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The definition of Thomae function is $$f(x) = \begin{cases} \dfrac{1}{q}, & \text{if $x=\dfrac{p}{q}$} \\[2ex] 0, & \text{if $x$ is irrational} \end{cases}$$ where $p\in \mathbb{Z}, q\in \mathbb{Z}^+$ and $\gcd(p,q)=1.$

The conclusion is $\lim_{x \to P}f(x)=0$ for any point P of R.

The proof is below. Unfortunately, I don't understand it even from the first sentence.

Proof: 1. Indeed, for each positive number $\epsilon$, the set $(P-1,P) \cup (P,P+1)$ contains only finitely many rational number $p/q$ with $1/q \geq \epsilon$ (namely, $q \leq 1/\epsilon$). Now I understand this. Thanks for @Momo's excellent explanation.

  1. Let $\delta$ be the distance between P and the closest such rational number. Then for all x satisfying $0<|x-P|<\delta$, we have $|f(x)-0|<\epsilon$.

My understanding is: Since such rational numbers are finite, we can get a minimum of such distances between P and p/q, namely $\delta$. But next? In this very tiny interval $0<|x-P|<\delta$, if x is irrational, then obviously $|f(x)-0|=0<\epsilon$. But what if x is rational in the form of p/q?

My rethinking: this specific $\delta$ can make sure our interval $0<|x-P|<\delta$ does not contain any rational p/q. But it seems wrong, because I know rational number is dense in any interval in R.

Mariana
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    What part of the sentence do you not understand? There are finitely many rational numbers with $q\leq\frac1\varepsilon$ in the set. – Rushabh Mehta Jun 15 '21 at 02:05
  • I don't understand the whole 1 part, including the part you said. – Mariana Jun 15 '21 at 02:24
  • Since $1/ \epsilon$ is positive, and q is positive, hence q is finitely many (because q has a fixed upper bound). I know only this. – Mariana Jun 15 '21 at 02:30
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    @Mariana Ok, so if $q$'s are finitely many and $P-1<\frac{p}{q}<P+1$, then $(P-1)q<p<(P+1)q$, so for each of the finitely many $q$'s you have only finitely many $p$'s satisfying the inequality. – Momo Jun 15 '21 at 02:48
  • @Momo excellent explanation! I also update my question. – Mariana Jun 15 '21 at 03:00
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    @Mariana Indeed $0<|x-P|<\delta$ (it is not an interval, but two open intervals as $P$ is excluded) contains an infinity of rational numbers $x=p/q$, but since we shrinked $\delta$ to exclude the (finitely many) points with $1/q\ge\epsilon$ we are going to have $0\le f(x)=f(p/q)=1/q\lt\epsilon$, so the limit in $P$ is zero. – Momo Jun 15 '21 at 11:10
  • Thank you. One follow-up question is whether this function is differentiable at these irrational numbers points (they are continuous, but are they also differentiable)? – Mariana Jun 20 '21 at 03:10

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