Continous function from $ \Bbb Q \rightarrow \Bbb R $, $ f = 1 $ for $x > \sqrt2$ and $ f = 0$ for $x < \sqrt2$

Question

I'm not really sure how to go about this problem.

Show that $h : \Bbb Q \rightarrow \Bbb R $, with
$$ h(x)=\begin{cases} 0 &\text{for $|x|< \sqrt{2}$} \\ 1 &\text{for $|x|>\sqrt{2}$} \end{cases} $$ is a continous function.

Answer 1

We have to show that for each point $x_0 \in \mathbb{Q}$ and each $\epsilon > 0$ there is a $\delta > 0$ such that: $$|x_0-x| < \delta \implies |f(x_0) - f(x)| < \epsilon. $$

It is obvious that we have continuity everywhere. The only point where things look bad is $\sqrt{2}$ but this is not a rational point.

Hint: You can always choose some $\delta$ such that $\delta < |x_0 - \sqrt{2}|$.