Somebody knows if there exists a function $f:\mathbb{R}\to\mathbb{R}$, with it's usual topology, such that the image (or preimage) of every open set is closed and the image (or preimage) of every closed set is open? If not, is possible find such a function between some topological spaces (without consider the discrete topology off course)?

Thank you.


For "image" this is clearly impossible, since every single point is a closed subset, its image must be a single point, and those are not open.

For inverse image, constant maps have this property because the preimage of anything is either the empty set or all of $\mathbb R$, and both are closed and open at the same time.

The only interesting question is, are constant functions the ONLY ones with this property (for pre-image)? The answer is yes; the pre-image of every point must be open, and $\mathbb R$ is connected, so it can't be the union of disjoint open sets.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.