# Thread: Connected subsets of {(z,w) in C^2: z not equal to w}?

1. ## Connected subsets of {(z,w) in C^2: z not equal to w}?

Find the connected components of $X = \{(z,w) \in C^2 : z \neq w \}$ with the topology induced from $C^2$.

The trouble with his is that I can't visualise the space X. I tried the same question with $\Re$ instead of $C$ and got the sets $\{(x,y) \in \Re^2 : x > y \}$ and $\{(x,y) \in \Re^2 : x < y \}$, but can't see how to do the "complex version."

2. Originally Posted by HenryB
Find the connected components of $X = \{(z,w) \in C^2 : z \neq w \}$ with the topology induced from $C^2$.

The trouble with his is that I can't visualise the space X. I tried the same question with $\Re$ instead of $C$ and got the sets $\{(x,y) \in \Re^2 : x > y \}$ and $\{(x,y) \in \Re^2 : x < y \}$, but can't see how to do the "complex version."
The set $X$ is connected.

This is obvious if you think the following way: Let $(x,y),(z,w)\in X$ (imagine four points on the complex plane). They are path-connected in $X$ if there is a path $\gamma_1:[0,1]\to\mathbb{C}$ from $x$ to $z$ and a path $\gamma_2:[0,1]\to\mathbb{C}$ from $y$ to $w$ such that, for every $t\in[0,1]$, $\gamma_1(t)\neq \gamma_2(t)$ (this assures that the path $(\gamma_1,\gamma_2)$ keeps inside $X$). There are plenty of such paths... On $\mathbb{R}$, if $x and $z>w$, the paths had to meet for some $t\in(0,1)$ due to the intermediate value theorem.