$\mathbb{C} \setminus \{z \in \mathbb{C} \mid \operatorname{Re}(z)=0\}$ is not simply connected.

Question

Let $X := \mathbb{C} \setminus \{z \in \mathbb{C} \mid \operatorname{Re}(z)=0\}$ be the space in question.

It suffices to show that $X$ is not path-connected. For this, I reckon that $-1+0\mathrm{i}$ and $1+0\mathrm{i}$ cannot be connected with a continuous path within $X$ (since such a path would have to cross the imaginary axis, which is not in $X$). But how do I write this down rigorously?

Answer 1

If $\gamma\colon[a,b]\longrightarrow\Bbb C$ is a path such that $\gamma(a)=-1$ and that $\gamma(b)=1$, let $f=\operatorname{Re}\circ\gamma$. Then $f(a)=-1$ and $f(b)=1$ and therefore, by the intermediate value theorem, $f(t)=0$, for some $t\in(-1,1)$. But then $\gamma(t)\notin X$.

Answer 2

$X$ is not even connected: $U = \{z \in \Bbb C; \text{Re}(z) >0\}$ is open and $V = \{z \in \Bbb C; \text{Re}(z) < 0\}$ too, and the form a disjoint cover of $X$.

Answer 3

It is easier to show that $X$ is not connected directly. For this note that the open sets $\{z\in\mathbb{C}|\Re(z)>0\}$ and $\{z\in\mathbb{C}|\Re(z)<0\}$ are open in $X$ and cover the complete set $X$. Therefore $X$ is not connected.