Closure of a set in weird Topological Space I am currently taking a Topology course and I try to find some problems online and try to solve them. The obvious downside is that, more often then not, I can not seem to find solutions to check my work or get a hint! So, hopefully someone here can give a hint or solution.
Here is the question:
Let $\mathbb{X}$ be a set and $A \mapsto \overline{A}$ be an operation on the subsets of $\mathbb{X}$ satisfying:
1.) $A \subseteq \overline{A}$
2.) $\overline{\overline{A}} = \overline{A}$
3.) $\overline{A \cup B} = \overline{A} \cup \overline{B}$
4.) $\overline{\emptyset} = \emptyset$
Let $\tau$ be defined by $A \in \tau \iff \overline{\mathbb{X} - A} = \mathbb{X} - A$. You may assume $\tau$ is a topology on $\mathbb{X}$. Show that $cl_{\tau}(A) = \overline{A}$ for all $A \subseteq \mathbb{X}$.
Here is my attempt that seems to be continuously leading me to dead ends:
By definition $cl_{\tau}(A) = \bigcap \{F: A \subseteq F \text{ and } F \text{ closed}\}$, so for any $F \in cl_{\tau}(A)$ the following inclusions and implications hold: $(A \subseteq F) \implies (F^c \subseteq A^c) \implies (\mathbb{X} - A^c \subseteq \mathbb{X} - F^c)$. However, $F \text{ closed} \implies F^c \text{ open(i.e. in the topology)}$. Hence, by the topology defined we have the following equality is true: $\overline{\mathbb{X} - F^c} = \mathbb{X} - F^c$. Now, by the last implication we have the following holds: $\mathbb{X} - A^c \subseteq \overline{\mathbb{X} - F^c}$. This is where I get stuck...
Thanks to whoever can help or give insight!
 A: Let $\mathscr{F}=\{F\subseteq\Bbb X:A\subseteq F\text{ and }F\text{ is closed}\}$. If $F\in\mathscr{F}$, then $\overline A\subseteq\overline F=F$. To see this, note that $A\cup F=F$, so
$$\overline F=\overline{A\cup F}=\overline A\cup\overline F\,,$$
and therefore $\overline A\subseteq\overline F$. Thus, $\overline A\subseteq\bigcap\mathscr{F}$. On the other hand, $\overline{\overline A}=\overline A$, so $\overline A$ is closed, and certainly $A\subseteq\overline A$, so $\overline A\in\mathscr{F}$, and therefore $\bigcap\mathscr{F}\subseteq\overline A$. This shows that $\overline A=\bigcap\mathscr{F}=\operatorname{cl}_\tau(A)$.
Note that the statement $F\in\operatorname{cl}_\tau(A)$ in your argument makes no sense, since both $F$ and $\operatorname{cl}_\tau(A)$ are subsets of $\Bbb X$; $F\subseteq\operatorname{cl}_\tau(A)$ would at least be a possible statement, but it appears from what follows that you actually meant that $F\in\mathscr{F}$ as I defined $\mathscr{F}$.
This isn’t actually weird: any topological space can be defined in this way. There are several provably equivalent ways to define topological spaces: in terms of topologies (i.e., families of open sets satisfying certain axioms), in terms of families of closed sets, in terms of neighborhood bases, and more. Here you’re showing that topological spaces can be defined in terms of a closure operator that satisfies certain axioms.
