closure map

Let $P$ be a poset. A function $c:P\to P$ is called a closure map if

• $c$ is order preserving,

• $1_P\le c$,

• $c$ is idempotent: $c\circ c=c$.

If the second condition is changed to $c\le 1_P$, then $c$ is called a dual closure map on $P$.

For example, the real function $f$ such that $f(r)$ is the least integer greater than or equal to $r$ is a closure map (see Archimedean property). The rounding function $[\cdot ]$ is an example of a dual closure map.

A fixed point of a closure map $c$ on $P$ is an element $x\in P$ such that $c(x)=x$. It is evident that every image point of $c$ is a fixed point: for if $x=c(a)$ for some $a\in P$, then $c(x)=c(c(a))=c(a)=x$.

In the example above, any integer is a fixed point of $f$.

Every closure map can be characterized by an interesting decomposition property: $c:P\to P$ is a closure map iff there is a set $Q$ and a residuated function $f:P\to Q$ such that $c=f^{+}\circ f$, where $f^{+}$ denotes the residual of $f$.

Again, in the example above, $f=g^{+}\circ g$, where $g:ℝ\to \mathbb{Z}$ is the function taking any real number $r$ to the largest integer smaller than $r$. $g$ is residuated, and its residual is $g^{+}(x)=x+1$.

Remark. Closure maps are generalizations to closure operator on a set (see the parent entry). Indeed, any closure operator on a set $X$ takes a subset $A$ of $X$ to a subset $A^c$ of $X$ satisfying the closure axioms, where Axiom 2 corresponds to condition 2 above, and Axiom 3 says the operator is idempotent. To see that the operator is order preserving, suppose $A\subseteq B$. Then $B^c={(A\cup B)}^c=A^c\cup B^c$ by Axiom 4, and hence $A^c\subseteq B^c$. Axiom 1 says that the empty set $\mathrm{\varnothing }$ is a fixed point of the operator. However, in general, this is not the case, for $P$ may not even have a minimal element, as indicated by the above example.