ag.algebraic geometry - subspace topology for functors

I don't think this is true even when $X$ and $V$ are schemes if you only require that the
map $Vto X$ 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
$Vto X$ (corresponding to stratification of $X$) is a categorical monomorphism: it induces an embedding of functors.

EDIT:

Now we add the assumption that $Vhookrightarrow X$ 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
$X_nhookrightarrow X_{n+1}$ 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 $Vsubset X$ is given by a compatible
family of locally closed $V_nsubset X_n$ (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 $X_nhookrightarrow X_{n+1}$ being the coordinate embedding.
Now let $V_n$ be the union of the origin $0$ and $n-1$ punctured lines

$$l_k:=lbrace(k,0,dots,0,x,0,dots,0)|xne 0rbrace,$$
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, $U_nsubset V_n$ 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 $W_nsubset X_n$
such that $U_n=V_ncap W_n$. Indeed, $W_1$ contains $0$, so it is non-empty, and thus contain
$nin{mathbb A}^1=X_1$ for some $n$. But then $W_n$ must contain $(n,0,dots,0)$ without meeting
$l_n$.

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