$, you can enumerate it, yielding $S=\{s_k\mid1\leqslant k\leqslant N\}$. Is this your question? $\endgroup$
    – Did
    Jul 31, 2017 at 14:13

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