Question 21

A poset $(X, R)$ is a lattice if and only if every pair of elements $x, y \in X$ has a unique greatest lower bound (meet, $\text{glb}(x, y)$) and a unique least upper bound (join, $\text{lub}(x, y)$).

The given Hasse diagram has 5 elements: $a$ is the unique bottom element, $e$ is the unique top element, and the two parallel paths from $a$ to $e$ are $a \to b \to c \to e$ and $a \to d \to e$. This is the standard pentagon poset $N_5$.

Check all incomparable pairs of elements in $(X, R)$:

1. Pair $\{b, d\}$:
   • Common lower bounds: $\{a\} \implies \text{glb}(b, d) = a$.
   • Common upper bounds: $\{e\} \implies \text{lub}(b, d) = e$.
2. Pair $\{c, d\}$:
   • Common lower bounds: $\{a\} \implies \text{glb}(c, d) = a$.
   • Common upper bounds: $\{e\} \implies \text{lub}(c, d) = e$.

For any comparable pair $x \le y$, $\text{glb}(x, y) = x$ and $\text{lub}(x, y) = y$, which always exist uniquely.

Because every pair of elements already has a unique GLB and LUB, $(X, R)$ is already a lattice! Therefore, the minimum number of ordered pairs that need to be added to $R$ is 0.