 Solution

If $S = \emptyset$ , every element of $A$ is an upper bound. There is no maximum, because $S$ has no elements, and no least upper bound, because the distinct elements $2, 3$ are incomparable minimal elements of $A$ .
If $S = {p}$ with $p \in P$ , its upper bounds are ${p} \cup C$ ; its maximum and least upper bound are both $p$ .
If $S \subseteq P$ has at least two elements, its upper bounds are exactly $C$ . There is no maximum in $S$ and no least upper bound, because $C$ has at least two incomparable elements.
If $S \cap C = {c}$ , its only upper bound is $c$ : no different element is above this nonprime element. Since every remaining element of $S$ is a prime number, $c$ belongs to $S$ and dominates all of it. Thus $c$ is both its maximum and its least upper bound.
If $S$ contains at least two elements of $C$ , these distinct nonprime elements have no common upper bound, and $S$ has neither a maximum nor a least upper bound.

Consequently an upper bound exists exactly when $∣ S \cap C ∣ \leq 1$ ; a maximum or least upper bound exists exactly when $∣ S \cap C ∣ = 1$ or $S$ is a singleton subset of $P$ . The exhaustive cases also justify why the last two existence criteria coincide here.

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