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Let $f(z)$ be an entire function that is not identically constant. Show that

$$\lbrace z : |f(z)| \geq \alpha \rbrace = \text{cl }\lbrace z : |f(z)| > \alpha \rbrace.$$

This question was in our exam and hinted that we had to apply the maximum modulus principle. I was wondering what that solution looked like as my proof used the open mapping theorem.

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1 Answer 1

Let's prove this by showing inclusion in both directions.

Let $w$ be a limit point of $E = \{z : |f(z)| > a\}$. This means that there is a sequence $\{z_k\} \subset E$ so that $z_k \to w$ as $k \to \infty$. Since $f$ is continuous, it follows that $|f(w)| \ge a$ and $\operatorname{cl}(E) \subset \{z : |f(z)| \ge a\}$.

Let $w$ be a point so that $|f(w)| = a$. By the maximum modulus principle, every neighborhood of $w$ contains a point $z$ so that $|f(z)| > |f(w)|$. Thus, $|f(z)| > a$ and $z \in E$. Hence, $\{z : |f(z)| \ge a\} \subset \operatorname{cl}(E)$.

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