Upper Closure in Ordered Subset is Intersection of Subset and Upper Closure
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Theorem
Let $L = \left({S, \preceq}\right)$ be an ordered set.
Let $\left({T, \precsim}\right)$ be an ordered subset of $L$.
Let $t \in T$.
Then $t^\succsim = T \cap t^\succeq$
Proof
By definition of ordered subset:
- $T \subseteq S$
We will prove that
- $t^\succsim \subseteq T \cap t^\succeq$
Let $x \in t^\succsim$
By definition of upper closure of element:
- $x \in T$ and $t \precsim x$
By definition of ordered subset:
- $t \preceq x$
By definition of upper closure of element:
- $x \in t^\succeq$
Thus by definition of intersection:
- $x \in T \cap t^\succeq$
$\Box$
We will prove that
- $T \cap t^\succeq \subseteq t^\succsim$
Let $x \in T \cap t^\succeq$
By definition of intersection:
- $x \in T$ and $x \in t^\succeq$
By definition of upper closure of element:
- $t \preceq x$
By definition of ordered subset:
- $t \precsim x$
Thus by definition of upper closure of element:
- $x \in t^\succsim$
$\Box$
By definition of set equality:
- $t^\succsim = T \cap t^\succeq$
$\blacksquare$