1. ## Function prove

Show that the function f defined by f(x)=0 if x is irrational and f(x)=1 if x is rational does not have a limit at any point.

Any suggestions?

2. Originally Posted by paperclip
Show that the function f defined by f(x)=0 if x is irrational and f(x)=1 if x is rational does not have a limit at any point.

Any suggestions?
Let $x_{n}=\bigg(1+\frac{1}{n}\bigg)^{n}$ for all $n\in\mathbb{N}$, then we know that $x_{n}\in\mathbb{Q}$ for all $n\in\mathbb{N}$ and $x_{n}\to\mathrm{e}\in\mathbb{Q}^{c}$ as $n\to\infty$.
Therefore, we have $f(x_{n})=1$ for all $n\in\mathbb{N}$ but $\lim\nolimits_{n\to\infty}f(x_{n})=f(\mathrm{e})=0 \neq1$.
Thus $f$ is not continuous at $\mathrm{e}$.

However, you can give a good proof for the general case since for any $\xi\in\mathbb{R}$, you may find $\{\sigma_{n}\}_{n\in\mathbb{N}}\subset\mathbb{Q}$ and $\{\zeta_{n}\}_{n\in\mathbb{N}}\subset\mathbb{Q}^{c }$ such that $\sigma_{n},\zeta_{n}\to\xi$ as $n\to\infty$.

(a) for any $s,t \in \mathbb{R}, \exists r \in \mathbb{Q} \ : \ s < r < t$
(b) for any $r \in \mathbb{Q}^c$, there exists a sequence $\left\{a_n\right\}_{n=1}^{\infty}$ such that $\forall n \in \mathbb{N}, a_n \in \mathbb{Q}$ and $a_n \rightarrow r$