Lower Closure is Lower Set

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Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.

Let $T$ be a subset of $S$.

Let $L$ be the lower closure of $T$.


Then $L$ is a lower set.


Proof

Let $a \in L$.

Let $b \in S$ with $b \preceq a$.

By the definition of lower closure, there is a $t \in T$ such that $a \preceq t$.

By transitivity, $b \preceq t$.

Thus, again by the definition of lower closure, $b \in L$.

Since this holds for all such $a$ and $b$, $L$ is a lower set.

$\blacksquare$


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