# Proving a function is discontinuous on Q and continuous on R\Q

• michael.wes
In summary, the conversation discusses using an epsilon-delta argument to prove continuity on R\Q and the need to consider all sequences, both rational and irrational. It is noted that there are examples of sequences that converge to an irrational number while alternating between rational and irrational numbers, and the question is raised if there is an error of omission in only considering rational sequences. The answer clarifies that it is necessary to consider all sequences and provides an explanation for how to use this fact in the proof.
michael.wes
Gold Member
I have cast my post in LaTeX here:

http://www.michaelwesolowski.com/asdjhf.pdf

Any and all help is appreciated!

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Don't use the hint directly if you find it's confusing to use it. If f(x)>=1/M then that means x=n/m where m<=M, right? For x in (0,1) that's a finite set of n/m for a given M. So min(|alpha-n/m|) is greater than 0 if alpha is irrational since the set is finite. Use that to construct an easy epsilon delta argument.

I'm still spinning my wheels for the first part.. but I will ask my next substantive question now.

We need to show that $$f(x_n)\rightarrow f(\alpha) = 0$$ for all sequences $$x_n, \ where \ x_n \rightarrow\alpha \ as \ n\rightarrow\infty$$ in order to prove continuity on R\Q. I didn't ask this in the original post: why is it sufficient to consider all rational sequences only? Irrational sequences obviously don't need treatment, as their image is zero and thus they converge to zero immediately.

But aren't there sequences that have not all rational and not all irrational for all natural numbers n? I can't think of any that would converge to an arbitrary irrational number, but consider $$x_n=1/(\sqrt{2})^n$$. With a little work we can show this converges to zero (rational), but alternates between rational and irrational numbers.

Another example: we know that the sum of a rational and an irrational is irrational. Consider the sequence $$x_n=\frac{1+((-1)^n+(-1)^{2n})\sqrt{2}}{n}$$. This sequence also converges to zero, but alternates between rational and irrational numbers.

My point is: is there an error of omission? I realize there are pure irrational and pure rational subsequences to each of these examples, but they still make me think about whether we only need consider rational sequences overall.

If you are going to do it by sequences, you need to consider ALL sequences. As I said before, there are are only a finite number of points of the form n/m in (0,1) where m<=M for some fixed M. For any other point x, f(x)<1/M be it rational or irrational. If you take delta to be the minimum of |alpha-n/m| then any point within a distance of delta from alpha must have a value of f less than 1/M. Does that help?

## 1. What does it mean for a function to be discontinuous on Q and continuous on R\Q?

A function is considered discontinuous at a point if it fails to meet the criteria for continuity at that point. In this case, a function is said to be discontinuous on Q if it is not continuous at any rational number, and continuous on R\Q if it is continuous at all irrational numbers.

## 2. How do you prove that a function is discontinuous on Q and continuous on R\Q?

To prove that a function is discontinuous on Q and continuous on R\Q, you must show that the function does not meet the criteria for continuity at any rational number, and meets the criteria for continuity at all irrational numbers. This can be done by using the definition of continuity and providing an example or counterexample at each type of number.

## 3. What is the importance of proving a function is discontinuous on Q and continuous on R\Q?

Proving a function is discontinuous on Q and continuous on R\Q is important because it helps to understand the behavior of the function at different types of numbers. It also allows for a more precise analysis of the function, as it can help identify where the function may have discontinuities and how it behaves near these points.

## 4. Can a function be discontinuous on Q and continuous on R\Q without having any points of discontinuity?

Yes, it is possible for a function to be discontinuous on Q and continuous on R\Q without having any points of discontinuity. This can occur if the function is defined differently at rational and irrational numbers, but still meets the criteria for continuity at these points. For example, a function that is defined as f(x) = x at all irrational numbers and f(x) = 0 at all rational numbers would be discontinuous on Q and continuous on R\Q.

## 5. How does proving a function is discontinuous on Q and continuous on R\Q relate to the concept of density?

The concept of density refers to the idea that between any two points on a number line, there is always another number. In the case of proving a function is discontinuous on Q and continuous on R\Q, the function is defined differently at rational and irrational numbers, which are both dense on the number line. This illustrates the importance of understanding the behavior of a function at these types of numbers and how it can affect the overall continuity of the function.

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