I'm trying to show that if $D$ is a countably complete lattice and $F:D \rightarrow D$ is a monotone function, then the least fixed point is $\sup\{ f^n\bot \mid n \geq 0\}$, where $\bot$ is the least element of the lattice.
In order to show that I would like to use induction on $n$ and at some point I have to say that:
$f(\sup\{f^n\bot \mid n \geq 0\}) = \sup\{f(f^n\bot) \mid n \geq 0\}$
that is true if $F$ if continuous. How can I demonstrate it?
Here's what may be the accurate description of the problem.
An (partial) order (S,<=) is countably complete when
for all countable A, sup A is in S.
A function f:S -> S, is continuous when for all nonempty chains C,
f(sup C) = sup f(C).
When S isn't complete but only countably complete, to makes sense
of the problem, let's define a countably continuous function
f:S -> S, as for all nonempty countable chains C, f(sup C) = sup f(C).
If S is countably complete, there's a bottom element b in S.
Proposition. If S is countably complete and f:S -> S, is
increasing and countably continuous, then C = { f^n(b) | n = 0,1,.. }
is a notempty countable chain and a = sup C is a fixed point of f.
Proceeding b <= f(b), f(b) <= ff(b),... by induction, it's apparent
that C is a chain. Thus f(sup C) = sup f(C). Since f in increasing,
sup C <= sup f(C). Since f(C) = { f^n(b) | n = 1,2,.. } is
a subset of C, sup f(C) <= sup f(C). Whence f(sup C) = sup C, QED.
Does this answer your question?