Preimage of Upper Section under Increasing Mapping is Upper
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Theorem
Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be preordered sets.
Let $f: S \to T$ be an increasing mapping.
Let $X \subseteq T$ be a upper subset of $T$.
Then $f^{-1} \sqbrk X$ is upper
where $f^{-1} \sqbrk X$ denotes the preimage of $X$ under $f$.
Proof
Let $x \in f^{-1} \sqbrk X$, $y \in S$ such that
- $x \preceq y$
By definition of increasing mapping:
- $\map f x \precsim \map f y$
By definition of preimage of set:
- $\map f x \in X$
By definition of upper section:
- $\map f y \in X$
Thus by definition of preimage of subset:
- $y \in f^{-1} \sqbrk X$
$\blacksquare$
Sources
- Mizar article WAYBEL17:2