Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 3 / c / ii

The displayed relation is not a partial order on the printed $A$. Since the empty set belongs to $A$, vacuous truth makeshold for every finite $T$, including nonempty $T$. This violates the defining condition of an antisymmetric relation. It also violates transitivity: $\{2\}{⪯}_2\varnothing {⪯}_2\{1\}$ holds, but $\{2\}{⋠}_2\{1\}$. Consequently the usual unique-maximum and unique-least upper bound terminology for a partially ordered set does not apply as claimed.

The available inputs still determine what happens if one applies the formulas in (b) literally to this relation:

For $S_1$, no nonempty finite set can lie above all triples of arbitrarily large numbers, while $\varnothing$ does so vacuously. Thus $\varnothing$ is its sole literal upper bound, but is not a member of $S_1$. For each of the two bounded families, the literal upper bounds are $\varnothing$ and all nonempty finite sets whose elements are at least $100$. Among these, exactly $\varnothing$ and $\{100\}$ lie below every such upper bound. Both belong to $S_2$, but neither belongs to its three-element subfamily.

For comparison, a natural correction is to retain the given rule for nonempty sets and declare $\varnothing$ to be a bottom element: no nonempty set is then below $\varnothing$. Under this explicitly corrected partial order, $S_1$ has neither a maximum nor a least upper bound; $S_2$ has maximum and least upper bound $\{100\}$; and $S_1\cap S_2$ has no maximum but has least upper bound $\{100\}$. A nonempty upper bound must have every element at least as large as every element of the bounded family. These corrected answers are conditional on the stated repair, not answers obtained by silently excluding the empty set from the source.