Lower Closure is Lower Section

Theorem
Let $(S, \preceq, \tau)$ 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 by the definition of lower closure, $b \in L$.

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

Also see

 * Upper Closure is Upper Set