Meet-Continuous and Distributive implies Shift Mapping Preserves Finite Suprema

Theorem
Let $\struct {S, \vee, \wedge, \preceq}$ be a meet-continuous distributive complete lattice.

Let $x \in S$.

Let $f: S \to S$ be a mapping such that
 * $\forall y \in S: \map f y = x \wedge y$

Then
 * $f$ preserves finite suprema

Proof
Let $X$ be finite subset of $S$ such that:


 * $X$ admits a supremum

By definition of complete lattice:


 * $f \sqbrk X$ admits a supremum

We will prove the result by induction on the cardinality of $X$.

Basis Case

 * $\forall X \subseteq S: \card X = 0 \implies \map \sup {f \sqbrk X} = \map f {\sup X}$

where:


 * $\card X$ denotes the cardinality of $X$

Let $X \subseteq S$ such that:


 * $\card X = 0$

By Cardinality of Empty Set:


 * $X = \O$

By definition of image of set:


 * $f \sqbrk X = \O$

By definitions of bottom and smallest element:


 * $\bot \preceq x$

Thus:

Induction Hypothesis

 * $\forall X \subseteq S: \card X = n \implies \map \sup {f \sqbrk X} = \map f {\sup X}$

Induction Step

 * $\forall X \subseteq S: \card X = n + 1 \implies \map \sup {f \sqbrk X} = \map f {\sup X}$

Let $X \subseteq S$ such that:
 * $\card X = n + 1$

Then $X = \set {x_1, \dots, x_n, x_{n + 1} }$.

Thus: