# Sequence proof for discontinuity

• tjackson3
In summary, the function h(x) = x for rational numbers and h(x) = 0 for irrational numbers is continuous at x=0 and nowhere else. This can be proven by considering sequences of rational and irrational numbers converging to a point a, and showing that they converge to different values for all a≠0. Therefore, the function is discontinuous at all other points in its domain.
tjackson3
[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!

tjackson3 said:

## 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.
Yes, that is basically what you need to do. For a not equal to 0, take a sequence of rational numbers, ri, converging to a. Because those are rational numbers, f(ri)= ri and those converge to what? Take a sequence of irrational numbers, si, converging to a. Because those are irrational numbers, f(si)= 0. What does that sequence converge to? For what a are those the same?

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!

Last edited by a moderator:
Great, thanks!

## 1. What is a sequence proof for discontinuity?

A sequence proof for discontinuity is a method used in mathematics to show that a function is discontinuous at a certain point. It involves finding two sequences that approach the point from different directions and showing that the limit of the function at that point does not exist.

## 2. Why is a sequence proof for discontinuity important?

A sequence proof for discontinuity is important because it provides a rigorous and concrete way to prove that a function is discontinuous. It is also useful in identifying the type of discontinuity (removable, jump, or infinite) at a particular point.

## 3. How do you construct a sequence proof for discontinuity?

To construct a sequence proof for discontinuity, you need to find two sequences (xn and yn) that approach the point of discontinuity from different directions (e.g. xn approaches from the left and yn approaches from the right). Then, you need to show that the limit of the function as xn and yn approach the point is different, or that it does not exist.

## 4. What are the limitations of a sequence proof for discontinuity?

A sequence proof for discontinuity can only be used to prove that a function is discontinuous at a point. It cannot be used to prove continuity or discontinuity on an interval or at multiple points. Additionally, it may not work for certain types of discontinuities, such as oscillating or non-monotonic functions.

## 5. Can a function be discontinuous at every point?

Yes, a function can be discontinuous at every point, meaning that it has no points of continuity. This is known as a nowhere continuous function. It can be proven using a sequence proof for discontinuity by showing that for every point, there exist two sequences that approach the point with different limits.

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