Complex open set proof

**Danneedshelp** · September 24th 2010, 09:47 AM

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 $|z-z_{0}|<r$ . 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

**Ackbeet** · September 24th 2010, 09:53 AM

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.

**Danneedshelp** · September 24th 2010, 10:08 AM

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

**Ackbeet** · September 24th 2010, 10:11 AM

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

**Defunkt** · September 24th 2010, 12:52 PM

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.

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