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