Question about open mapping theorem Let, $f:\Omega\to \mathbb C$ be a non constant anlytic function on an open set $\Omega \subset \mathbb C$.For $r>0$ let $\mathbb D_r=\{z\in \mathbb C:|z|<r\}$ and let $\bar{ \mathbb D_r}$ be its closure. Which of the followings are true ?
(A) If  $\bar{ \mathbb D_1}\subset f(\Omega)$ then $\mathbb D_r\subset f(\Omega)$ for some $r>1$.
(B) If $\bar{ \mathbb D_1}\subset f(\Omega)$ then $\mathbb D_r= f(\Omega)$ for some $r>1$.
(C) If $\bar{ \mathbb D_1}\subset f(\Omega)$ then $\bar{\mathbb D_r}\subset f(\Omega)$ for some $r>1$.
(D) $f(\Omega)$ is open.
As, $\Omega$ is an open subset of $\mathbb C$ , so by open mapping theorem $f(\Omega)$ is open. So, option (D) is true.
But I have no idea about the other options.
Please help...
 A: First note that (A) and (C) are equivalent.
The implication “(C) ⇒ (A)” is trivial, and the other direction “(A) ⇒ (C)” is almost trivial: If $\mathbb D_r ⊂ f(Ω)$, then for $r’ = r/2$, you have $\overline{\mathbb D_{r’}} ⊂ \mathbb D_r$.
For (B), take a very trivial map $ℂ → ℂ$ and restrict it to a non-disc shape $Ω$ containing $\overline{\mathbb D_1}$.
A: Option b : We can give a counter example , $\Omega= {|z|<2} $  union  $  |z-10|<1$ (not connected) with $f(z)= z$. So, this works only because we don't need to have the domain to be connected. 
Option d : If the domain is not connected, then we could have a function that gives different constants on the different components as a counterexample.  If the domain is connected, then we apply the open mapping theorem to imply that it is true. 
Option a and c : ( Since thay are equivalent as mentioned in another answer )  Suppose $f(\Omega)$ is open. We have that the closed unit disc is contained in it. For every point on the boundary, there is an open disc centred at that point which lies completely inside $f(\Omega)$. The collection of all such open discs is an open cover for the boundary $|z|=1$. Since this is compact, this open cover has a finite cover. From the finite cover (which is a finite number of open discs) choose the radius of the smallest disc. This radius works for every point on the boundary. From here we can conclude that a ( and therefore c ) are true. 
