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.



Is the functor of open subschemes representable?
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.


