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

Put $J=\{k\in \mathbb{N}:k\le 100\}$, avoiding any dependence on whether the convention includes $0$ among the natural numbers. Under set inclusion, the results are

The set union of all three-element subsets of $\mathbb{N}$ is $\mathbb{N}$. An upper bound for $S_1$ under set inclusion would have to contain this entire infinite set, whereas members of $A$ are finite. Thus $S_1$ has no upper bound and hence no maximum or least upper bound.

For $S_2$, every member is a subset of the finite set $J$, and $J$ itself belongs to $S_2$, making it both maximum and least upper bound. The union of the three-element subsets of $J$ is again $J$, so $J$ is the least upper bound of $S_1\cap S_2$. It is not a member of that family, and no three-element member contains all the others, so there is no maximum.

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.