Intersection of Subset with Lower Bounds

Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $T \subseteq S$.

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

Then $T_* \cap T \ne \O$ if and only if:

and

$T_* \cap T$ is a singleton such that $T_* \cap T = \set m$

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

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

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

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

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

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

From Intersection is 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 $T$.

$\blacksquare$