I am aware that the existence of a Kuratowski closure $\mathtt{c} : \wp(M) \to \wp(M)$ naturally induces a cotopology on $M$ (that's just my name for the family of closed sets w.r.t. a topology): the proof is here. However, suppose we are given a cotopology $\kappa$: can we naturally construct a closure operator $\mathtt c$ satisfying Kuratowski's axioms?
My idea was that yes, this can be done – just adopt the usual definition of topological closure: for all subsets $B \subseteq X$,
$$\overline B = \bigcap \{C \in \kappa\ |\ B \subseteq C\} $$
This closure is trivially idempotent: the closure of a closed set is itself. This holds for $\varnothing$ too, because it is closed, so this operation preserves the null set. By definition of intersection, $B \subseteq \overline B$, so it is extensive too.
Now I'm having trouble proving that $\overline{A \cup B} = \overline A \cup \overline B$. How does one go by showing that?
Addendum. If $A \mapsto \overline{A}$ is a Kuratowski closure, is the cotopology that it induces the same as the original $\kappa$?
Best Answer
Suppose we have a co-topology $\kappa$ and we define
$$c(A) = \bigcap \{C \in \kappa: A \subseteq C\}$$
so that $c(A) \in \kappa$ ($\kappa$ is closed under intersections) and $A \subseteq c(A)$. The intersection is non-void because $X \in \kappa$ is always one of the sets in the intersection.
We want to see that $c(A \cup B) = c(A) \cup c(B)$.
It's already clear from the definition that $A \subseteq B$ implies $c(A) \subseteq c(B)$. (the sets from $\kappa$ that contain $B$ also contain $A$, so $c(A)$ is the intersection of possibly more sets than $c(B)$ hence smaller).
So $A \subseteq c(A), B \subseteq c(B)$ and so $A \cup B \subseteq c(A) \cup c(B)$. The right hand side is in $\kappa$ as this is closed under finite unions, and as it contains $A \cup B$, $c(A) \cup c(B)$ is one of the sets being intersected in the definition of $c(A \cup B)$ and so
$$c(A \cup B) \subseteq c(A) \cup c(B)$$
On the other hand $A,B \subseteq A \cup B$ so $c(A), c(B) \subseteq c(A \cup B)$ and it follows that
$$c(A) \cup c(B) \subseteq c(A \cup B)$$
Hence we have equality.
We also have that $c(A) = A$ iff $A \in \kappa$, to answer the final question affirmatively. $c(A) = A$ implies $A \in \kappa$ because $c(A) \in \kappa$ always. And if $A \in \kappa$, $A$ itself is in the intersecting family defining $c(A)$ so that $A \subseteq c(A) \subseteq A$ and $c(A) = A$.