# Function in which every inverse image of open "interval" is open interval but not continuous.

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.

• An open set is a union of open intervals, and the inverse image of a union of the union of the inverse images, so... Oct 8, 2018 at 18:08
• On the other hand, it's easy to come up with examples of functions that are continuous, but the inverse image of an open interval is not necessarily an open interval. e.g. $f(x) = \sin x$, $A = (-\frac{1}{2}, \frac{1}{2})$. Oct 8, 2018 at 22:34

## 1 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.