For $f : X \to Y$, the following S1 and S2 are equivalent.
S1. $f$ is continuous on $X$
S2. if $A \in Y$ is open in $Y$, then $f^{-1}(A)$ is open in $X$
Since open interval is only a special case of open set, I guess there must be an example of a function $f$ that satisfies S2 only in "interval" case, and $f$ is not continuous (because otherwise, S2 must reduce to "interval" statement).
But I'm having hard time coming up with such function. Any help will be appreciated.
Best Answer
There is another equivalent condition
S3. For a fixed base $\mathcal{B}$ of the topology on $Y$, for all $B \in \mathcal{B}$ : $f^{-1}[B]$ is open in $X$.
S2. implies S3. because basic open sets are open in particular.
S3. implies S2. because if $O$ is open in $Y$, $O = \bigcup \mathcal{B}'$ for some subfamily $\mathcal{B}'$ of $\mathcal{B}$ by the definition of a base.
But then $f^{-1}[O] = \bigcup \{f^{-1}[B] : B \in \mathcal{B}'\}$ which by S3. is a union of open sets in $X$, hence open, as required.
As open intervals form a base for the topology on $\mathbb{R}$, if a function has the property that inverse images of intervals are open (or even open intervals in particular) then $f$ is continuous.