Here are my puzzles about closed elements: Let \\(f: A\to B\\) and \\(g: B\to A\\) be the left and right adjoints in a Galois connection between posets \\(A\\) and \\(B\\), so that \\(f(a)\leq b \iff a \leq g(b)\\). Given \\(a\in A\\), we'll write \\(g(f(a))\\) as \\([a]\\) for short, and \\([b]=f(g(b))\\) for each \\(b\in B\\).

**Puzzle OB1 (Kenobi):** Show that for all \\(a\in A\\) and \\(b\in B\\), we have \\(a\leq [a]\\) and \\(b\geq [b]\\).

We'll say that an element \\(x\\) (of \\(A\\) or \\(B\\)) is _closed_ if \\(x = [x]\\).

**Puzzle OB2:** Show that for all \\(a\in A\\) and \\(b\in B\\), the elements \\(f(a)\in B\\) and \\(g(b)\in A\\) are both closed. (Hint: You can deduce both \\(f(a)\leq f(g(f(a)))\\) and \\(f(g(f(a)))\leq f(a)\\) from OB1.) Deduce that \\([a]\\) and \\([b]\\) are themselves closed.

**Puzzle OB3:** Show that if \\(a\leq a'\\) in \\(A\\) then \\([a]\leq [a']\\). Deduce that if \\(a\leq a'\\) and \\(a'\\) is closed, then \\([a]\leq a'\\). We say that \\([a]\\) is the _smallest_ closed element that is at least as big as \\(a\\). Show that for any \\(b\in B\\), \\([b]\\) is the _biggest_ closed element that is at least as _small_ as \\(b\\).

**Puzzle OB4:** Let's write \\(A_\text{closed}\\) and \\(B_\text{closed}\\) for the sets of closed elements in \\(A\\) and \\(B\\). Show that \\(f\\) and \\(g\\) restrict to functions \\(A_\text{closed}\leftrightarrow B_\text{closed}\\), and that on these subsets they are inverses of each other! Deduce that \\(A_\text{closed}\\) and \\(B_\text{closed}\\) are isomorphic posets.

(John, this all ended up being longer than I meant it to be. I hope you don't mind me plopping it all in here!)