My question is best stated by using an example:(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Continuity and Domain of Function

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