Let $f:[0, 1]\rightarrow\mathbb{R}$ be defined by $$f(x)=\left\{ \begin{array}{ll} 0, & \hbox{if}\,\,\, x\in \mathbb{R\backslash Q} \\ \frac{1}{q}, & \hbox{if}\,\,\, x=\frac{p}{q}\in\mathbb{Q},\,\,\, \gcd(p, q)=1 \end{array} \right.$$
How do you prove that $f$ is continuous in $\{0\}\cup (\mathbb{R\backslash Q}\cap [0, 1])$?
Let x be in $[0,1]$ and in $\mathbb{R \backslash Q}$.
Let $\epsilon > 0$. We want to find a neighbourhood of $x$ where $f(y)-f(x) = f(y) < \epsilon$.
If $y \in \mathbb{R \backslash Q}$, $f(y)=0$, it's clearly the case.
If $y \in \mathbb{Q}$, $f(y)$ has to be small so $y$'s denominator has to be big (actually bigger than $1/\epsilon$). And we want that to be true in a neighbourhood of $x$. Let us consider the rationals we don't want. Their denominator is smaller than $ceil(1/\epsilon):=p$. They are also between 0 and 1. So they are 0/1, 1/1 ; 0/2, 1/2, 2/2 ; 0/3, 1/3, 2/3, 3/3, ; ... ; 0/p, 1/p, ..., p/p. The important thing is that there's a finite number of them. So they can't be arbitrarily close to $x$. So there is a small open interval around $x$ that excludes them. And on that interval, $f(y) < \epsilon$.
