# Smallest Element is Lower Bound

## Theorem

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

Let $T \subseteq S$.

Let $T$ have a smallest element $m \in T$.

Then $m$ is a lower bound of $T$.

It follows by definition that $T$ is bounded below.

## Proof

Let $m \in T$ be a smallest element of $T$.

By definition:

$\forall y \in T: m \preceq y$

But as $T \subseteq S$, it follows that $m \in S$.

Hence:

$\exists m \in S: \forall y \in T: m \preceq y$

Thus $T$ is bounded below by the lower bound $m$.

$\blacksquare$