Do I need to prove rational number is discontinued on the number axis (irrational numbers are filled in between) and solve this question? We define the Riemann function $R:\Bbb R\to\Bbb Q$ by $$R(x)=\begin{cases} \frac{1}{q} &\text{if }x=\frac pq\text{ for }\ p\in\Bbb Z\setminus\{0\},\ q\in\Bbb N\setminus\{0\}\text{ and }\operatorname{gcd}(p,q)=1\\
 0&\text{if }x\in\Bbb R\setminus\Bbb Q\\
 1&\text{if }x=0\end{cases}$$
Prove that for any $a \in\Bbb R$ ，$\lim_{x\to a}R(x)=0$
I tried to solve is using the fact that the irrationals are dense in real number, but I had no idea how to write it. While searching on internet, I wonder would this idea contradicts Dirichlet function?
 A: No.  You do not need to prove or assume any such thing. Although it is absolutely true.
You need to prove that for any $a \in \mathbb R$ and any $\epsilon > 0$ that there is a $\delta$ so that 1) either $0 \not \in (a-\delta,a+\delta)$ or $a = 0$. and 2) that if there are any rational numbers (other than $0$) in $(a-\delta, a+\delta)$[$*$] they are such that their denominators are all larger than $\frac 1{\epsilon}$
That will be enough to prove it even without proving that irrationals and the rationals are mutually dense.
Hint: If $a \ne 0$ find $q$ so that $\frac jq < |a| < \frac {j+1}q$.  Any rational $r$ so that $\frac jq < r <\frac {j+1}q$ will have denominators that are larger than $q$ and $R(r) < \frac 1q$.  ..... So if $q > \frac 1{\epsilon}$......
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[$*$] Of course, there always will be rational numbers in $(a-\delta,a+\delta)$.  But you don't have to prove there will be.  You just have to assume there might be.  As for the irrationals since $|R(x) - 0| = 0 < \epsilon$ they are trivial to deal with.  You don't have to prove there will be but just demonstrate that if there are $|R(x) - 0| = 0 < \epsilon$.
