Upper Closure of Coarser Subset is Subset of Upper Closure

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Theorem

Let $L = \left({S, \preceq}\right)$ be a preordered set.

Let $A, B$ be subsets of $S$ such that

$A$ is coarser than $B$.


Then $A^\succeq \subseteq B^\succeq$


Proof

Let $x \in A^\succeq$

By definition of upper closure of subset:

$\exists y \in A: y \preceq x$

By definition of coarser subset:

$\exists z \in B: z \preceq y$

By definition of transitivity:

$z \preceq x$

Thus by definition of upper closure of subset:

$x \in B^\succeq$

$\blacksquare$


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