How to define a finite topological space?
You need an additional property to guarantee that the closure operator will be idempotent. Requiring $$x \in c(y) \Rightarrow c(x) \subseteq c(y)$$ in addition to $x \in c(x)$ is necessary and sufficient.
It's sufficient to define a relation $x \le y$, the specialisation order, (so just a subset of $X^2$), which is a pre-order, so $x \le x$ for all $x$,and $x \le y$ and $y \le z$ implies $x \le z$. ($\le$ is reflexive and transitive).
This relation is defined in a topological space by $x \le y$ iff $x \in \overline{\{y\}}$. But if we start with a preorder $\le$, we can define a topology: $\overline{\{y\}}$ is then defined by $\downarrow y = \{x \in X: x \le y\}$ and $\overline{A} = \cup_{x \in A}\downarrow x$.
The set of finite topological spaces is just the set of finite pre-orders.
No. First of all, it must be a function $c\colon\mathcal{P}(X)\longrightarrow\mathcal{P}(X)$. Besides, it must satisfy these conditions:
- $A\in\mathcal{P}(X)\Longrightarrow A\subset c(A)$;
- $A,B\in\mathcal{P}(X)\wedge A\subset B\Longrightarrow c(A)\subset c(B)$;
- $A\in\mathcal{P}(X)\Longrightarrow c\bigl(c(A)\bigr)=c(A)$.