algebraic closure  and real closure are closure operators? Are the algebraic closure (of a field) and the real closure (of a totally ordered field) closure operators (when restricted to appropriate sets of fields, so that they are maps on a set instead of a class)?
From what I see, yes, but I just need a confirmation from someone knowledgeable of the subject. I'm in doubt because I haven't found one citation of the real closure being a closure operator. 
 A: The algebraic closure and real closure are closures in the sense that $\overline{\overline{K}} = \overline{K}$ and $(K^r)^r = K^r$ (where '=' might not literally be equality, but is definitely a canonical, natural isomorphism), and there are canonical, natural inclusions, e.g. $K \mapsto \overline{K}$.  And given an inclusion $K \subseteq L$, you can choose algebraic closures with $\overline{K} \subseteq \overline{L}$. You'd have to jump through some hoops to turn it into a closure of the form you describe.
On the partially ordered set of subfields of a field $K$, you can define a closure operator $F \mapsto \overline{F} \cap K$. Similarly for a formally real field and its set of formally real subfields.
EDIT: the unbounded version is best stated in terms of category theory. Let $\bf Fld$ be the category of fields and homomorphisms. There is a functor $C : {\bf Fld} \to {\bf Fld}$ and a natural transformation $i : {\bf 1} \to C$ such that


*

*$C(F)$ is the algebraic closure of $F$

*$i_{C(F)} : C(F) \to C(C(F))$ is an isomorphism


A similar version should work for real closure too, on the category of formally real fields, or maybe ordered fields.
EDIT2: as the comments show, I got the above part wrong. You can define $C$ as a graph homomorphism, but it won't be a functor. You could switch to a suitable subcategory of $\bf Fld$ to make it a functor, but I'm not sure how informative that would be.
