The sequential definition is the following: $f$ is continuous if whenever $x_n$ is a sequence, $x_n \to c$, then $f(x_n) \to f(c)$. You can show these are equivalent, but first read them up. The is called the epsilon-delta definition of continuity, while the definition being used there is called the sequential definition of continuity. This shows discontinuity at $c \in \mathbb Q$. Then, since $x \notin \mathbb Q, f(x) = 0$, so $|f(x)-f(c)| = 1 > \epsilon$ although $|x - c| < \delta$. You want to find an $\epsilon>0$ such that $\forall \delta > 0$, there is an $x \in I$, such that $|x-c| \epsilon$.įor 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). For irrationals, a similar argument follows. I'll prove what you are asking, the way you are asking, but only for $c \in \mathbb Q$. The function $f$ is continuous at a point $c$ if and only if Let $I\subseteq \Bbb R$ be an open interval, $c\in\ I$ and a function $f:I\rightarrow \Bbb R$. What's the idea behind all this? I don't see how this proves that a function is not continuous. Then $f(c_n)=1$ so the sequence $(f(c_n))_n$ doesn't converge to $0.$
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 $f(c_n)=0$ so the sequence $(f(c_n))_n$ doesn't converge to $1$.
Let a sequence $(c_n)_n$ from $\Bbb R\setminus \Bbb Q$ which converges to $c$. The function $f:\Bbb R\rightarrow \Bbb R$ defined with:
0 kommentar(er)
