Set is Coarser than Image of Mapping of Infima
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Theorem
Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $f, g:\N \to S$ be mappings such that:
- $\forall n \in \N: \map g n = \inf \set {\map f m: m \in \N \land m \le n}$
Then $f \sqbrk \N$ is coarser than $g \sqbrk \N$
where $f \sqbrk \N$ denotes the image of mapping $f$.
Proof
Let $x \in f \sqbrk \N$.
By definition of image of mapping:
- $\exists n \in \N: x = \map f n$
By definition of $g$:
- $\map g n = \inf \set {\map f m: m \in \N \land m \le n}$
By definition of reflexivity:
- $n \le n$
Then
- $\map f n \in \set {\map f m: m \in \N \land m \le n}$
By definitions of infimum and lower bound:
- $\map g n \preceq x$
By definition of image of mapping:
- $\map g n \in g \sqbrk \N$
Thus
- $\exists y \in g \sqbrk \N: y \preceq x$
$\blacksquare$
Sources
- Mizar article WAYBEL12:31