# Continuity and Domain of Function

1. Sep 30, 2009

### JG89

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 $$|f(x) - f(1)| < \epsilon$$ if $$|x-1| < \delta$$.

My question is, every open interval about x = 1 contains irrational x values where f isn't defined. If we consider only rational x, $$f(x) - f(1)| = 0 < \epsilon$$ 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 $$|f(x) - f(1)| < \epsilon$$ must hold for all x in the interval $$(1-\delta, 1 + \delta)$$, 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?

2. Sep 30, 2009

### Office_Shredder

Staff Emeritus
I don't know in this case.... where in the definition of continuity does it say you have to look at real values for x? You don't test complex values of x when determining continuity, but nothing says you shouldn't.

They may be referring to this function not as a function from the reals to the reals only defined on the rational functions (these functions are generally discontinuous, for example f(x) = x-pi fails to be 0 anywhere), but instead as a function defined on the rationals not embedded in the reals. So you lose a lot of properties of continuous functions (like the intermediate value theorem) but on the bright side you can still characterize it as continuous.

A lot of it depends on the context of the question

3. Sep 30, 2009

### lurflurf

Yes, you consider only values in the domain.
We can say f is continuous, or f is rational continuous.
That said rationals are a horrible field to do calculus on as they are not complete.
This is an artifact of all the garbage books with exercises like
What is the domain of f=log(1+x)
You tell me book it is your function.