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This proof takes place in the complex space, and $C$ is a set of complex numbers. Also, let $\rightarrow$ denote the $\varepsilon$ definition of convergence. I have some experience with real analysis but this is my first attempt at complex analysis, I figure the proofs should be similar but I am not sure my proof is formal enough any suggestions or corrections are greatly appreciated! Also, I haven't learned any theorems so this will be a barebones proof using only definitions of open, closed and convergence.

Show that $C$ is closed if and only if $\{z_n\} \in C$ and $z_n \rightarrow z$ implies $z$ is in $C$.

I will prove the claim by contradiction.

Suppose $C$ is closed, consider a sequence $\{z_n\} \in C$ such that $z_n \rightarrow z$ for $z \notin C$. Thus, $z \in C^c.$ Since $z_n \rightarrow z$ then for every $\varepsilon$ there is an integer $N$ so that $|z_n - z| < \varepsilon$ for all $n \geq N$ such that all $\{z_n\} \in C$. Thus any disk of radius $\varepsilon$ centered around $z$ will contain some elements of $z_n$. So $z \in C^c$ and there is no $\varepsilon$ such that $D(z, \varepsilon) \subset C^c$, however, $C^c$ is open by definition. Thus, we have arrived at a contradiction.

Conversely, suppose $z_n \rightarrow z$ and $\{z_n\} \in C$ implies that $z \in C$.

Let $z_1 \in C^c$, I claim that there exists some $r > 0$ so that $D(z_1, r) \subset C^c$. Suppose not, then for any choice of $r$ $D(z_1, r) \cap C \neq \varnothing.$ In particular, consider $D(z_1, \frac{1}{n})$ for $n = 1, 2, 3 \dots$. Suppose $a_n \in D(z_1, \frac{1}{n}) \cap C$, then $a_n \in C$ and $|a_n - z_1| < \frac{1}{n}$, therefore $a_n \rightarrow z_1$ as $n \rightarrow \infty$. Then by the hypothesis $z_1 \in C$, however this is a contradiction. Therefore, there exists some $r > 0$ so that $D(r, z_1) \subset C^c$ for arbitrary $z_1 \in C^c$ and thus $C^c$ is open and $C$ is closed.

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    $\begingroup$ The proof is correct but the converse implication is missing. The if is an iff here. $\endgroup$ – Gribouillis Sep 2 '17 at 20:16
  • $\begingroup$ Yes haha I am still working on that side figured I'd see if I got this direction down first. Also, in the text the $C$ is meant to denote a set of complex numbers! $\endgroup$ – student_t Sep 2 '17 at 20:17
  • $\begingroup$ @Gribouillis Added the converse! $\endgroup$ – student_t Sep 3 '17 at 1:46
  • $\begingroup$ It is correct. Your proof works in any metric space, not just ${\mathbb C}$. Metric spaces are sequential spaces. $\endgroup$ – Gribouillis Sep 3 '17 at 5:46
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The other direction is straightforward. Suppose $z \notin C$. Then consider the sequence of balls $B(z,{1 \over n})$. If an infinite number of these balls intersect $C$ then we have $z\in C$, hence at most a finite number of these balls intersect $C$. Hence there is some $n$ such that $B(z,{1 \over n}) \cap C = \emptyset$, and so $C^c$ is open.

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