Intersection of Subset with Upper Bounds
Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T^*$ be the set of all upper bounds of $T$ in $S$.
Then $T^* \cap T \ne \O$ if and only if:
- $T$ has a greatest element $M$
and
- $T^* \cap T$ is a singleton such that $T^* \cap T = \set M$
Proof
Suppose $T^* \cap T = \O$, where $\O$ denotes the empty set.
That means $T$ contains none of its upper bounds, if indeed it has any.
From Greatest Element is Upper Bound, if $T$ had a greatest element, it would be an upper bound contained in $T$.
It follows that $T$ can have no greatest element.
Otherwise $T^* \cap T \ne \O$.
That means $T$ contains at least one of its upper bounds.
Suppose $\exists a, b \in T^* \cap T$.
From Intersection is Subset it follows that $a, b \in T$.
Then:
- $\forall y \in T: y \preceq a$
- $\forall y \in T: y \preceq b$
Thus both $a$ and $b$ fulfill the criteria for being a greatest element of $T$.
From Greatest Element is Unique it follows that $a = b$ and so $T^* \cap T$ is a singleton containing the greatest element of $T$.
$\blacksquare$
Also see
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 14$: Order