Below is a proof of the Open Mapping Theorem:
I understand the proof, except I am unsure how we can guarantee that there exists a $\delta >0$ such that $f(z)\neq w_0$ for all $\delta = \mid(z-z_0)\mid$.
Here's my reasoning why this might be the case:
Suppose there was no such $\delta >0$. Then in particular there is a point $z_n$ on each circle $\frac{1}{n}= \mid(z-z_0)\mid$ for $n=N,N+1,...$ such that $f(z_n)=w_0$ (for some sufficiently large integer N). This is a bounded sequence in the closed disk $\frac{1}{N}=\mid(z-z_0)\mid$. Hence it has a convergent subsequence. f is constant on this subsequence and hence f must be constant everywhere in the disk because f is holomorphic.
Is this reasoning correct? Am I making it too complicated. Is there simpler explanation.