This is a question in the topic of Supremum/Maximum/Infimum/Minimum.
Showing that $ O \cap M = \emptyset $ or $ O \cap M = \{b\} $ is true. Where $M$ is a set and $O$ is the set of Upper Bounds. And $b \in \mathtt R $. Also how does one show that the second condition is only true if $M$ has a Maximum (uniqueness not required).
My attempt: If:$$ \exists m\in M: a\le m, \forall a\in M \rightarrow m\in O \cap M $$ $$ \exists m\notin M: a\lt m, \forall a\in M \rightarrow m\in O \rightarrow O \cap M= \emptyset $$ The first line of the proof also constitutes that $m$ must be the $max(M)$. However, I am not sure if this suffices for a formal proof, and if not, what does?
You want to show that $O\cap M$ has at most one element.
Suppose $a,b\in O\cap M$. Then $a\le b$ (because $a\in M$ and $b$ is an upper bound for $M$), and $b\le a$ (because $b\in M$ and $a$ is an upper bound for $M$). From $a\le b$ and $b\le a$ it follows that $a=b$. This shows that all elements of $O\cap M$ are equal, i.e., $O\cap M$ has at most one element.