# Intersection of Subset with Upper Bounds

## Contents

## Theorem

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

Let $T \subseteq S$.

Let $T^*$ be the set of all upper bounds of $T$ in $S$.

Then $T^* \cap T \ne \varnothing$ if and only if:

- $T$ has a greatest 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 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 $T^* \cap T \ne \varnothing$.

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 $b$ fulfil 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