real analysis - Show that the following function is continuous on Q:

Let $h : \mathbb{Q}\rightarrow\mathbb{R}$ be given by $$ h(x)= \begin{cases} 0, &\ x^2<2\\ \ \\ 1,&\ x^2>2 \end{cases} $$

see why this function is continuous on $\mathbb{Q}$, but I can't show it. Maybe we could work with the epsilon-delta criterion?

So for every $\epsilon> 0$ we find a $\delta> 0$ with $| x - x_0 | <\delta$ implying $ | h(x) - h(x_0) | < \epsilon$ ? Am I right? How do I have to choose my $\delta$ then?

Answer

Let $q\in \mathbb Q$.

Since $q\ne\sqrt2$, then there exists $\delta>0$ such that $\sqrt2\not\in(q-\delta,q+\delta)$. So $h$ is constant on $(q-\delta,q+\delta)$, and thus $$ |h(x)-h(q)|=0<\varepsilon $$ for any $\varepsilon>0$ that you choose.