Proposition 3

Let E be a right directed preordered set (resp. a lattice, resp. a totally ordered set). Then every non-empty finite subset of E is bounded above (resp. has a least upper bound and a greatest lower bound, resp. has a greatest element and a least element).

The proof is by induction on the number $n$ of elements of the subset under consideration. The result is trivial for $n = 1$. Let X be a subset of $n + 1$ elements of E (with $n \geqslant 1$), and put $\mathrm{X} = \mathrm{Y} \cup \{x\}$, where Y has $n$ elements and is therefore not empty. The inductive hypothesis implies the existence of an upper bound (resp. least upper bound, resp. greatest element) $y$ of Y. Since E is right directed (resp. a lattice, resp. totally ordered), $\{x, y\}$ has an upper bound (resp. least upper bound, resp. greatest element) which is evidently an upper bound (resp. least upper bound, resp. greatest element) of X.