**Puzzle 73**
> I claimed that every finite subset of a poset have a join if and only if every pair of elements has a join and the empty set has a join. Why is this true?

The empty set has a join because that is stipulated in the requirement. For any non-empty subset \$$A \in (P,\le) \$$ we must show that for \$$a,b,c,...,z \in A \$$, \$$\bigvee A = \bigvee ( a,b,c...,z ) = a \vee b \vee c \vee d \vee ... \vee y \vee z = (a \vee b) \vee (c \vee d) \vee ... \vee (y \vee z) \$$, i.e that we can split the join into seperate pairs, for which the joins exist. To do this we need to show that we can freely set parentheses, which is equivalent to showing that joins are associative. Therefore, we want to prove that \$$( a \vee b) \vee c = a \vee (b \vee c) \$$

A join is the least upper bound. Since it is an upper bound \$$a \le a \vee b \le (a \vee b) \vee c\$$ and \$$c \le b \vee c \le a \vee (b \vee c)\$$.

Since it is the least upper bound \$$c \le a \vee (b \vee c) \$$ implies \$$(a \vee b) \vee c \le a \vee (b \vee c) \$$. Likewise, \$$a \le (a \vee b) \vee c \$$ implies \$$a \vee (b \vee c) \le (a \vee b) \vee c \$$.

By proving the inequality from both sides we have shown that \$$(a \vee b) \vee c = a \vee (b \vee c) \$$ as desired \$$\square \$$

The proof that meets are associative follows by duality.