Prove that $f$ is discontinuous on $\mathbb{R}$

Question

$$f(x)=\begin{cases} x+1, & \text{if $x$ is irrational} \\ x, & \text{if $x$ is rational} \end{cases}$$

Show that $f$ is discontinuous on $\mathbb{R}$

edit: It would be preferable if this could be proved without the use of sequences or epsilon-delta

Answer 1

Let $r$ be a rational.

we have $\;f(r)=r$

and $\;\; \forall n>0$,

$$f(r+\frac{\sqrt{2}}{n})=r+\frac{\sqrt{2}}{n}+1$$

thus

$$\lim_{n\to+\infty}f(r+\frac{\sqrt{2}}{n})=r+1\neq f(r)$$

which means that $\;f\;$ is not continuous at $x=r$.

Let $s$ be an irrational.

we have $f(s)=s+1$ and

$$s=\lim_{n\to+\infty}\frac{\lfloor 10^n s \rfloor}{10^n}$$

but

$$\lim_{n\to+\infty}f(\frac{\lfloor 10^n s \rfloor}{10^n})=\lim_{n\to+\infty}\frac{\lfloor 10^n s \rfloor}{10^n}$$

$=s\neq f(s)$ .

$f$ is not continuous at $x=s$.

Answer 2

Regardless of delta we can find a y within delta of x that is more than distance 1 from x (there are rationals and irrationals arbitrarily close to x). But then there exists no delta corresponding to epsilon less than 1.