Let $d$ be the usual Euclidean metric on $\Bbb R$ and $f: (\Bbb R, d) \mapsto ([-1,1],d)$ be the map given by $f(x) = \left\{ \begin{matrix} 0, & x=0 \\ \sin \frac 1 x, & x \neq 0 \end{matrix} \right.$.
I need to show that $f$ is not continuous at $0$.
My attempt:
Suppose $f$ is continuous at $0$.
Since $f$ is continuous at $0$ then $\exists$ $N(x,\epsilon)$ such that $N(x,\epsilon)$ $\subset$ $f^{-1}([-1,1])$. Take $\epsilon = \frac 1 2$. Then $N(x,\frac 1 2) \subset f^{-1}([-1,1])$. However, the $N(-\frac 1 2, \frac 1 2)$ contains points outside of $f^{-1}([-1,1])$.
My question is this: What is the best method to prove this? By best I mean intuitively, not necessarily concisely. My goal was to find a neighborhood that was not included in the inverse but I feel that I messed up somewhere on the way.
Any advice is appreciated!
There is a much simpler way. Consider the two sequences $x_n = \dfrac 1 {2n \pi}$ and $y_n = \dfrac 1 {2n\pi + \frac \pi 2}$. They both tend to $0$, but $f(x_n) = \sin 2n\pi = 0 \to 0$ and $f(y_n) = \sin (2n\pi + \frac \pi 2) = 1 \to 1$, which shows that $f$ doesn't even have a limit in $0$, let alone be continuous there.