It's probably best to use the characterization of continuity in terms of closed sets, which are just finite sets in this case, and use the injectivity.
let f: R -> R be injective where R is given by the cofinite topology. Show that f is continuous.
so i know that the cofinite topology consists of sets that are either empty or have finite complements. in the case of R, the open sets in the topology are intervals containing + and - infinite while leaving out finitely many terms. to prove that f is continuous i have to show that given any open set V in the codomain, the preimage f^-1 (V) in the domain is also open. i also know that in order for a set to be open in R, there must exist a ball centered at each point and contained within the set for every point in that set.
i can see intuitively that since f is injective, the cardinality of V is less than or equal that of the preimage of V. and since V is open the preimage of V should be open as well. however that was based on my intuition and probably does not constitute a formal proof. how would you show that the preimage of any V is open?
Yes, it is. Suppose is continuous and let be a closed set of , then is open, and so is open in , therefore is closed.
Conversely, suppose that closed in implies closed in . Consider an open set of , then is closed in and so, is closed in . Therefore is open in i.e. is continuous.
Fernando Revilla
thanks for the proof!
now to prove this i need to show that any preimage of a closed set V in R is closed as well. the closed sets in this case are finite sets and since f is injective, the preimage of f must contain the same number of elements or less than the closed sets in the codomain. so the preimages are finite as well and since a ball around any point in a finite set in R necessarily is not contained in the finite set, the preimages must be closed sets. therefore f is continuous.
would this be the correct reasoning?