Exercise 7, § 2

Let E be an ordered set and let F be the set of all closures (§ 1, Exercise 13) in E. Order F by putting $u \leqslant v$ whenever $u(x) \leqslant v(x)$ for all $x \in E$. Then F has a least element $e$, the identity mapping of E onto itself. For each $u \in F$ let $I(u)$ denote the set of elements of E which are invariant under $u$.

(a) Show that $u \leqslant v$ in F if and only if $I(v) \subset I(u)$.

(b) Show that if every pair of elements of E has a greatest lower bound in E, then every pair of elements of F has a greatest lower bound in F. If E is a complete lattice, then so is F (§ 1, Exercise 11).

(c) Show that if E is inductive (with respect to the relation $\leqslant$), then every pair $u$, $v$ of elements of F have a least upper bound in F. (Show that if $f(x) = v(u(x))$ and if $w(x)$ denotes the greatest element of the chain of $x$, relative to $f$ (Exercise 6), then $w$ is a closure in E and is the least upper bound of $u$ and $v$.)