Kuratowski's closure-complement problem for the upper/lower bound functions This is an order theory problem in the flair of Kuratowski's closure-complement problem. Let $(X,\le)\newcommand{\up}{{\uparrow}}\newcommand{\dn}{{\downarrow}}$ be a set with a reflexive order on it, and define the operations 
$$A\up=\{x\in X \mid\forall y \in A,y\le x \}$$
$$A\dn=\{x\in X \mid\forall y \in A,y\ge x \}$$
called the upper and lower bound functions, respectively. For example, $\emptyset\up=X$, $X\up=1$ (the set of top elements), $A\up\cap A\up\dn$ is the set of least upper bounds of $A$. How many different sets are accessible by applying $\up$ and $\dn$ to some set $A$? In particular, does the sequence $A\up\dn\up\dn\dots$ ever stabilize?
The easiest case is that of a complete total order, such as the set $\Bbb R^*$ of extended real numbers. In this case we have $A\up=[\sup A,\infty)$ and $A\dn=(-\infty,\inf A]$ for any $A$, so there is immediate stabilization: $A\dn\up=A\up$ and $A\up\dn=A\dn$. The same is apparently true of the non-complete order $\Bbb Q$ (where the intervals are now $[a,\infty)\cap\Bbb Q, (-\infty,a]\cap\Bbb Q$ for some $a\in\Bbb R^*$), so I expect that the above argument can be extended to arbitrary total orders.
If $\le$ is not known to be transitive, it is fairly easy to come up with examples with infinitely many different sets: if we consider $(\Bbb N,\preceq)$ where $m\preceq n$ if $m\le n\le m+1$, we have $\{0\}\up=\{0,1\}, \{0,1\}\up=\{1\}$, and in general $\{0\}\up^{2n}=\{n\}$, so we can get infinitely many distinct sets.
If we suppose that $\le$ is transitive, i.e. a preorder, then it is less trivial. Starting from a preorder, we can reduce to the case of a partial order, because the equivalence $x\sim y\leftrightarrow x\le y\land y\le x$ is compatible with the $\up,\dn$ functions (so after the zeroth stage all the sets are unions of equivalence classes), so we can safely quotient it out.
For the interested: The source of this problem is the apparent asymmetry in the proof that a least upper bound of lower bounds is a greatest lower bound, but not conversely. This translates to the assertion $A\dn\up\cap A\dn\up\dn\subseteq A\dn\cap A\dn\up$, but I don't know enough about the behavior of these alternating sequences of $\up,\dn$ to establish anything nontrivial.
 A: There are two main properties of interest for the $\newcommand{\up}{{\uparrow}}\newcommand{\dn}{{\downarrow}}\up,\dn$ functions (note that since these functions are dual to each other there are also dualized versions of the theorems below), which hold for any relation $\le$ at all:


*

*$A\subseteq B\to B\up\subseteq A\up$. Suppose that $A\subseteq B$ and $x\in B\up$; then given any $y\in A$ we have $y\in B$ so $y\le x$. Thus $x\in A\up$.

*$A\subseteq A\dn\up$. Suppose that $x\in A$, and $y\in A\dn$. Then $y\le x$ by definition of $\dn$, so $x\in A\dn\up$ by definition of $\up$.
We can combine these to prove that $A\dn\up\dn=A\dn$: on the one hand we have $A\dn\subseteq A\dn\up\dn$ by the second theorem applied to $A\dn$, but on the other hand we also have $A\subseteq A\dn\up\to A\dn\up\dn\subseteq A\dn$. Thus we can eliminate any $\dn\up\dn$ or $\up\dn\up$ sequence.
Now consider the case where $\le$ is a bounded partial order. Then $A\up$ contains $1$ for any $A$, while $x\in\{1\}\up\to 1\le x\to x=1$, so $1\in A\up\up\subseteq \{1\}\up\subseteq\{1\}$ and hence $A\up\up=\{1\}$. Dually, $A\dn\dn=\{0\}$. Thus we get the following list (plus the duals of listed sets):
$$\begin{array}{ll} 
A&A\up\\
A\up\up=\{1\}&A\up\dn\\
A\up\up\dn=X&\require{cancel}\cancel{A\up\dn\up=A\up}\\
\cancel{A\up\up\up=A\up\up}&\cancel{A\up\dn\dn=A\dn\dn}\\
\cancel{A\up\up\dn\up=A\up\up}&\cancel{A\up\up\dn\dn=A\dn\dn}\\
\end{array}$$
Thus we have a total of $8$ sets generated (note that $A$ and $A\up\up\dn$ are their own duals), and this bound is saturated even in the complete total order example: Let $X=[0,3]$ and $A=[1,2]$ with the standard order. Then $$\begin{array}{ll} 
A=[1,2],&A\up=[2,3],&A\up\up=\{3\},&A\up\dn=[0,2],\\
X=[0,3],&A\dn=[0,1],&A\dn\dn=\{0\},&A\dn\up=[1,3]\end{array}$$
are the required distinct sets.
If $\le$ is a total order, we can make progress in a different direction. If $x\in A\up\up$, then either $A\up$ is empty (so $A\up\up=X$), or there is some $y\in A\up$, and then for any $z\in X$ either $z\le y\le x$ or $y\le z$ hence $z\in A\up$ and $z\le x$, so $x$ is the top element. Thus we have either that $A$ is bounded (and hence follows the same pattern as above), or $A\up\up=\emptyset$ replaces $\{1\}$ above; since $A\up\up\dn=\emptyset\dn=X$ is still true this does not generate any new sets over the BPO case.
This leaves open the case of $\le$ an unbounded partial order.
