My question is best stated by using an example: Suppose f is a function defined only for rational x, and for rational x f(x) = 1. Say we want to prove that f is continuous at x = 1. Then we want to show that for every positive epsilon there exists a delta > 0 such that [tex] |f(x) - f(1)| < \epsilon [/tex] if [tex] |x-1| < \delta [/tex]. My question is, every open interval about x = 1 contains irrational x values where f isn't defined. If we consider only rational x, [tex] f(x) - f(1)| = 0 < \epsilon [/tex] and so it seems to be continuous. But what about the irrational values? I'd say that my function f actually is indeed continuous, because [tex] |f(x) - f(1)| < \epsilon [/tex] must hold for all x in the interval [tex] (1-\delta, 1 + \delta) [/tex], where x is IN THE DOMAIN OF F. Since only rational values are in the domain of f, we consider only those x values. Is this right?