- #1

tjackson3

- 150

- 0

**[SOLVED] Sequence proof for discontinuity**

## Homework Statement

Given a function h(x) = x for all rational numbers x and h(x) = 0 for all irrational numbers, prove that h(x) is continuous at the point x=0 and nowhere else.

## Homework Equations

A function is continuous at a point x0 if and only if for all sequences xn converging to x0, f(xn) converges to f(x0). Therefore, a function is discontinuous if there exists a sequence xn converging to x0 that does not cause f(xn) to converge to f(x0).

Also, because either a sequence proof or an epsilon-delta proof works here:

A function is continuous at a point x0 in dom(f) if |x-x0| < delta implies |f(x) - f(x0)| < epsilon

## The Attempt at a Solution

I'm having a hard time even figuring out where to begin on this. The Denseness of Q theorem states that for every a < b, where a and b are real numbers, there exists a rational number r such that a < r < b. The same applies for irrationals. Therefore, h(x) looks like two different functions, f(x) = 0 and g(x) = x, since Q and not Q are so densely populated. This means that it appears that there is an intersection between these two functions at x = 0.

For a sequence proof, I really don't have a grasp of how sequence proofs for discontinuity (or continuity, for that matter) work. I guess if you let xn = x and took the limit as x->0, that implies that the limit as h(x) approaches 0 is 0 just by looking at the graph, but I don't think that's analytical enough.

As far as an epsilon-delta proof goes, I'm not sure what to use as f(x) [or h(x) in this case]. I figure that you could say |x - 0| < delta implies |x - 0| < epsilon, meaning if you let delta = epsilon, |x| < delta implies |x| < epsilon. Is that sufficient?

That being said, there's also the part where I have to prove that it's not continuous for any other point in dom(h), but it's not defined enough to do that with epsilon-delta, and again, I don't understand the sequence definition enough to apply it.

Thanks for your help!