Can we describe a topology when generated by subset of the power set? Let $X$ be a set on which we wish to define a topology. Let $\mathcal{A}$ be a subset of the power set of $X$, i.e. elements of $\mathcal{A}$ are subsets of $X$. This question arose when studying the Zariski topology, so I'm approaching it by closed subsets, not open ones. Consider the intersection of all topologies for which elements of $\mathcal{A}$ are all closed : it is itself a topology. Equip $X$ with that topology and call $\mathcal{T}$ the set of $X$'s closed elements.
From this Wikipedia page I get the impression that it is easy to explicitly describe $\mathcal{T}$ and to do so I introduce the notation $\mathcal{P}_f(\mathcal{A})$ to denote the finite elements of the power set of $\mathcal{A}$. Then, my understanding is that :
$$\mathcal{T} = \lbrace{\bigcap_{\omega \in \Omega} \bigcup_{F \in \omega} F \quad | \quad \Omega \subset \mathcal{P}_f(\mathcal{A}) \quad \rbrace} $$
I tried to prove this, convinced that it would be straightforward but I'm actually really stuck. To make notation simpler, notice that a supposedly closed subset of $X$ is entirely determined by $\Omega$ so I defined :
$$ \Phi : \mathcal{P}(\mathcal{P}_f(\mathcal{A})) \rightarrow \mathcal{P}(X) $$ by $$ \Phi(\Omega)=\bigcap_{\omega \in \Omega} \bigcup_{F \in \omega} F$$
My claim now simply turns into $\quad\mathcal{T}=\text{Im}\Phi$.
It is clear that $\text{Im}\Phi \subset \mathcal{T}$ from the properties of closed subsets.
I need to prove that also $\mathcal{T} \subset \text{Im}\Phi$, for which it suffices to prove that $\text{Im}\Phi$ is a topology (or rather, the set of closed sets of a topology).


*

*$X=\Phi(\emptyset)$ and $\emptyset=\Phi(\lbrace \emptyset \rbrace)$ so $X$ and $\emptyset$ are given as closed subsets by $\Phi$

*The intersection property should be easy to prove

*Let $I \subset \mathcal{P}(\mathcal{P}_f(\mathcal{A}))$ be finite. We need to prove that $\quad \bigcup_{\Omega \in I} \Phi(\Omega)\in \text{Im}\Phi$ that is exactly : $$ \bigcup_{\Omega \in I}\bigcap_{\omega \in \Omega} \bigcup_{F \in \omega}F $$ can be written in the form $$\bigcap_{\omega \in \Omega'} \bigcup_{F \in \omega}  F $$ and I've drawn diagrams and all sorts but I can't swap that left-hand side cup with the cap.

 A: We need to prove that $H=\bigcup_{\Omega\in I}\bigcap_{\omega\in\Omega}\bigcup_{F\in\omega}F$ can be written in the form $\bigcap_{\omega\in\Omega'}\bigcup_{F\in\omega}F$ for some $\Omega'\in\mathcal{P}(\mathcal{P}_f(\mathcal{A}))$.
Denote $\mathcal{F}_I$ the family of all functions $\varphi\colon I\to\mathcal{P}_f(\mathcal{A})$ satisfying the property $(\forall \Omega\in I)\,\varphi(\Omega)\in\Omega$. For every $\varphi\in\mathcal{F}_I$ denote $\omega_\varphi=\bigcup_{\Omega\in I}\varphi(\Omega)$. Since $I$ is finite and $\varphi(\Omega)$ is a finite subset of $\mathcal{A}$ for every $\Omega\in I$, we have $\omega_\varphi\in\mathcal{P}_f(\mathcal{A})$.
Put $\Omega'=\{\omega_\varphi:\varphi\in\mathcal{F}_I\}$.
Now, we have $x\in X\setminus H$ iff $$x\in\bigcap_{\Omega\in I}\bigcup_{\omega\in\Omega}\bigcap_{F\in\omega}(X\setminus F)$$ iff $$(\forall\Omega\in I)(\exists\omega\in\Omega)\ x\in\bigcap_{F\in\omega}(X\setminus F)$$ iff $$(\exists\varphi\in\mathcal{F}_I)(\forall\Omega\in I)\ x\in\bigcap_{F\in\varphi(\Omega)}(X\setminus F)$$ iff $$x\in\bigcup_{\varphi\in\mathcal{F}_I}\bigcap_{\Omega\in I}\bigcap_{F\in\varphi(\Omega)}(X\setminus F),$$ hence $$H=\bigcap_{\varphi\in\mathcal{F}_I}\bigcup_{\Omega\in I}\bigcup_{F\in\varphi(\Omega)}F=\bigcap_{\omega\in\Omega'}\bigcup_{F\in\omega}F.$$
A: In this answer I describe a proposition for the open sets. 
This can of course easily be dualised, using de Morgan and complements.
If we have a collection of sets $\mathcal{C}$ that we want to be closed, 
the minimal collection of a topology (described by closed sets) is
$$\hat{\mathcal{C}} = \left(\mathcal{C}^{\cup ,< \infty}\right)^{\cap}$$
and the open sets are their complements. The notation is analogous to the one used in the linked answer. It boils down to the same thing as your $\mathcal{T}$.
