Proof that $\sin$ is continuous by open sets on $\mathbb{R}$ with the usual topology The open set definition of continuity is:
$f:A \to B$ is continuous $\iff$ $U_{B}\in\tau_{B} \implies f^{-1}U\in\tau_{A}\ \forall U_B$, where $\tau_A$ and $\tau_B$ are the topologies of $A$ and $B$.
I believe that in the usual topology on $\mathbb{R}$ this reduces to:
$f:\mathbb{R}\to\mathbb{R}$ is continuous $\iff$ $\forall\epsilon\exists\delta$ s.t. $|f(x)-f(y)|<\epsilon \implies |x-y|<\delta$,
since $U_{B} = \{f(x) : |f(x)-f(y)|<\epsilon\}$ and $U_A = \{x : |x-y|<\delta\}$.
My textbook (Nakahara) makes it very clear the the converse definition is not true; i.e., you can't just show that open sets in the domain map to open sets in the range, you must show that the inverse image of an open set in the range is an open set in the domain.
I'm trying to prove that $\sin:\mathbb{R}\to\mathbb{R}$ is continuous. Now, I can easily do the following by choosing $\delta = \epsilon/2$:
$$
|x-y|<\frac{\epsilon}{2} \\
2\left|\frac{x-y}{2}\right| < \epsilon \\
2\left|\sin\left(\frac{x-y}{2}\right)\right| < \epsilon \\
2\left|\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\right| < \epsilon \\
\left|\sin x - \sin y\right| < \epsilon
$$
But I'm operating under the impression that I must show the reverse, and it's not at all clear to me that the step:
$$
2\left|\sin\left(\frac{x-y}{2}\right)\right| < \epsilon \implies 2\left|\frac{x-y}{2}\right| < \epsilon
$$
is true in this direction (although it naturally follows the other way around).
So my questions are:


*

*Am I correct in thinking that I need to show the converse?

*If so, how do I do it?

*If not, what is the meaning of Nakahara's statement?


If you want to use the "inverse image of open sets are open" definition of continuity you could begin by noting that there are three types of basic open intervals in $[-1,1]$


*

*$(a,b)$ where $-1<a<b<1$

*$(a,1]$

*$[-1,b)$
Then point out the the inverse image under $\arcsin$ of each of these three types of basic open set is a countable union of open intervals in $\mathbb{R}$.
Then let $U$ denote an open set in $[-1,1]$.
Then for each $y\in U$ there is a basic open set $V$ of one of the three types above such that $p\in V\subseteq U$, and the inverse image of $V$ under the $arcsin$ function is an open set containing the inverse images of $y$.
Thus the inverse image of $U$ is open in $\mathbb{R}$.
