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 $\displaystyle x_{n}=\bigg(1+\frac{1}{n}\bigg)^{n}$ for all $\displaystyle n\in\mathbb{N}$, then we know that $\displaystyle x_{n}\in\mathbb{Q}$ for all $\displaystyle n\in\mathbb{N}$ and $\displaystyle x_{n}\to\mathrm{e}\in\mathbb{Q}^{c}$ as $\displaystyle n\to\infty$.
Therefore, we have $\displaystyle f(x_{n})=1$ for all $\displaystyle n\in\mathbb{N}$ but $\displaystyle \lim\nolimits_{n\to\infty}f(x_{n})=f(\mathrm{e})=0 \neq1$.
Thus $\displaystyle f$ is not continuous at $\displaystyle \mathrm{e}$.
However, you can give a good proof for the general case since for any $\displaystyle \xi\in\mathbb{R}$, you may find $\displaystyle \{\sigma_{n}\}_{n\in\mathbb{N}}\subset\mathbb{Q}$ and $\displaystyle \{\zeta_{n}\}_{n\in\mathbb{N}}\subset\mathbb{Q}^{c }$ such that $\displaystyle \sigma_{n},\zeta_{n}\to\xi$ as $\displaystyle n\to\infty$.
I hope this can help you...
You can solve this using the facts that:
(a) for any $\displaystyle s,t \in \mathbb{R}, \exists r \in \mathbb{Q} \ : \ s < r < t$
(b) for any $\displaystyle r \in \mathbb{Q}^c$, there exists a sequence $\displaystyle \left\{a_n\right\}_{n=1}^{\infty}$ such that $\displaystyle \forall n \in \mathbb{N}, a_n \in \mathbb{Q}$ and $\displaystyle a_n \rightarrow r$
That is to say, for any irrational number there is a sequence of rationals that converges to it.