Proof Irrational number $x$, given $N\in{\mathbb{N}}$ exists $\epsilon>0$ so all rationals $\in{V_\epsilon{(x)}}$ have denominator larger than N.

Question

I am trying to show that Thomae's function is continuous at every irrational point (I have already shown discontinuous at every rational point), and for that - following this answer Prove continuity/discontinuity of the Popcorn Function (Thomae's Function). - I would like to show that for an irrational number $x$, given $N\in{\mathbb{N}}$, there exists $\epsilon>0$ so that all rationals $\in{V_\epsilon{(x)}}$ have denominator larger than N.

Answer 1

Since there are only finitely many rational points in $[0,1]$ whose denominator is not larger than $N$ there has to be a $\varepsilon$ so small that $(x-\varepsilon,x+\varepsilon)$ contains none of them; just take$$\varepsilon=\min\{|x-y|\,:\,y\text{ is rational and its denominator is}\leqslant N\}.$$