# The closure of $\{ z:|f(z)|<c\}$ is the set $\{z:|f(z)|\leq c\}$.

Exercise: Let $f$ be entire and non-constant. For any positive real number $c>0$ show that the closure of $\{ z:|f(z)|<c\}$ is the set $\{z:|f(z)|\leq c\}$.

Solution: Let $\varepsilon>0$ and define $A:=\{ z:|f(z)|<c\}$. Then, because $f$ is continuous (via entire) and non-constant, there exists $z\in A$ such that

$$B(c;\varepsilon)\cap A \neq \varnothing .$$

That is to say that $c$ is a limit point of $A$. Since $A^-$ must contain all of its limit points, we have that $\{z:|f(z)|\leq c\}$.

Is my solution correct? I have another way of doing it using a sequence $z_n\to z$, but I would prefer to use the above if it correct. The reason I am feeling unsure about my solution is because I am not sure how we know (geometrically) that $|f(z)|$ goes all the way up to the boundary (i.e. $c$). Is it safe to say that $|f(z)$ gets infinitely close to $c$ but does not touch because $f$ is entire & non-constant? FYI I am new at these concepts so please be a detailed as possible. Thank you in advance!

Note: I am using the notation $A^-$ to indicate the closure of $A$. This is the notation used in my text (Complex, Conway) and I suspect it is used as to not be confused with the complex conjugate.

• You need to use the open mapping theorem. – Pedro Tamaroff Apr 16 '16 at 1:58

First, something is wrong with your ball $B(c,\epsilon)$. Did you mean $z$ instead of $c$? Without some kind of correction your argument unfortunately doesn't make sense.
Second, what you would be showing with your argument is just that $\{z\in \mathbb{C} : |f(z) \leq c\} \subseteq A^-$. You need to justify the other inclusion as well.
Also, you can usually turn a sequence argument into a ball argument by noting that convergence of a sequence to a limit $L$ from within a set is equivalent to arbitrarily small balls around L intersecting nonemptily (I dunno if this is a word) with your set.