Let $f$ be a non-constant entire function and $\Omega$ be a bounded open subset of $\mathbb{C}$. Let $S=\{Re(f(z)))+Im(f(z)) : z\in \Omega\}$. Then which of the following is/are necessarily correct ?
$(1)S$ is an open set in $\mathbb{R}$
$(2)S$ is a closed set in $\mathbb{R}$
$(3)S$ is a open set in $\mathbb{C}$
$(4)S$ is a discrete set in $\mathbb{R}$
My attempt
Some general observations:-
By Open Mapping Theorem ,$f(\Omega)$ is also open in $\mathbb{C}$. So
$f(\Omega )=\displaystyle\bigcup_{\alpha \in I }B_{\alpha }$ where $I$ is some index set and $B_{\alpha}=B(z_{\alpha},r_{\alpha})$ by the metric topology
I claim $S_{\alpha}=\{Re(w)+Im(w) : w\in B_{\alpha}\}$ is open in $\mathbb{R}$
A probable Proof :-
Let $w \in B_{\alpha}$. Then $w=z_{\alpha}+r(cos(\theta)+isin(\theta))$ for some $r$ and $\theta$ satsifying $0\le r \lt r_{\alpha}$ and $0\le \theta \lt 2\pi$
Let $z_{\alpha}=x_{\alpha}+iy_{\alpha}$
Then $S_{\alpha}=\displaystyle\bigcup_{r,\theta}\{x_{\alpha}+y_{\alpha}+rcos(\theta)+rsin(\theta)\}$ with $r$ and $\theta$ satisfying the above conditions
$\Rightarrow S_{\alpha}=\displaystyle\bigcup_{r} \displaystyle\bigcup_{\theta}\{ x_{\alpha}+y_{\alpha}+ r \sqrt(2)sin(\theta +\pi/4)\}$
$\Rightarrow S_{\alpha}= \displaystyle\bigcup_{r} [ x_{\alpha}+y_{\alpha} -r \sqrt(2), x_{\alpha}+y_{\alpha}+ r\sqrt(2) ]$
$\Rightarrow S_{\alpha}= (x_{\alpha}+y_{\alpha} -r_{\alpha} \sqrt(2), x_{\alpha}+y_{\alpha} +r_{\alpha}\sqrt(2))$
Hence the claim follows .
What really makes doubtful is the transition from second last step to last step in above proof. This is the same question as
$\displaystyle\bigcup_{0\le r \lt R}[a-br,a+br]=^{?}(a-bR,a+bR) $ for some non-zero $a$ and $b$.
Of course, if this proof is correct then it is not difficult to see $S$ is open in $\mathbb{R}$ since then it would be (arbitary) union of open intervals.
Again assuming the proof to be true, I can see (at least intuitively) that $S$ is neither an open nor clossed subset of $\mathbb{C}$.
So basically ,my answer is depending on the truthfulness of the above proof.
Do you have any suggestions ? Any alternative ideas are appreciated.
Thanks for your time!!
I'll assume $\Omega$ is a connected open set and $f$ is holomorphic and nonconstant on $\Omega.$
Lemma 1: Let $f=u+iv.$ Then $u+v$ is nonconstant on $\Omega.$
Proof: Suppose otherwise. Then $u+v\equiv c,$ where $c$ is a real constant. The C-R equations then imply
$$u_x=-u_y,\,\, u_y= u_x.$$
This implies $u_x=0=u_y$ in $\Omega.$ Since $\Omega$ is connected, $u$ is constant in $\Omega.$ Hence so is $v.$ That implies $f$ is constant in $\Omega,$ contradiction, and the lemma is proved.
Lemma 2: If $w$ is a nonconstant real harmonic function on $\Omega,$ then $w(\Omega)$ is an open interval in $\mathbb R.$
Proof: Because $\Omega$ is connected, $w(\Omega)$ is connected, hence is an interval. Now go through the cases: Could $w(\Omega)=[a,b]$ for some $a,b\in \mathbb R?$ If so, then $w$ has a maximum in $\Omega,$ which by the maximum principle implies $w$ is constant, contradiction. All cases other than $(a,b)$ can be similarly excluded (here $-\infty \le a<b\le \infty$).
Conclusion: $S$ open in $\mathbb R$ must hold. The other options need not hold.
Note: It seems irrelevant that $f$ is entire or $\Omega$ is bounded. But assuming $\Omega$ is connected is crucial.