Intersection of Subset with Lower Bounds

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

Let $T \subseteq S$.

Let $T_*$ be the set of all lower bounds of $T$ in $S$.

Then $T_* \cap T \ne \varnothing$ iff:
 * $T$ has a smallest element $m$

and
 * $T_* \cap T$ is a singleton such that $T_* \cap T = \left\{{m}\right\}$

Proof
Suppose $T_* \cap T = \varnothing$, where $\varnothing$ denotes the empty set.

That means $T$ contains none of its lower bounds, if indeed it has any.

From Smallest Element is Lower Bound, if $T$ had a smallest element, it would be a lower bound contained in $T$.

It follows that $T$ can have no smallest element.

Otherwise $T_* \cap T \ne \varnothing$.

That means $T$ contains at least one of its lower bounds.

Suppose $\exists a, b \in T_* \cap T$.

From Intersection Subset it follows that $a, b \in T$.

Then:
 * $\forall y \in T: a \preceq y$
 * $\forall y \in T: b \preceq y$

Thus both $a$ and $b$ fulfil the criteria for being a smallest element of $T$.

From Smallest Element is Unique it follows that $a = b$ and so $T_* \cap T$ is a singleton containing the smallest element of $T$.