Continuity of $f(x)=\begin{cases}x, & x \in\Bbb{Q}\\0 & x \in\Bbb{Q}^c\end{cases}$

Question

I want to prove that the function $f:\Bbb{R}\to\Bbb{R}$ defined by $$f(x)=\begin{cases}x, & x \in\Bbb{Q}\\0 & x \in\Bbb{Q}^c\end{cases}$$

is not continuous at $x_0\in \Bbb{R},\;x_0\neq 0,$ but do not know how to solve the problem. Please, can anyone help me out?

Answer 1

Note that you can always find a sequence of rational numbers converging to an irrational number. For example

$$x_n = \frac{F_{n+1}}{F_n}$$

where $F_n$ is the $n^{\text{th}}$ Fibonacci number. We have $$x_n \rightarrow \phi=\frac{1+\sqrt{5}}{2}$$ where $\phi$ is the golden ratio

Therefore, if $f$ was continuous then $f(x_n) \rightarrow f(\phi)$, but $$f(x_n)=x_n, \ \ \ \ \ f(\phi)=0$$ therefore $f$ isn't continous

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Similarly, you can find a sequence of irrational numbers converging to a rational number, for example $x_n = \sqrt[n]{n} \rightarrow 1$, but on applying $f$ we get $f(\sqrt[n]{n}) = 0$ and $f(1)=1$

Answer 2

note that $\mathbb{Q}$ is dense in $\mathbb{R}$. You can always find a sequence of rational numbers that converge to a irrational number