subspace topology for functors let $X : Ring \to Set$ be a functor (a Z-functor in the language of demazure, gabriel) and $V \subseteq X$ a locally closed subfunctor. assume that $U \subseteq V$ is an open subfunctor. does then exist an open subfunctor $W \subseteq X$ such that $U = V \cap W$? if $X$ and $V$ are schemes, this should be true.
 A: I don't think this is true even when $X$ and $V$ are schemes if you only require that the
map $V\to X$ is an embedding of functors (rather than a locally closed embedding). Example:
Take $X=Spec(k[x])$, $V=Spec(k[x,x^{-1}]\times k)$, $U=Spec(k)$.
That is, $X$ is an affine line, $V$ is the disjoint union of a punctured line and origin,
and $U$ is a point, which we view as the connected component of $V$. The natural map
$V\to X$ (corresponding to stratification of $X$) is a categorical monomorphism: it induces an embedding of functors.
EDIT:
Now we add the assumption that $V\hookrightarrow X$ is locally closed. I think the statement still fails,
here's a counterexample:
$X$ will be an ind-scheme: so there is a sequence of schemes $X_n$ related by closed embeddings
$X_n\hookrightarrow X_{n+1}$ and
$$X(A)=\lim_{\to} X_n(A).$$
In other words, every point of $X(A)$ factors through one of $X_n$'s. 
For ind-schemes, a locally closed subfunctor $V\subset X$ is given by a compatible
family of locally closed $V_n\subset X_n$ (so that it is a locally closed sub-ind-scheme). Problem
is, the statement fails for ind-schemes. 
Let's take
$X_n={\mathbb A}^n$, with the embedding $X_n\hookrightarrow X_{n+1}$ being the coordinate embedding.
Now let $V_n$ be the union of the origin $0$ and $n-1$ punctured lines 
$$l_k:=\lbrace(k,0,\dots,0,x,0,\dots,0)|x\ne 0\rbrace,$$
where $x$ is in the $k$-th position, and $k$ varies from $2$ to $n$. (Let's say I work over
a field of characteristic $0$, so all integers are distinct.)
Finally, $U_n\subset V_n$ is the origin, which is a component of each $V_n$, so it is both open
and closed.
It is easy to see that it is impossible to find a compatible family of open subsets $W_n\subset X_n$
such that $U_n=V_n\cap W_n$. Indeed, $W_1$ contains $0$, so it is non-empty, and thus contain
$n\in{\mathbb A}^1=X_1$ for some $n$. But then $W_n$ must contain $(n,0,\dots,0)$ without meeting
$l_n$. 
