© 1996, Topology Atlas

Last spring I taught topology, and one of the first theorems I proved was that a function from R to R is continuous in the epsilon-delta sense iff the inverse image of every open set is open. A student asked if you could characterize continuity using images instead of inverse images, which got me thinking about the following question:

Is there a family F of sets of reals such that for every function f : R -> R, f is continuous iff for every X in F, f(X) is also in F?

The answer is NO, but it took me a long time to prove it, and the proof I came up with is a bit tricky. Have anyone ever seen this question before?

Is there an easy way to prove that the answer is no? If you use sequences of reals rather than sets of reals, then the answer is easily seen to be yes: a function is continuous iff the image of every convergent sequence is convergent. Also, if you're willing to use TWO families of sets then the answer is yes:

a function f : R -> R is continuous iff the image of every interval is an interval AND the image of every compact set is compact.

One could also ask about various generalizations of the question to R^n and other topological spaces.

We believe that this is an ideal question to attract discussion on the topic. Indeed, the question is so basic that even an undergraduate student can understand it, and nevertheless the solution found by the author is somewhat tricky. We would like to hear any "easy proofs" which might exist, as well as remarks about generalizing this question to other topological spaces, as suggested in the end of the article.

Please address your remarks, suggestions and "easy proofs" to one of the editors via e-mail.

We will post all relevant comments and proofs in our Discussion Forum, and we will later publish the author's original proof that the answer is no.