# Concerns about proof using sequences

Im working out this proof needed for the caratheodory - koebe theory.

The idea is quite simply to understand but there is an argument using sqeunces which im questioning about.

The statement is the following:

Let $$D \subset \mathbb{C}$$ a domain with $$0 \in D$$. $$f,g$$ holomorphic in $$D$$ with $$f,g$$ not constant. $$g$$ is injective, $$f(0) = 0$$ and $$\mid f(z) \mid \geq \mid g(z) \mid$$ for all $$z \in D$$.

Then $$\sup\lbrace r > 0 \mid B_r(0) \subset f(D) \rbrace \geq \sup\lbrace r > 0 \mid B_r(0) \subset f(D) \rbrace$$

This seems quite confusing on the first look but is actually quite easy to understand (at least when drawing a picture).

(Note: $$\sup\lbrace r > 0 \mid B_r(0) \subset f(D) \rbrace$$ is called the inner radius of $$f(D)$$ and just represents the supremum of the largest disk you could fit into $$f(D)$$ around $$0$$.

Proof:

We proove this by taking a disk $$B = B_r(0) \subset g(D)$$. This is actually possible because $$g(D)$$ has to contain $$0$$ because of $$f(0) = 0$$ and $$\mid f(z) \mid \geq \mid g(z) \mid$$. Also $$g(D)$$ has to be a domain because of the open mapping theorem ($$g \neq$$const). This means there is a disk fitting around $$0$$ in $$g(D)$$. We show the statement by showing $$B \subset f(g^{-1}(B))$$ (therefore $$g$$ has to be injective). This will follow from: Is $$b \in \partial f(g^{-1}(B))$$ then $$\mid b \mid \geq r$$.

After defining $$B$$ there has to be a sequence $$a_n \in B$$ with $$a_n \rightarrow a \in \partial B \quad (n \rightarrow \infty)$$ and $$f(g^{-1}(a_n)) \rightarrow b \quad (n \rightarrow \infty)$$.

That is my question. Why can we say that such sqequence has to exist?

Using this sequence we continue:

$$\mid b \mid = \lim \mid f(g^{-1}(a_n)) \mid \geq \lim \mid g(g^{-1} ( a_n)) \mid = \mid a_n \mid = r$$.

How can we say that there is such a sequence $$a_n$$ which converges in $$B$$ to $$a \in \partial B$$ AND $$\lim f(g^{-1}(a_n)) = b \in \partial f(g^{-1} (B))$$?