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!)