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My complex analysis textbook states the following:

A set is closed if its complement $\Omega^c = \mathbb{C} - \Omega$ is open. This property can be reformulated in terms of limit points. A point $z \in \mathbb{C}$ is said to be a limit point of the set $\Omega$ if there exists a sequence of points $z_n \in \Omega$ such that $z_n \not= z$ and $\lim_{n \to \infty} z_n = z$. The reader can now check that a set is closed if and only if it contains all its limit points.

I want to prove the following:

A set is closed if and only if it contains all its limit points.

Proof 1:

I begin by assuming that the set $\Omega$ contains all of its limit points.

$$\therefore \Omega = \{ z \in \mathbb{C} : (z_n \not = z) \cap (\lim_{n \to \infty} z_n = z) \},$$

where $z_n \in \Omega$ is a sequence of points.

And so we have that

$$\begin{align} \Omega^c &= \{ z \in \mathbb{C} : (z_n = z) \cup (\lim_{n \to \infty} z_n \not= z) \} \\ &= \{ z \in \mathbb{C} \} - \{ z \in \mathbb{C} : (z_n \not= z) \cap (\lim_{n \to \infty} z_n = z) \} \\ &= \mathbb{C} - \Omega \end{align}$$

Proof 2:

Assume that the set $\Omega$ is closed.

Therefore, $\Omega^c = \mathbb{C} - \Omega$ is open, since a set is closed if its complement $\Omega^c = \mathbb{C} - \Omega$ is open.

A set is open if every point in that set is an interior point.

Therefore, every point $z \in \Omega^c$ is an interior point.

Therefore, every point in the set $(\Omega^c)^c = \Omega$ is not an interior point.

This is where I'm unsure of how to proceed. Given what the author wrote, it is implied that they have provided all of the information necessary for this proof in the preceding sections. However, I have been unable to see how such a proof can be done.

Preceding the above textbook excerpt, the author also give descriptions of closed and open discs, as well as interior points as follows:

The interior of $\Omega$ consists of all its interior points. Finally, a set $\Omega$ is open if every point in that set is an interior point of $\Omega$.

I attempted to use this relationship between interior points and open sets, along with the definition of closed sets above, to form a proof. However, it's not clear to me whether this is fruitful.

I would greatly appreciate it if people could please take the time to review my work and help me with proof 2.

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If $\Omega$ is closed in the sense that $\Omega^\complement$ is open, then let $z$ be a limit point of $\Omega$ and we will show $z \in \Omega$. So in your definition of limit point we have a sequence of $z_n \neq z$ such that $z_n \in \Omega$ and $z_n \to z$. If $z \notin \Omega$ (striving for a contradiction), $z \in \Omega^\complement$ which is open by assumption, so we have some $r>0$ and an open disk $D(z,r) \subseteq \Omega^\complement$. Applying the definition of convergence we can find $N$ such that for all $ n \ge N$ we have that $|z_n -z| < r$. But then $|z_N -z| < r$, so $z_N \in D(z,r)$, so $z_N \in \Omega^\complement$ and $z_N \notin \Omega$, contradicting that all $z_n$ are in $\Omega$. This contradiction thus shows $z \in \Omega$ and we are done with one direction.

Now suppose $\Omega$ is closed in the limit point sense. Let $z \in \Omega^\complement$ and we want to show there is some disk $D(z,r)$ that is a subset of $\Omega^\complement$. Suppose this were not the case. Then for every $n \in \Bbb N$

$$D(z,\frac1n) \nsubseteq \Omega^\complement \text{ or } \exists z_n \in \Omega: |z_n - z| < \frac1n$$

Then all $z_n \neq z$ and $z_n \to z$ by construction, and as $\Omega$ is closed in the limit point sense we conclude that $z \in \Omega$ too, a contradiction.

So our supposition is wrong and every $z \in \Omega^\complement$ is an interior point of it, and so $\Omega^\complement$ is open and $\Omega$ is closed in that sense.

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  • $\begingroup$ Hi again, Henno. Thanks for the answer. With regards to the second proof, you claim that $z_n \neq z$ and $z_n \to z$ by construction; how is this the case? $\endgroup$ Jun 10, 2019 at 16:05
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    $\begingroup$ @ThePointer $z_n$ are in $\Omega$ and $z$ is not, so the first is clear. The second follows as $\frac1n \to 0$. $\endgroup$ Jun 10, 2019 at 16:06

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