Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Discrete mathematics, probability and continuous mathematics / 1 / c

The greatest element of $S$, also called its maximum, is an element $m\in S$ such that $s⪯m$ for every $s\in S$. Membership in $S$ is essential; an upper bound need not belong to $S$.

A least upper bound of $S$ is an element $u\in A$ such that $s⪯u$ for every $s\in S$, and such that $u⪯v$ for every other upper bound $v\in A$ of $S$. It need not belong to $S$. The antisymmetric relation axiom makes it unique whenever it exists.

Write each member as $3-2/(n+1)$. Every member is strictly below $3$, so $3$ is an upper bound. If $u<3$, the Archimedean property supplies $n\in \mathbb{N}$ with $n+1>2/(3-u)$; then $3-2/(n+1)>u$. Thus no smaller number is an upper bound, proving $\sup S=3$. Since $3\notin S$, there is no greatest element. Equivalently, each member is followed by a larger member because $2/(n+2)<2/(n+1)$.