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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?

user3658307
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Pranita Gupta
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  • Finiteness of $A_n$ follows from the fact that $1/n\to 0$ – Naive Jul 31 '17 at 13:05
  • No but it has to be related with $\epsilon$ in $\epsilon \delta $ definition – Pranita Gupta Jul 31 '17 at 13:07
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    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 $n=1,...N-1$ – Naive Jul 31 '17 at 13:12
  • Ok so this is N. What about $p_k and q_k$ appeared suddenly? There are other links like this https://math.stackexchange.com/questions/530097/proof-of-continuity-of-thomae-function-at-irrationals but none of them contains this type of arguments. – Pranita Gupta Jul 31 '17 at 13:14
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    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? – Did Jul 31 '17 at 14:13
  • yes,this is one of the ques,this is clear. – Pranita Gupta Jul 31 '17 at 18:00

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