Is the functor of open subschemes representable? First some simple observations in order to motivate the question:
The functor $Set^{op} \to Set, X \to \{\text{subsets of }X\}, f \to (U \to f^{-1}(U))$ is representable. The representing object is $\{0,1\}$ with the universal subset $\{0\}$. Also the functor $Top^{op} \to Set, X \to \{\text{open subsets of }X\}$ is representable: Endow $\{0,1\}$ with the topology such that $\{0\}$ is the unique nontrivial open subset (Sierpinski space), then it is again the representing object.
But what about schemes. Is the functor $Sch^{op} \to Set, X \to \{\text{open subschemes of }X\}$ representable? Of course, we could also talk about open subsets of $X$. My first idea was to endow the Sierpinski space above with a scheme structure, using DVR, but this does not work properly.
 A: No, it is not representable. If it were, the functor of open subset for schemes over a field $k$ would also be representable by the base change to $\mathop{\rm Spec} k$ of the representing object. Suppose that this last functor is represented by an open embedding of $k$-schemes $S_1 \subseteq S$. If $T$ is a $k$-scheme, there is a unique morphism $\phi\colon T \to S$ such that $\phi^{-1}S_1 = T$; this means that there a unique morphism $T \to S_1$, so $S_1 = \mathop{\rm Spec}k$. On the other hand, it is easy to see that a $k$-rational point on any $k$ is always closed. So $S_1$ is closed, which implies that every open subschems on a $k$-scheme is also closed; and this is patently false.
This example must be somewhere in my notes on descent theory.
A: It is representable by an fpqc sheaf (SGA I.VIII.4.4).  Also, since the inverse limit of affine schemes has the inverse limit topology (EGA IV.8), the sheaf is locally of finite presentation.  Since the open subsets of an infinitesimal thickening of a scheme are the same as those of the original scheme, this functor is even formally étale over $\mathrm{Spec} \:\mathbf{Z}$.  It's therefore just about as close as it could be to being an algebraic space without actually being one.
What goes wrong is algebraization.  If $X$ is an algebraic space and you have a complete local ring $R$ then $\mathrm{Hom}(\mathrm{Spec}\:R,X) = \varprojlim_n  \mathrm{Hom}(\mathrm{Spec}\:R_n,X)$ where the $R_n$ are the quotients of $R$ by powers of the maximal ideal.  It's pretty clear that this equality doesn't hold for the sheaf of open subsets.
