Image under Increasing Mapping equal to Special Set is Complete Lattice

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.

Let $f:S \to S$ be an increasing mapping.

Let $P = \left({M, \precsim}\right)$ be an ordered subset of $L$ such that
 * $M = \left\{ {x \in S: x = f\left({x}\right)}\right\}$

Then $P$ is complete lattice.

Proof
We will prove that
 * $\forall X \subseteq M: \forall Y \subseteq S: Y = \left\{ {x \in S: x}\right.$ is upper bound for $\left.{X \land f\left({x}\right) \preceq x}\right\} \implies \inf_L Y \in M$

Let $X \subseteq M$, $Y \subseteq S$ such that
 * $Y = \left\{ {x \in S: x}\right.$ is upper bound for $\left.{X \land f\left({x}\right) \preceq x}\right\}$

We will prove that
 * $f\left({\inf Y}\right)$ is lower bound for $Y$.

Let $y \in Y$.

By definition of $Y$:
 * $f\left({y}\right) \preceq y$

By definitions of infimum and lower bound:
 * $\inf Y \preceq y$

By definition of increasing mapping:
 * $f\left({\inf Y}\right) \preceq f\left({y}\right)$

Thus by definition of transitivity:
 * $f\left({\inf Y}\right) \preceq y$

We will prove that
 * $f\left({f\left({\inf Y}\right)}\right)$ is upper bound for $X$.

Let $m \in X$.

We will prove that
 * $m$ is lower bound for $Y$.

Let $y \in Y$.

By definition of $Y$:
 * $f\left({y}\right)$ is upper bound for $X$ and $f\left({y}\right) \preceq y$

By definition of upper bound:
 * $m \preceq f\left({y}\right)$

Thus by definition of transitivity:
 * $m \preceq y$

By definition of infimum:
 * $m \preceq \inf Y$

By definition of increasing mapping:
 * $f\left({m}\right) \preceq f\left({\inf Y}\right)$

By definition of subset:
 * $m \in M$

By definition of $M$:
 * $m = f\left({m}\right)$

By definition of increasing mapping:
 * $m \preceq f\left({f\left({\inf Y}\right)}\right)$

By definition of infimum:
 * $f \left({\inf Y}\right) \preceq \inf Y$

By definition of increasing mapping:
 * $f \left({f \left({\inf Y}\right)}\right) \preceq f \left({\inf Y}\right)$

By definition of $Y$:
 * $f \left({\inf Y}\right) \in Y$

By definitions of infimum and lower bound:
 * $\inf Y \preceq f \left({\inf Y}\right)$

By definition of antisymmetry:
 * $\inf Y = f \left({\inf Y}\right)$

Thus by definition of $M$:
 * $\inf Y \in M$

Define $Y_0 = \left\{ {y \in S: f \left({y}\right)}\right.$ is upper bound for $\left.{M \land f \left({y}\right) \preceq y}\right\}$

By lemma:
 * $\inf Y_0 \in M$

We will prove that
 * $\forall X \subseteq M: X$ admits a supremum in $P$.

Let $X \subseteq M$.

Define $Y = \left\{ {y \in S: f \left({y}\right)}\right.$ is upper bound for $\left.{M \land f \left({y}\right) \preceq y}\right\}$

Define $z = \inf Y$.

By lemma:
 * $z \in M$

We will prove that
 * $z$ is upper bound for $X$ in $P$.

Let $m \in X$.

By analogy:
 * $m$ is lower bound for $Y$ in $L$.

By definition of infimum:
 * $m \preceq \inf Y$

Thus by definition of ordered subset:
 * $m \precsim z$

We will prove that
 * $\forall x \in M: x$ is upper bound for $X$ in $P \implies z \precsim x$

Let $x \in M$ such that
 * $x$ is upper bound for $X$.

By definition of $M$:
 * $x = f\left({x}\right)$

By definition of ordered subset:
 * $f \left({x}\right)$ is upper bound for $X$ in $L$.

By definition of $Y$:
 * $x \in Y$

By definitions of infimum and lower bound:
 * $\inf Y \preceq x$

Thus by definition of ordered subset:
 * $z \precsim x$

Hence $X$ admits a supremum in $P$.

Thus by {{Complete iff Admits All Suprema]]:
 * $P$ is a complete lattice.

{{qed}}