Increasing Mapping Preserves Upper Bounds
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Theorem
Let $L = \left({S, \preceq}\right)$, $L' = \left({S', \preceq'}\right)$ be ordered sets.
Let $f:S \to S'$ be an increasing mapping.
Let $x \in S$, $X \subseteq S$ such that
- $x$ is upper bound for $X$.
Then $f\left({x}\right)$ is upper bound for $f\left[{X}\right]$.
Proof
Let $y \in f\left[{X}\right]$.
By definition of image of set:
- $\exists z \in X: y = f\left({z}\right)$
By definition of upper bound:
- $z \preceq x$
Thus by definition of increasing mapping:
- $y \preceq' f\left({x}\right)$
$\blacksquare$
Sources
- Mizar article YELLOW_2:14