What is $\sin^{-1}(]\sin(x)-\varepsilon,\sin(x)+\varepsilon[)$?

Question

The subject is : prove that $\sin:\mathbb{R}\to\mathbb{R}$ is continuous using this definition: $$ \forall W\in \mathcal{V_{\sin(x)}}, \sin^{-1}(W)\in \mathcal{V}_x $$

let $\varepsilon>0, W=]\sin(x)-\varepsilon,\sin(x)+\varepsilon[$

So I have to find $ \sin^{-1}(]\sin(x)-\varepsilon,\sin(x)+\varepsilon[)$ and see if it is a neighborhood of $x$ in $(\mathbb{R},|.|)$

$$ \sin^{-1}(]\sin(x)-\varepsilon,\sin(x)+\varepsilon[)= \{y\in \mathbb{R}, \sin(y)\in ]\sin(x)-\varepsilon,\sin(x)+\varepsilon[\}$$

can I say that $\sin^{-1}(\sin(x)-\varepsilon)<y<\sin^{-1}(\sin(x)+\varepsilon)$ ?

But how to continue?

Thank you

Answer 1

Because $|\sin(x)-\sin(y)| \leq |x-y|$, $\sin^{-1}((\sin(x)-\varepsilon,\sin(x)+\varepsilon))$ contains $(x-\varepsilon,x+\varepsilon)$, and thus is a neighborhood of $x$ according to the definition of "neighborhood" I have in mind. Does this match the definition you have in mind? If not then I will edit.