Working with complex numbers, let $f$ be an entire and non-constant function.
What I want to show is that for any positive real number c>0, the closure of $\{z:|f(z)|<c\} = \{z:|f(z)|≤c\}$. I've managed to show the inclusion going from left to right using balls argument. However, I'm struggling with the other inclusion.
I've found a previous discussion about this question here: The closure of $\{ z:|f(z)|<c\}$ is the set $\{z:|f(z)|\leq c\}$.
But there's only some indication about using the open map theorem. I'm familiarizated with it and I've tried to apply it on this question but I can't find the way it goes. I would really appreciate if someone could give me the manner of doing it. Thanks for help!
So you want to show that $\{z : |f(z)| \leq c\} \subseteq \overline{\{z : |f(z)| < c\}}$ (where $\overline{S}$ denotes closure of $S$). Clearly, we only need to worry about the set $\{z : |f(z)|= c\}$. So let $z$ s.t. $|f(z)| = c$. We need to show that $z$ is a limit point of $\{z : |f(z)| < c\}$. Let $U$ an open set containing $z$. We need to show that $U \cap \{z : |f(z)| < c\} \neq \emptyset$. But by the open mapping theorem $f(U)$ is an open set containing $f(z) = c$. Convince yourself that $\overline{\{w : |w| < c\}} = \{w : |w| \leq c\}$. Then since $f(U)$ is an open set containing $c$, $f(U) \cap \{w : |w| < c\} \neq \emptyset$. I.e. there is some $a \in U$ s.t. $|f(a)| < c$. Then $a \in U \cap \{z : |f(z)| < c\}$, so $U \cap \{z : |f(z)| < c\} \neq \emptyset$.