# Which of the following are necessarily true?

Let $f : \Omega \rightarrow : \mathbb C$ be an analytic function on an open set $\Omega \subseteq \mathbb C$. For $r > 0, \mathbb D_r = \{ z \in \mathbb C : |z| <r \}$ and $\overline { \mathbb D }$ be its closure . Which of the followings are necessarily true ?

1. If $\overline {\mathbb D_1} \subset f(\Omega)$ , then $\mathbb D_r \subset f(\Omega )$ for some $r >1$.

2. If $\overline {\mathbb D_1} \subset f(\Omega)$ , then $\mathbb D_r = f(\Omega )$ for some $r >1$.

3. If $\overline {\mathbb D_1} \subset f(\Omega)$ , then $\overline {\mathbb D_r} \subset f(\Omega )$ for some $r >1$.

4. $f(\Omega )$ is open.

By open mapping theorem $f(\Omega )$ is open (4) is true. Please help me how to verify other options. Thank you

• The set $\overline {\mathbb D_1}$ is compact and $f(\Omega)$ is open. This helps with 3 and hence 1. You should be able to create a simple counterexample for 2. – copper.hat Jun 10 '15 at 16:49
• how to prove (3). – user120386 Jun 10 '15 at 17:05

Suppose for all $n$, we have $\overline {\mathbb D_{1+{1 \over n}}} \not\subset f(\Omega )$. Since they overlap, we see that the sets $K_n = \overline {\mathbb D_{1+{1 \over n}}} \setminus f(\Omega )$ are non empty, closed, bounded and nested.
Then we can find some $x_n \in K_n$, and since $K_1$ is compact, we can find a convergent subsequence $x_{n_k} \to x$. Since $x_m \in K_n$ for all $m \ge n$, we see that $x \in K_n$ for all $n$, and so $x \in{\mathbb D_{1}}$. Since $f(\Omega)^c$ is closed, we have $x \in f(\Omega)^c$, which contradicts $\overline {\mathbb D_{1}} \subset f(\Omega)$.
• You have to prove that $x \in \overline{ \mathbb D_1 } \subset f(\Omega)$ by nested property ., How can you say that $x \in f(\Omega)^c$ – user120386 Jun 10 '15 at 23:46
• By definition of $K_n$. The $\overline {\mathbb D_{1+{1 \over n}}}$ are nested, hence so are the $K_n$. if $x_m \in K_n$ then $x_m \in \overline {\mathbb D_{1+{1 \over n}}}$ and $x \in f(\Omega)^c$. – copper.hat Jun 10 '15 at 23:59
• You need to actually try. Take $\Omega = \mathbb{D}_2$ and $f(z) = z+1$. Then $f(\Omega) = \mathbb{D}_2 + \{1\}$. – copper.hat Jun 11 '15 at 0:35