Smallest Element is Infimum

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

Let $T \subseteq S$.

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

Then $m$ is the infimum of $T$ in $S$.

Proof
Let $M$ be the smallest element of $T$.

Then by definition:
 * $\forall x \in T: m \preceq x$

By definition of infimum, it is necessary to show that:


 * $(1): \quad m$ is a lower bound of $T$ in $S$
 * $(2): \quad L \preceq m$ for all lower bounds $L$ of $T$ in $S$.

By Smallest Element is Lower Bound, $m$ is a lower bound of $T$ in $S$.

It remains to be shown that:
 * $L \preceq m$ for all lower bounds $L$ of $T$ in $S$.

Let $L \in S$ be a lower bound of $T$ in $S$.

By definition of lower bound:


 * $\forall t \in T: L \preceq t$

We have that $m \in T$.

Therefore:
 * $L \preceq m$

Hence the result.

Also see

 * Greatest Element is Supremum