Non existence of function which is continuous only at rationals

Question

I want to understand why there does not exist a function continuous only at rationals. By Google search I know that the subset of the domain on which a function can be continuous is $G_{\delta}$. And since set of rationals is not a $G_{\delta}$ set such a function does not exist. Can anyone please explain it more clearly, i.e. why the set on which a function can be continuous should be a $G_{\delta}$ set. Thank you

Answer 1

Given a function $f:\mathbb R\to\mathbb R$, for each $n\in\mathbb N$ let $A_n$ be the union of all open sets $U$ whose image set $f(U)$ has diameter at most $\frac1n$. Then $A_n$ is an open set, and $\bigcap_{n\in\mathbb N}A_n$ is a $G_\delta$ set. Observe that $f$ is continuous at a point $a$ in $\mathbb R$ if and only if $a\in\bigcap_{n\in\mathbb N}A_n$, that is, for each $n\in\mathbb N$ the point $a$ has an open neighborhood whose image has diameter at most $\frac1n$.

The same goes for a function $f:X\to Y$ where $X$ is any topological space and $Y$ is any metric space.

Answer 2

Let $A_n=\inf \{x: g(x) <\frac 1 n\}$ where the oscillation $g$ is defined by $g(x)=inf _{\delta >0} \sup \{|f(x)-f(y)|: |y-x | \leq \delta\}$. Then each $A_n$ is open and $f$ is continuous at $x$ iff $g(x)=0$ iff $g(x) <\frac 1 n$ for all $n$ iff $x \in \cap_n A_n$.

[$g(x) <t$ implies there exists $\delta >0$ such that $\sup_{|y-x| \leq \delta} |f(x)-f(y)|<t$. You can verify that $|x-x'|<\delta/2$ implies $g(x') <t$. This proves that each $A_n$ is open].