Smallest Element is Lower Bound

Theorem

Let $\struct {S, \preceq}$ 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 $m$.

$\blacksquare$

Also see

• Greatest Element is Upper Bound

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: $\S 2.7$: Maximum and Minimum