Prove that $f(x)$ is not continuous at any point Let $f:R->R$ be defined by 
\begin{equation} f(x)= \left\{ \begin{array}{cc} [x] & \text{if} & x \quad \text{rational}\\\\ [x]+3 & \text{if} &x \quad \text{irrational}\\ \end{array} \right. \end{equation}
where $[x]$ is the greatest integer less than or equal to $x$. 
(Example: $[0.5]=0$, $[2]=2$, $[-0.2]=-1$)
Prove that $f(x)$ is not continuous at any point. I know I have to use $Sn$ as a sequence and $C$ as an integer. The sequence I'm using is $Sn=C+\frac{(-1)^n}{n}$. Case 1 will be when $C$ is an integer and case 2 will be when $C$ is an arbitrary point. So far I have:
Let $Sn$ be a sequence such that $lim Sn=C$ where $C$ is not an integer. Then $lim[Sn]=[C]$.
 A: If your function is continous then:
$\forall \epsilon >0, \exists \delta > 0 : |x-a|<\delta \implies |f(x) - f(a)|<\epsilon$ must be true.
Let $\epsilon = \frac 12$
for all $x,$ and all $\delta>0$ there is both an irrational and a rational value for $a\in (x, x+\delta)$
And for one of those $|f(x) - f(a)| > \epsilon$
$f(x)$ is not continuous at any $x.$
A: Perhaps you are looking for something that involves a sequence $S_n \rightarrow C$, for which it is not the case that the sequence $f(S_n) \rightarrow f(C)$. In this case, I reckon that the two situations to deal with are $C \in \mathbb{Q}$ and $C \not\in \mathbb{Q}$. The result then follows from the following fact: for every real number $C$, there is both a sequence of only rational numbers that converges to $C$, and a sequence of only irrational numbers that converges to $C$. For example, if $C \in \mathbb{Q}$, then the constant sequence $C, C, C, \ldots \rightarrow C$ consists of all rational elements, whereas the sequence $C + \sqrt{2}/n \rightarrow C$ consists of all irrational elements. If $f$ were continuous, then we would need $f(C)$ to be both $[C]$ and $[C] + 3$, a contradiction. I will leave the case of $C \not\in \mathbb{Q}$ to you.
