Proof that $\sin$ is continuous by open sets on $\mathbb{R}$ with the usual topology

Question

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:

1. Am I correct in thinking that I need to show the converse?
2. If so, how do I do it?
3. If not, what is the meaning of Nakahara's statement?

Answer 1

Let $f:X\rightarrow Y$.

The "open set definition" says that $f$ is continuous if for all open $U\subset Y$, $f^{-1}U$ is open.

The "$\epsilon-\delta$ definition" says that $f$ is continuous if for all $a$ in $X$, and for all $\epsilon > 0$, there exists $\delta > 0$ such that $|a-b|<\delta \Rightarrow |f(a) - f(b)|<\epsilon$ for $b$ in $X$.

This means your first line (where you defined continuity) is incorrect.

However, your proof that $\sin:\mathbb{R}\rightarrow \mathbb{R}$ is continuous looks correct!

There is no need to show the converse, as it is clearly not true because $\sin(x) - \sin(n\pi x) = 0$ for any $n\in \mathbb{N}$. The converse has nothing to do with continuity of $\sin$ anyway.

EDIT Using the open set definition:

Suppose $U\subset \mathbb{R}$ is open. This is not necessarily an interval in $\mathbb{R}$ (although it is a countable union of intervals). We know that arbitrary unions of open sets are open. Therefore, suppose we can show that the inverse image $f^{-1}U_i$ of every open interval $U_i$ is open. Then any open set $U$ in $\mathbb{R}$ is equal to $\bigcup_{i\in K} U_i$ for some set $K$. Hence the inverse image of $U$ is $\bigcup_{i\in K} f^{-1}U_i$ which is a union of open sets, which is open. Therefore it suffices to just consider open intverals.

Let $U\subset \mathbb{R}$ be an open interval. I will show that $\{x: \sin(x) \in U\}$ (the inverse image of $U$) is open. Remember that a set is open if there is an open ball contained in the set, centered at any point in your set. For $\mathbb{R}$ this is saying a set is open if given any point in the set, you can find an open interval containing that point which is also contained in the set.

Let $c\in \{x: \sin(x) \in U\}$. This means $\sin(c) \in U$. Because $U$ is open, there is an interval $(\sin(c) - \epsilon, \sin(c) + \epsilon)$ contained in $U$. But now this is equivalent to $|\sin(c) - k|=|\sin(c) - \sin(y)| <\epsilon$, and the question is "can you find a $\delta$ so that $|c-y|<\delta$ makes this true?" The proof ends up becoming the same even though I started with a different approach.

Answer 2

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 $[-1,1]$

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

2. $(a,1]$

3. $[-1,b)$

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 $\mathbb{R}$.