# 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.

$\blacksquare$