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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.

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    $\begingroup$ You need to use the open mapping theorem. $\endgroup$ – Pedro Tamaroff Apr 16 '16 at 1:58
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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.

Lastly, if you want to know whether your functions will actually get big enough in norm, you can apply the result that entire nonconstant functions are unbounded, and so will take arbitrarily large norms somewhere.

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