# 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 $\displaystyle X = \{(z,w) \in C^2 : z \neq w \}$ with the topology induced from $\displaystyle C^2$.

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

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

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

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