Continuity of Dirichlet-type function

[iloveannaw]

Homework Statement

$$f: ]0, \infty[ \rightarrow \mathbb{R}$$ is defined as

$$f = 0$$ if x is irrational and

$$f = \frac{1}{n}$$
if $$x = \frac{m}{n}$$
where m and n are co-prime

Show that f is only then continuous about x₀ when $$x \in \frac{\mathbb{R}}{\mathbb{Q}}$$

The Attempt at a Solution

with $$x, y \in \mathbb{Q}, f\left(x\right) = \frac{1}{n}$$

with $$\left|x - y\right| < \delta$$

$$\forall x,y \in \mathbb{Q} \exists z$$ such that

$$x < z < y$$ or $$x > z > y$$

such that

$$f\left(z\right) = 0$$ ,e.g. z is irrational

with the result that

$$\left|x - z\right| < \left|x - y\right| < \delta$$

and

$$\left|f\left(x\right) - f\left(z\right)\right| > \left|f\left(x\right) - f\left(y\right)\right| < \epsilon$$

for all $$x \in ]0, \infty[$$

is this on the right lines?

 

[mighty2000]

Yes, your solution is on the right track. To show that f is continuous at x0, you need to prove that for any given ε > 0, there exists a δ > 0 such that for all x ∈ ]0, ∞[, if |x - x0| < δ, then |f(x) - f(x0)| < ε. Your proof shows that for any given ε > 0, there exists a δ > 0 such that for all x, y ∈ ℚ, if |x - y| < δ, then |f(x) - f(y)| < ε, which is exactly what we need to show.

However, you should clarify that the existence of z is dependent on the choice of x and y, and that it is not necessarily true that there exists a single z that satisfies the conditions for all x and y. Also, you may want to mention that since x and y are both rational, their difference (x - y) is also rational, which means that the irrational z that you choose will always be between two rational numbers. This is important because it ensures that the function f is well-defined at z, since f is only defined for rational numbers.

Additionally, you could mention that for any given x ∈ ℚ, there exists a sequence of irrational numbers that converges to x, which means that for any ε > 0, there exists a δ > 0 such that for all irrational numbers z with |x - z| < δ, we have |f(x) - f(z)| < ε. This is important because it shows that f is also continuous at x when x is irrational, which is necessary for the function to be continuous on the entire domain, as stated in the problem.

Overall, your solution is correct, but it could benefit from some clarification and further explanation. Good job!