# Thread: Complex open set proof

1. ## Complex open set proof

Q: Let $\displaystyle U=\{z\in{\mathbb{C}}:0<|z|<1\}$. Prove that $\displaystyle U$ is an open set.

A: Let $\displaystyle z_{0}\in\\U$ by arbitrary. We need to find a $\displaystyle r>0$ such that the disk centered at $\displaystyle z_{0}$, $\displaystyle D(z_{0},r)$ is completely contained in $\displaystyle U$. So, let $\displaystyle r=min\{z_{0},1-z_{0}\}$. Then for any $\displaystyle z\in\\D(z_{0},r)$ we have that $\displaystyle |z-z_{0}|<r$. Therefore, $\displaystyle z\in\\U$ and we are done.

Is my choice of $\displaystyle r$ legit? It seams reasonable that disk with $\displaystyle r$ equal to the smaller of the two distance from the edge of the set will be in the set, since the neighborhood itself is open.

Thanks

2. Well, I assume you meant $\displaystyle r=\min\{|z_{0}|,1-|z_{0}|\},$ since you can't compare complex numbers. Other than that, and assuming you're using open disks there, I think your general idea works.

3. Originally Posted by Ackbeet
Well, I assume you meant $\displaystyle r=\min\{|z_{0}|,1-|z_{0}|\},$ since you can't compare complex numbers. Other than that, and assuming you're using open disks there, I think your general idea works.
Oh yeah, I forgot the vertical bars. Also, I am assuming that the disk is open.

Thanks

4. Great. I think you've got it. You're welcome, and have a good one!

5. Another way to do this using basic properties:

Let $\displaystyle U = \{z\in\mathbb{C}:0<|z|<1\}, \ \mathbb{S}^1 = \{z \in \mathbb{C}: |z|<1\}, \ \mathcal{O} = \{(0,0)\}$

Then $\displaystyle U = \mathbb{S}^1 \setminus \mathcal{O} = \mathbb{S}^1 \cap \mathcal{O}^c$, where $\displaystyle A^c$ is the complement of A.