# Thread: Complex open set proof

1. ## Complex open set proof

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

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

Is my choice of $r$ legit? It seams reasonable that disk with $r$ 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 $r=\min\{|z_{0}|,1-|z_{0}|\},$ 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 $r=\min\{|z_{0}|,1-|z_{0}|\},$ 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 $U = \{z\in\mathbb{C}:0<|z|<1\}, \ \mathbb{S}^1 = \{z \in \mathbb{C}: |z|<1\}, \ \mathcal{O} = \{(0,0)\}$

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