Understanding some points in the proof that the Thomae function is continuous at irrationals (pg.74 in Petrovic book)

Question

Here is the proof as it is given in the book:

My questions are:

1- Why the author assumed that $a \in (0,1),$ what about the case of $a=0$?

2- Why if $f(x) < \epsilon$ then $|f(x)| < \epsilon$ in our case here?

Could anyone help me in answering those questions please?

Answer 1

• The autor says that he is only considering the cases where $a$ is an irrational number, this is why he's pick $a \in (0,1)$.

• Well, since $f(x) \geq 0$ for any $x$, $f(x) < \varepsilon$ if and only if $-\varepsilon < f(x) < \varepsilon$, that is, $f(x) < \varepsilon$ if and only if $|f(x)| < \varepsilon$.