Let $f:\Omega \to \mathbb C$ be an analytic function such that $\Omega $ is an open set.Define $D_r=\{z:|z|<r\}$ for some $r>0$.
Prove that if $\overline {D_1}\subset f(\Omega ) $ then $\overline {D_r} \subset f(\Omega )$ for some $r>1$.
I thought that since $f$ is an analytic function ,it must map an open set into an open set ,so $f(\Omega )$ is an open set.
Now $\overline {D_1}\subset f(\Omega ) $ but how can I figure out the $r>1$ for which $\overline {D_r} \subset f(\Omega )$.
This is my first question,I have tried to abide my the rules of this site as far as I could.Hope I will get some help on this problem.
For $z \in \overline{D}_1$ define a function $\alpha(z) = d(z,f(\Omega)^c) = inf \lbrace d(z,w) | w\in f(\Omega)^c \rbrace $ on $\overline{D}_1$. Now $\alpha$ is a continuous function on $\overline{D}_1$(This just follows from triangle inequality). Observe that $f(\Omega)^c$ is closed and hence $\alpha(z) > 0$(if $\alpha(z) = 0$ then \exists $w_1 \in f(\Omega)^c$ such that $d(z,w_i) \rightarrow 0$. triangle inequality says that $w_i$ form a cauchy sequence and hence converge to a point in $f(\Omega)^c$ say $w \in f(\Omega)^c$ since $f(\Omega)^c$ is a closed subset of a complete metric space that is $\mathbb{C}$. Now $d(z,w) = 0$ and thus $z = w \in \overline{D}_1 \cap f(\Omega)^c = \emptyset$, a contradiction). Now since $\overline{D}_1$ is a compact set, the function $\alpha$ has a minima say $\delta > 0$ since $\alpha$ is a strictly positive function. Let $r = 1 + \delta/2$. Then it is easy to see that $\overline{D}_r \cap f(\Omega)^c = \emptyset$. Hence $\overline{D}_r \subset f(\Omega)$.