Let $P$ be a poset with a least element $\bot$ ($\forall x \in P.\ \bot \le x$).
Let $M$ be a set of *monotone* maps $P \to P$.
Call $x \in P$ reachable if $x = f\_1(f\_2(...f\_n(\bot)...))$ for some sequence $f\_i \in M$.
Call $x$ fixed if $x = f(x)$ for all $f \in M$.


If $x$ reachable, $y$ fixed, then $x \le y$. For then $x = F(\bot)$ for $F$ some composition of functions in $M$. Since $F$ is monotone and $\bot \le y$, we have $x = F(\bot) \le F(y) = y$. Thus there is at most one reachable fixed point.



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> **Question #1**: If there is a reachable fixed point $x$, will applying functions in $M$ uniformly at random (starting from $\bot$) eventually find $x$ with probability 1?
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This requires a uniform probability distribution over $M$, so is ill-defined if $M$ is infinite. So:



> 
> **Question #2**: If there is a reachable fixed point $x$, will any schedule which eventually includes every sequence of $M$s infinitely often eventually find $x$ (starting from $\bot$)?
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For our purposes, $M$ is at most countable. I'd even be interested in a solution to the case where $M$ is finite.



Random iteration of a set of monotone maps until fixed point
I think yes, to Question #2, which implies Question #1.


Does this work?


Suppose $x$ is reachable and fixed. Pick $F$ with $x = F(\bot)$ to witness $x$'s reachability. Let $s$ be an arbitrary reachable element, with $F' \in M^\*$ witnessing its reachability, i.e. $F'(\bot) = s$.


Observe that $F(s)$ is reachable, since $F(s) = F(F'(\bot))$. So $F(s) \le x$ because $x$ is fixed, by your own reasoning above. But $\bot \le s$ so $x = F(\bot) \le F(s)$ by monotonicity. Hence $x = F(s)$.


That is, if we start from $\bot$ and hit it with some functions $F'$ from $M$, $x$ will always remain reachable, *even from that point $F'(\bot)$*, by exactly the sequence of functions $F$.


