Show that $\left\{z:|z|\leq 3, |Re(z)|\geq 1\right\}$ is not connected.

Question

Show that $\left\{z:|z|\leq 3, |Re(z)|\geq 1\right\}$ is not connected.

I already showed this set was compact by the Heine-Borel theorem. Visually, I imagine this set to look like a "half circle" centered at the origin with $Re(z)\in [1,3]$. Am I even right? Anyway, how do I show this set is not connected? Do I use the closure of its open version and path-connectedness? Any tips are appeciated!

Answer 1

Write

$$C:=\{\;z\in\Bbb C\;|\;|z|\le3\,,\,\,|\text{Re}\,z|\ge1\;\}$$

and observe for example that $\;-1,\,1\in C\;$, yet any path connecting this two points must necessarily intersect the imaginary axis (the $\;y\,-$ axis) and there the real part of numers equals zero and thus cannot be in $\;C\;$ , meaning $\;C\;$ isn't path connected.

Now, you can also take $\;A:=\{\,z\in C\;|\;\text{Re}\,z\ge1\,\}\;$ and $\;B:=\{\,z\in C\;|\;\text{Re}\,z\le1\,\}\;$ . Then both $\;A,B\;$ are non-empty closed in $\;C\;$ and $\;A\cap B=\emptyset\;,\;\;A\cup B=C\;$ , so $\;C\;$ isn't connected, either.

Answer 2

Let $C$ be your set. Consider a function $f: C \to \{-1, 1\}$, $f(z) = \frac{\operatorname{Re} z}{|\operatorname{Re} z|}$. $f$ is well defined, since $\operatorname{Re}z \ne 0$ for all $z \in C$. It is continuous, and onto a disconnected discrete set, hence the domain cannot be continuous, since the image of connected set is connected.