The function $f:\Bbb R\rightarrow \Bbb R$ defined with:
$$f(x)= \left\{ \begin{array}{}
1, &x\in \mathbb{Q} \\
0, &x\not\in\mathbb{Q}
\end{array}\right.$$
is not continuous.Let $c\in\Bbb Q$ and $f(c)=1$. Let a sequence $(c_n)_n$ from $\Bbb R\setminus \Bbb Q$ which converges to $c$. Then $f(c_n)=0$ so the sequence $(f(c_n))_n$ doesn't converge to $1$. Analogously, for $c\not\in \Bbb Q$ and $f(c)=0$ let $(c_n)_n$ be a sequence from $\Bbb Q$ which converges to $c$. Then $f(c_n)=1$ so the sequence $(f(c_n))_n$ doesn't converge to $0.$
What's the idea behind all this? I don't see how this proves that a function is not continuous. I have the following definition:
Let $I\subseteq \Bbb R$ be an open interval, $c\in\ I$ and a function $f:I\rightarrow \Bbb R$. The function $f$ is continuous at a point $c$ if and only if
$$(\forall \varepsilon>0)(\exists \delta>0)(\forall x\in I)((|x-c|<\delta)\Rightarrow(|f(x)-f(c)|<\varepsilon)).$$
Then if we want to prove that a function is not continuous, doesn't it mean we should prove that:
$$(\exists \varepsilon>0)(\forall\delta>0)(\exists x\in I)((|x-c|<\delta)\land(|f(x)-f(c)|\geq \varepsilon))?$$
Best Answer
I'll prove what you are asking, the way you are asking, but only for $c \in \mathbb Q$. For irrationals, a similar argument follows.
You want to find an $\epsilon>0$ such that $\forall \delta > 0$, there is an $x \in I$, such that $|x-c| < \delta$ but $|f(x)-f(c)| > \epsilon$.
So we will take $\epsilon = \frac 12$.
For any $\delta > 0$, pick an irrational number $x \in (c, c - \delta)$ (we can do this however small $\delta$ is, as long as it is positive). Then, since $x \notin \mathbb Q, f(x) = 0$, so $|f(x)-f(c)| = 1 > \epsilon$ although $|x - c| < \delta$. This shows discontinuity at $c \in \mathbb Q$.
The is called the epsilon-delta definition of continuity, while the definition being used there is called the sequential definition of continuity. You can show these are equivalent, but first read them up.
The sequential definition is the following: $f$ is continuous if whenever $x_n$ is a sequence, $x_n \to c$, then $f(x_n) \to f(c)$. ($\to$ means "converges to").
So in this case, you can find an sequence of irrationals $c_n$ converging to $c$, however $f(c_n) = 0$ in that case, but $f(c) = 1$, and clearly the constant sequence $0$ doesn't converge to $1$, which is a contradiction. This is what the author is trying to say.
Hope you have understood. Do reply back.