Proving $f : \Bbb R \to S^1 , f (t) = e^{2iπt}$ is an open map.

Question

Let $S^1=\{z\in\mathbb{C}\mid |z|=1\}$. Consider the map $f : \Bbb R \to S^1 , f (t) = e^{2iπt} = \cos(2πt) + i \sin(2πt).$

The map $f$ is onto and continuous. I was reading an online lecture note about proving $f$ is an open map and came across the following observation.

For every $x \in \Bbb R$ and $\varepsilon < \frac 12$ , the image $f(x − \varepsilon, x + \varepsilon)$ is the intersection of an open half-plane with the unit circle $S ^1$ .

I'm having trouble understanding the above statement. The definition of open-half plane found on Wolfram MathWorld is

A half-plane is a planar region consisting of all points on one side of an infinite straight line, and no points on the other side.

If the points on the line are included, then it is called an closed half-plane. If the points on the line are not included, then it is called an open half-plane.

But I cannot understand the importance of $\varepsilon$ to be less than $\frac12$ and the comment about $f(x − \varepsilon, x + \varepsilon).$

Any help is appreciated. Thank you.

Answer 1

Let $f(\nu)=e^{2\pi i\nu}$ and identify $\mathbb{C}\cong\mathbb{R}^2$. Our goal is to show $f(\theta-\epsilon,\theta+\epsilon)$ is the intersection of an open half plane and the unit circle $\mathbb{S}^1$. Assume $\theta=0$. Then we can see that $$f\left(-\epsilon,\epsilon\right)=\{e^{2\pi i\nu}\in\mathbb{S}^1:|\nu|<\epsilon\}\subset \{(x,y)\in\mathbb{R}^2:x>\cos(2\pi\epsilon)\}\cap \mathbb{S}^1$$ since $\Re [e^{2\pi i \nu}]=\cos(2\pi \nu)>\cos(2\pi\epsilon)$ for all $|\nu|<\epsilon(<\frac{1}{2})$. Since $H=\{(x,y)\in\mathbb{R}^2:x>\cos(2\pi\epsilon)\}$ is an open half plane, this proves the claim for $\theta=0$. Now, let $\theta\in\mathbb{R}$ be arbitrary and $R_\theta$ be the rotation of angle $2\pi \theta$. Then $R_\theta$ is an isometric homeomorphism on $\mathbb{R}^2$. Also $R_\theta$ maps $H$ to some open half plane $H_\theta$ and $\mathbb{S}^1$ to itself. Thus it holds $$ f(\theta-\epsilon,\theta+\epsilon) = R_\theta\left(f\left(-\epsilon,\epsilon\right)\right)\subset H_\theta \cap \mathbb{S}^1. $$ This shows the claim for general $\theta\in\mathbb{R}$.

Answer 2

Let $\epsilon=\frac{1}{2}$. Then $f(x-\epsilon,x+\epsilon)=f(x-\frac{1}{2},x+\frac{1}{2})$ has a pair of antipodal points on $\mathbb{S}^1$. Thus by the definition of closed-half plane, $f(x-\frac{1}{2},x+\frac{1}{2})$ is intersection of a closed half-plane with the unit circle $\mathbb{S}^1$ due to the antipodal points. I hope this is clear.