Operators on the family of all subsets of a topological space that maybe generates a base for these family. I will try to do at least something of my first question.
Given a topological space $\langle X,\tau\rangle$, we define two operators on $2^X = \{ A : A \subseteq X \}$ as follows. For $\alpha \subseteq 2^X$ and $M\in 2^X$, define 


*

*$M\in\tilde{\alpha} \;\Leftrightarrow\; \exists L\in\alpha: \overline{L}\cap \overline{M}\ne\emptyset.$

*$M\in\tilde{\alpha} \;\Leftrightarrow\;
\exists L\in\alpha: L\cap M\ne\emptyset.$
I think the operators define closed bases for two topologies on $2^X$, and that the first operator defines a topology which is a refinement of the second. That is, 
$\{\beta\in 2^X|\exists \alpha\in\ 2^X:\beta=\tilde{\alpha}\}$ is a base for a topology given by a set of all closed sets in $2^X$.
(See also on MO)
 A: Neither of these proposed operators are "closure operators" on $\mathcal{P} ( X ) = 2^X$. Recall that given a set $Y$, a mapping $\operatorname{cl} : \mathcal{P}(Y) \to \mathcal{P}(Y)$ is the closure operator of a topology on $Y$ if and only if it satisfies the Kuratowski closure axioms:


*

*$\operatorname{cl} ( \varnothing ) = \varnothing$;

*$A \subseteq \operatorname{cl} ( A )$;

*$\operatorname{cl} ( A \cup B ) = \operatorname{cl} ( A ) \cup \operatorname{cl} ( B )$; and

*$\operatorname{cl} ( \operatorname{cl} ( A ) ) = \operatorname{cl} ( A )$.


The proposed operators both fail (2) and usually both fail (4).


*

*For (2), note that $\varnothing \notin \overline{\alpha}$ for all $\alpha \subseteq \mathcal{P} ( X )$, so in particular $\{ \varnothing \} \not\subseteq \overline{\{ \varnothing \}}$. (This holds for both proposed operators.) If you instead consider your $2^X$ to be the family of nonempty subsets of $X$ this problem can be removed, so is in some sense inessential.

*(4) is a more substantial problem. Note that if $\tau$ is the discrete topology on the set $X$, then the two proposed closure operators coincide. Considering $X = \{ 0 , 1 \}$ with the discrete topology $\tau = \{ \varnothing , \{ 0 \} , \{ 1 \} , \{ 0 , 1 \} \}$, it is easy to calculate $\overline{ \{\,\{ 0 \}\,\} } = \{\,\{ 0 \} , \{ 0 , 1 \}\,\}$, and $\overline{\{\,\{0\},\{0,1\}\,\}} = \{\,\{0\},\{1\},\{0,1\}\,\}$, so $\overline{ \overline{ \{\,\{0\}\,\} } } \neq \overline{ \{\,\{0\}\,\} }$.
More generally, given any topological space $X$, if $\alpha \subseteq \mathcal{P}(X)$  contains a nonempty element, then $X \in \overline{ \alpha }$, and it follows that $\overline{ \overline{ \alpha } } = \{ A \subseteq X : A \neq \varnothing \}$. (Again, this holds for both proposed operators.)
