Define the "model theoretic" notion of a closure function as follows:

Definition (1): Let $D$ be a non-empty set. A function $cl:P(D)\longrightarrow P(D)$ called a closure function iff it has the following properties:

$(1)~\forall A\subseteq D~~~~A\subseteq cl(A)$

$(2)~\forall A,B\subseteq D~~~~A\subseteq B\Longrightarrow cl(A)\subseteq cl(B)$


We say that $cl$ is a "good" closure on $D$ if it has the following property too:

$(4)~\forall A\subseteq D~~~~cl(A)=\bigcup_{B\in P_{<\omega}(A)}cl(B)$

and a "pregeometry" if we have:

$(5)~\forall A\subseteq D~~~~\forall a,b\in D~~~~~~a\in cl(A\cup \lbrace b\rbrace)\setminus cl(A)\Longrightarrow b\in cl(A\cup \lbrace a\rbrace)$

There are many trivial good closures on a given set which are not related to any structure, but even there are some non-trivial natural good closures on the domain of an arbitrary $\mathcal{L}$-structure $\mathcal{M}$ like well known "algebraic closure" ($acl_{\mathcal{M}}$), "definable closure" ($dcl_{\mathcal{M}}$) and "structural closure" ($scl_{\mathcal{M}}$) which is defined as follows:

$\forall A\subseteq Dom(\mathcal{M})~~~~~scl_{\mathcal{M}}(A):=Dom(\langle A\rangle_{\mathcal{M}})$

and $\langle A\rangle_{\mathcal{M}}$ is the substructure of $\mathcal{M}$ generated by the set $A$.

Now the main question is about the "essential" properties of these closure functions on an arbitrary structure. In the other words is "goodness" the unique essential property of $acl_{\mathcal{M}}$, $dcl_{\mathcal{M}}$ and $scl_{\mathcal{M}}$ on an arbitrary $\mathcal{L}$ - structure $\mathcal{M}$? Precisely:

Question (1): Let $D$ be an arbitrary non-empty set, and $cl:P(D)\longrightarrow P(D)$ be a good closure function on $D$, is there a first order language $\mathcal{L}$ and an $\mathcal{L}$-structure $\mathcal{M}$ such that: $Dom(\mathcal{M})=D$ and $scl_{\mathcal{M}}=cl$?

Question (2): What is the answer of question (1) for $acl_{\mathcal{M}}$ and $dcl_{\mathcal{M}}$?

Remark (1): Note that producing a negative answer for the above questions needs finding a property $P$ different from "being a good closure" and proving that the functions $acl$, $dcl$ and $scl$ have the property $P$ on "any" structure and so there is no such language and structure for an arbitrary function $cl:P(D)\longrightarrow P(D)$ because it is possible that such function have the property $\neg P$.So one can re ask the above questions by the following re stating:

"Let $D$ be an arbitrary non-empty set, and $cl:P(D)\longrightarrow P(D)$ be a good closure function with property $P$ on $D$, is there a first order language $\mathcal{L}$ and an $\mathcal{L}$-structure $\mathcal{M}$ such that: $Dom(\mathcal{M})=D$ and $scl_{\mathcal{M}}=cl$ ($acl_{\mathcal{M}}=cl$ or $dcl_{\mathcal{M}}=cl$)?

So we can go further and try to characterize "all" essential properties of $scl, acl, dcl$ on an arbitrary structure and ask the following question:

Question (3): What is the property $P$ such that for any non-empty set $D$ and any function $cl:P(D)\longrightarrow P(D)$ which is a good closure function with property $P$, the answer of the question (1) or (2) be true?

In the other direction It is well known that $acl_{\mathcal{M}}$ is a pregeometry on "strongly minimal" structures. So:

Question (4): Is there a known type of $\mathcal{L}$-structures which the functions $scl$ or $dcl$ be a pregeometry on them?


$\textbf{Question 1}$ The answer to this question is affirmative. The closure operators that you call good closure operators are normally called algebraic closure operators. Furthermore, each algebraic closure operator is of the from $\mathrm{scl}_{\mathcal{A}}$ for some algebra $\mathcal{A}$. A proof of this fact is given here on Theorem 3.5 in Chapter 2. Suppose that $C$ is an algebraic closure operator on some set $A$.

Let $\mathcal{F}_{n}$ be the set of all $n$-ary functions $f:A^{n}\rightarrow A$ such that $f(a_{1},...,a_{n})\in C(a_{1},...,a_{n})$ whenever $a_{1},...,a_{n}\in A$. Let $\mathcal{A}=(A,\bigcup_{n}\mathcal{F}_{n})$. Then it is easy to see that $C=\mathrm{scl}_{\mathcal{A}}$.

  • $\begingroup$ Good answer. Have you any idea about other questions? $\endgroup$ – user36136 Sep 21 '13 at 4:28

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