Are fixpoints of monotone functions monotone?

Question

Edited!
I work with power sets and functions on power sets. These are by definition partially ordered by set inclusion. Therefore I have limited the used definitions to those (typically, these definitions are much more general).

Def. A complete lattice is a partially ordered set $(\wp(A),\subseteq)$ where every subset ($X \subseteq \wp(A)$) has a supremum (join) $\bigcup X \in A$ and an infimum (meet) $\bigcap X \in A$.

Def. A function $f\colon \wp(A) \rightarrow \wp(A)$, where $(\wp(A),\subseteq)$ is a partially ordered set, is a monotone function if $X \subseteq Y \Rightarrow f(X) \subseteq f(Y)$ whenever $X,Y \in \wp(A)$.

Def. A fixpoint of a function $f\colon \wp(A) \rightarrow \wp(A)$ is an element $X \in \wp(A)$ where $X = f(X)$. Let $X\in \wp(A)$, we denote $n$ applications of $f$ to $X$ by $f^n(X)$, eg. $f(f(X)) = f^2(X)$.

Theorem (Knaster-Tarski). Let $(\wp(A),\subseteq)$ be a complete lattice and let $f\colon \wp(A) \rightarrow \wp(A)$ be a monotone function. Then the set of fixed points of $f$ in $\wp(A)$, i.e., $\mathsf{Fix}\, f =\{X \in\wp(A)\mid X = f(X)\}$ is also a complete lattice.

Question 1: Is the following correct?
Since $(\wp(A),\subseteq)$ be a complete lattice then $\wp(A)$ has a supremum $\bigcup \wp(A) \in \wp(A)$. So, because $f$ is monotone and there is no bigger element than $\bigcup \wp(A)$ then $f(\bigcup \wp(A))=\bigcup \wp(A)$.

Question 2: Is the following correct?
Let $X \in \wp(A)$. Because $f$ is monotone then there exists a $k$ such that $f^k(X) \in \mathsf{Fix}\, f$. This seems somehow related to the monotone convergence theorem.

Question 3: We will need an extra definition (can only be introduced if Question 2 is ok).
Def. Let $\mathsf{lfp}_f(X)$ denote the fixpoint reached by $f^k(X)$ whenever $X\in \wp(A)$.
Is the following correct and if so, I'm looking for clues to why.
$X \subseteq Y \Rightarrow \mathsf{lfp}_f(X) \subseteq \mathsf{lfp}_f(Y)$ ?

Answer 1

1. Incorrect. Incidentally, note that $\bigcup \wp(A) = A$. It is indeed the greatest element of $\wp(A)$, but it doesn't mean that $f(A) = A$, because there may be no $B \subseteq A$ such that $f(B) = A$ at all. Consider for example $f(B) = \varnothing$ for all $B \in \wp(A)$.

2. Wrong. Consider a function $f : \wp(\mathbb{N}) \to \wp(\mathbb{N})$ given by $\wp(B) = \{ b+1 : b \in B \}$. It is clearly monotone but the sequence $f^k(\{ 0 \}) = \{ k \}$ doesn't stabilize. Also it's an interesting example to test any other conjecture you might have.

3. Invalidated, but can be partially answered. Suppose $X \subseteq Y$. Then by monotonicity for each $n \in \mathbb{N}$ we must have that $f^n(X) \subseteq f^n(Y)$, so if both $f^k(X)$ and $f^k(Y)$ stabilize at some point, then there is common $n \in \mathbb{N}$ such that $\mathsf{lfp}_f(X) = f^n(X)$ and $\mathsf{lfp}_f(Y) = f^n(Y)$, so $\mathsf{lfp}_f(X) \subseteq \mathsf{lfp}_f(Y)$.