Is there a name for a function that maps a set into a subset of itself? Say $X$ is a set of subsets of some arbitrary set. Is there a name for a function $f:X\to X$ satisfying $f(A)\subseteq A$ for all $A\in X$?
More specifically, is there a name for an $f:X\to X$ with $f(A)\subseteq A$ and $f(f(A)) = f(A)$? In decision theory, this is how choice functions behave on sets of acts, but I wonder if there is a mathematical name for these properties.
 A: (I'm stating the following definitions for function defined on the entire power set, but they make sense for any subset of the power set...)
A function $f\colon \mathcal{P}(X)\to\mathcal{P}(X)$ such that $f(A)\subseteq A$ for all $A$ is said to be decreasing. (If $A\subseteq f(A)$ for all $A$, we say the function is increasing).
If the function satisfies $A\subseteq B\Rightarrow f(A)\subseteq f(B)$ for all $A$ and $B$, then we say $f$ is isotone.
If the function satisfies $f(f(A)) = f(A)$, then we say the function is idempotent.
A function that is increasing, isotone, and idempotent is called a closure operator. If, in addition, 
$$f(A) = \bigcup_{B\subseteq A,\ B{\rm\ finite}} f(B)$$ for all $A$, then we say the closure operator $f$ is algebraic.  If $f(A\cup B) = f(A)\cup f(B)$ for all $A$ and $B$, then we say the closure operator is topological. 
A function that is decreasing, isotone, and idempotent is called an interior operator. If in addition $f(A\cap B)=f(A)\cap f(B)$ for all $A$ and $B$, then we say the interior operator $f$ is topological. If $$f(A) = \bigcap\limits_{B\subseteq A, B\text{ finite}} f(B)$$ for all $A$, then we say the interior operator $f$ is algebraic.
So it looks like you might have an interior operator; it is certainly decreasing and idempotent, but you don't say enough to tell whether it is also isotone. 
You can find some of this in George Bergman's Invitation to General Algebra and Universal Constructions, Section 5.3, pages 134-139.
A: If you have a function with the properties in your question, then the function $g:A\mapsto X\setminus f(X\setminus A)$ satisfies $A\subseteq g(A)$ and $g(g(A))=g(A)$ for all $A$.  This is called a hull operator.  What you are looking for is the dual, and I am not aware of a name for this.  Maybe we should call it core-operator.  Also see my comment:  The interior operator on a topological space satisfies your conditions.
