Prove a set is a region

Question

Let $\Bbb C$ and $S$ be the set of complex numbers and $\{z:z=yi, |y|\geq b>0\}$ respectively, where $y$ and $b$ are real numbers. If $A=\Bbb C-S$ then how do I prove that $A$ is a region? A region should be connected and open but is there any theorem to prove it quicker?

Answer 1

I think it's easiest to do this directly; there's no need for any fancy theorems. It shouldn't be too hard to see from the picture that this is a region, and if it's not clear, it would probably be better to write out a direct argument based on the picture to get some intuition (this seems to be largely an exercise in working with the complex topology and mathematical formalism in general; i.e., checking all the conditions in the definition of "region" are satisfied and so on).

$A$ is open because it's the complement of two closed [infinite] line segments. It isn't too hard to explicitly find an open ball contained within $A$ about any point $p\in A$, if you're not satisfied with this. Say $p = x + iy$, $x = 0$. Then $\left|y\right| < b$, so take $B(p,b - \left|y\right|)$ ($B(p,r)$ is the open ball of radius $r$ centered at $p$ in $\Bbb C$; i.e., $B(p,r) = \{z\in\Bbb C\mid \left|z - p\right| < r\}$). If $x < 0$ and $\left|y\right| < b$, take $B(p, \min(\left|x + iy - ib\right|,\left|x + iy + ib\right|))$. If $x < 0$ and $\left|y\right|\geq b$, take $B(p, \left|x\right|)$. You can treat the other cases similarly, and I leave it to you to verify that these open balls are contained within $A$ in each case.

To see that $A$ is connected, you can show that $A$ is path-connected. To connect two points in the same (left or right) half plane (or if both lie on the line $x = 0$), just draw the line segment connecting them. If they're in different half planes, draw the line segment from each to $0$ -- the combination of the segments will give you a path from one to the other.