I am trying to prove the following theorem in the context of complex analysis:
A set $\Omega$ is closed if its complement $\Omega^c = \mathbb{C} - \Omega$ is open.
Preceding this theorem, the textbook does define open/closed sets and interior points, but it is only after this theorem that the textbook defines limit points. Therefore, I'm guessing that this should be provable without limit points (using the information conveyed up till this point in the textbook). The textbook later gives an equivalent theorem in terms of limit points, and I will attempt to prove this one later.
My proof proceeds as follows:
We begin by assuming that $\Omega^c = \mathbb{C} - \Omega = \{ z \in \mathbb{C} : z \not\in \Omega \}$ is an open set.
A set is open if every point in that set is an interior point. Therefore, if $z_0 \in \mathbb{C}$, there exists $r > 0$ such that
$$D_r (z_0) = \{ z \in \mathbb{C} : |z - z_0| < r \} = \Omega^c$$
This is the open set.
Now here's where I get confused.
Next, I take the complement of the set $\Omega^c$. As I understand it, the complement is the negation, and so when we take the complement of a set, we take the negation of the set condition:
$$(\Omega^c)^c = \{ z \in \mathbb{C} : |z - z_0| \ge r \} = \Omega$$
This should be a closed set, but it isn't. A closed set is defined as
$$\bar{D}_r (z_0) = \{ z \in \mathbb{C} : |z - z_0| \le 0 \}$$
But the negation of $<$ is $\ge$, which is why I have $\Omega = \{ z \in \mathbb{C} : |z - z_0| \ge r \}$
I would greatly appreciate it if people could please tell me what I'm misunderstanding and how this should be done.
EDIT:
The closed disc $\bar{D}_r(z_0)$ of radius $r$ centred at $z_0$ is defined by
$$\bar{D}_r(z_0) = \{ z \in \mathbb{C} : |z - z_0| \le r \}$$
and the boundary of either the open or closed disc is the circle
$$C_r(z_0) = \{ z \in \mathbb{C} : |z - z_0| = r \}$$
Given a set $\Omega \subset \mathbb{C}$, a point $z_0$ is an interior point of $\Omega$ if there exists $r > 0$ such that
$$D_r(z_0) \subset \Omega$$
Finally, a set $\Omega$ is open if every point in that set is an interior point of $\Omega$.
You want $D_r(z_0)\color{blue}{\subseteq}\Omega^c$.
I think you went off track after that.
How bout the fact that each $z_0\in\Omega^c$ having an open set containing it which is disjoint from $\Omega$ implies that $z_0$ is not in the frontier* of $\Omega$. Thus all of $\Omega$'s frontier points are in $\Omega$. Thus $\Omega$ is closed.
*frontier of $X$ is defined as the intersection of the closure of $X$ and the closure of $X^c$.
Alternatively, consider that $D_r(z_0)^c$ is closed and contains $\Omega$. Thus $z_0\not\in\bar\Omega$. The closure is the smallest closed set containing $\Omega$. Thus $\bar\Omega\subset(\Omega^c)^c=\Omega$.
This second way seems more like what you were trying to do.