Lower Bound for Subset

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

Let $L$ be a lower bound for $S$.

Let $\left({T, \preceq}\right)$ be a subset of $\left({S, \preceq}\right)$.

Then $L$ is a lower bound for $T$.

Proof
By definition of lower bound:
 * $\forall x \in S: L \preceq x$

But as $\forall y \in T: y \in S$ by definition of subset, it follows that:
 * $\forall y \in T: L \preceq y$.

Hence the result, again by definition of lower bound.

Also see

 * Upper Bound for Subset