Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / E / 1 / c

Every open interval is open in the usual metric topology. Conversely, if $U\subset \mathbb{R}$ is open and $x\in U$, there exists ${ϵ}_x>0$ with $(x-{ϵ}_x,x+{ϵ}_x)\subseteq U$. Taking the union of these intervals over all $x\in U$ recovers $U$. Thus open intervals form a basis of the usual topology.

If $f$ is continuous, the two sets are inverse images of the open rays $(-\infty ,\lambda )$ and $(\lambda ,\infty )$, so they are open. Conversely, suppose both families of inverse images are open. For any interval $(a,b)$,which is open. By part (c)(i) and the basis of a topology criterion from part (b), all inverse images of open sets are open. Hence $f$ is continuous.