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What is the equivalent to generating a topology with a basis for closure operators satisfying the Kuratowski closure operators.

Let $X$ be a topological space. Let $c : 2^X \to 2^X$ be its closure operator.

The set of open sets definition of a topology, as well as the set of closed sets definition, has a notion of generating a topology via a basis or subbasis.

I'm wondering if there are similar construction for closure operators.

Here's one I can think of off the top of my head that sends $f : 2^X \to 2^X$ to $f^* : 2^X \to 2^X $. I think it works but I'm not sure.

Then we define $f^*$ as follows:

Let $g : 2^X \to 2^X$ be a closure operator on $X$.

Let's say that $g$ is compatible with $f$ if and only if:

$$ f(A) \subset g(A) \;\; \text{for all $A \subset X$ where $A \neq \varnothing$} $$

$f$ is always compatible with the $g$ that sends every non-empty set to $X$, the whole space. Therefore there is always at least one $g$ that is compatible with $f$.

We compute $f^*$ in the following way.

$$ f^*(A) = \bigcap \bigg\{ g(A) : \text{$g$ is a closure operator and $g$ is compatible with $f$} \bigg\} $$

I will call each $g$ in the comprehension a component of $f^*$.

$f^*$ sends $\varnothing$ to $\varnothing$.

$f^*$ is extensive because all of its components are extensive.

$f^*$ probably respects binary unions. Let $\varphi$ be the sentence $g$ is compatible with $f$.

$$ f^*(A \cup B) \\ \bigcap\{g(A \cup B) : \varphi \} \\ \bigcap \{ g(A) \cup g(B) : \varphi \} \\ \textit{This step is suspicious} \\ \big( \bigcap \{g(A) : \varphi \} \big) \;\bigcup\; \big(\bigcap \{g(B) : \varphi \} \big) $$

I am pretty sure that there can't any elements of $f^*(A \cup B)$ that are not members of $f^*(A)$ or $f^*(B)$, but I'm not sure how to prove it.

Here is my attempt to prove idempotence. It has one suspicious step.

$$ f^*(f^*(A)) $$

Replace the inner $f^*$ with its definition.

$$ f^*\left(\bigcap\{g(A) : \varphi\}\right) $$

Replace the outer $f^*$ with its definition.

$$ \bigcap \left\{ g' \left( \bigcap \left\{ g(A): \varphi \right\} \right) : \varphi' \right\} $$

I don't trust this step. I am pretty sure the outermost closure operator distributes over the intersection by the properties of closed sets, but I am calling it out here.

$$ \bigcap \left\{ g' \left( g \left( A \right) \right) : \varphi \land \varphi' \right\} $$

For a single closure operator, $g \circ g = g$ holds. Additionally, the composition of any two closure operators is a closure operator. Therefore, this construction hits all and only the closure operators.

$$ \bigcap \left\{ g \left( A \right) : \varphi \right\} $$

which gives us the following as desired.

$$ f^*(\varphi) $$

Here is the original definition I had for $f^*$. It definitely does not work.

$$ f^*(A) = \bigcap \bigg\{ {\cup}\big\{ f^k(Z) : k \in \mathbb{N} \big\} : A \subset Z \subset X \bigg\} $$

Greg Nisbet
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    I'm not convinced that $f^\ast(A \cup B)= f^\ast(A) \cup f^\ast(B)$. Could you add a proof? Also of the idempotence (i.e. $f^\ast(f^\ast(A)) = f^\ast(A)$ for all $A$)? – Henno Brandsma Jun 07 '21 at 05:49
  • @HennoBrandsma. You are right. The binary union for the thing I had originally definitely doesn't work. I replaced the definition of $f^*$ with a different definition where I take all the closure operators, remove the ones that are incompatible with $f$, and then flatten the remaining ones into a closure operator. I called out the steps that I couldn't completely justify. – Greg Nisbet Jun 08 '21 at 00:54
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    Here's a proof of idempotence. Check first that $f^$ is monotone, which is easy. Then since $f^(A)\subseteq g(A)$ for any $g\in G$, you have $$f^(f^(A)) \subseteq f^(g(A)) \subseteq g(g(A))=g(A).$$ Hence $(f^)^2(A)\subseteq f^*(A)$. The other comparison is given by monoticity as well.

    Actually, you can see that this argument works for the pointwise intersection of any set of closure operators $G$. BUT I'm almost certain that $f^*$ does not distribute over $\cup$. It does, however, if $G$ is downwards directed.

     – Jackozee Hakkiuz Jun 09 '21 at 15:51
  • @JackozeeHakkiuz Thanks for the proof of idempotence. If $f^$ in fact works, then I think the compatible $g$'s have a unique "minimal" element anyway, so I can pick it rather than trying to make one via intersection (if $f^$ works in the first place). In the original question, I was mostly curious what the real construction is for taking $f$ and shaping it into a closure operator. I can create a set of sets, inflate it into a cotopology, and then fashion a closure operator, but I'm curious if there's a more direct way to closure-operator-ify a function that's relatively standard or common. – Greg Nisbet Jun 09 '21 at 16:03
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    @GregoryNisbet I think what you are trying to do is completely dual to what is done in frame theory for getting the nucleus associated to an inflator. A Kuratowski closure is the same as an interior operator, which is exactly a conucleus over the frame $PX$. At this level of generality, the task of explicitely constructing nuclei is rather difficult and the constructions I've seen have been done with transfinite induction. It may be possible to do better for the case of a power set, and perhaps the spatial case is not so bad, but I haven't looked at that. Good question, btw. I forgot to upvot – Jackozee Hakkiuz Jun 10 '21 at 02:36

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