$f(x) = \begin{cases} 0 & x\in( \mathbb{Q}\cap [0,1])^c \\ \frac{1}{q} & x=\frac{p}{q} \in \mathbb{Q} \cap [0,1], (p,q)=1 \end{cases}.$

Question

let $f$ be the function defined by

$$f(x) = \begin{cases} 0 & x\in( \mathbb{Q}\cap [0,1])^c \\ \frac{1}{q} & x=\frac{p}{q} \in \mathbb{Q} \cap [0,1], (p,q)=1 \end{cases}.$$

Prove that $f$ is continuous at every irrational number in $[0,1]$, and it discontinuous at every rational number in $[0,1]$.

Is this function? I mean that $0=\frac{0}{5}=\frac{0}{6}$, so $f(0)$ may be equals $\frac{1}{5}$ or $\frac{1}{6}$, so $0$ have more than $1$ image!

“Sorry, I don’t speak English well”

Answer 1

This is a classic counter-example in mathematics, demonstrating that it is possible for the sets of continuous points and discontinuous points to both be dense.

Yes, it is a function. You are correct that $\frac 05 = \frac 06$, but $(0,5) = 5 \ne 1$ and $(0,6) = 6 \ne 1$, so neither one of them gives the value of $q$ needed. (If you are not aware, here "$(p,q)$" means the greatest common divisor of $p$ and $q$. This was once a very common notation, but I think it is losing popularity because it is too confusing, particularly in this context.)

The only $q > 0$ for which $(0,q) = 1$ is $q = 1$, so $f(0) = 1$.