Let $k$ be a field complete with respect to a non-archimedean absolute value, and $X$ a separated algebraic space of finite type over $k$.
If $X$ is a scheme then $X(k)$ inherits a natural (Hausdorff) topology from $k$ (uniquely determined by functoriality, compatibility with open immersion, closed immersions, fiber products, and the special case of the affine line), locally compact
when $k$ is locally compact. There is even a natural $k$-analytic manifold structure if $X$ is smooth. This can be seen in (at least) two ways: algebraically by using Zariski-open covers by affines, and for smooth case take them to be "standard etale" over affine spaces (and then the $k$-analytic inverse function theorem can be applied), or analytically by identifying $X(k)$ with $X^{\rm{an}}(k)$ for the analytification $X^{\rm{an}}$ in the sense of rigid-analytic spaces (so then affinoid open covering of $X^{\rm{an}}$ does the job).
Now suppose $X$ is not a scheme (as often happens with moduli spaces which only exist as algebraic spaces; e.g., Rapoport's thesis, etc.). The 2nd method above can be used, but constructing $X^{\rm{an}}$ as a rigid-analytic space is really hard (the only method I know is to take a very long detour through Berkovich spaces to make the required rigid-analytic quotients from an etale scheme chart for $X$).
My question is this: is there a known simple procedure (much as the method in the scheme case is simple), bypassing the use of $X^{\rm{an}}$, to make a functorial Hausdorff topology on $X(k)$ (and $k$-analytic manifold structure when $X$ is smooth) which is locally compact
when $k$ is and recovers the usual topology when $X$ is a scheme and shares the same basic properties (good behavior for open immersions, closed immersions, and fiber products; carries etale maps to local homeomorphisms as well as local $k$-analytic isomorphisms in case $X$ is smooth; and $X(k) \rightarrow X(k')$ is a closed embedding any extension
$k'/k$ of such fields)?
A natural idea for the topology aspect is to choose an etale scheme cover $U \rightarrow X$ and identify $X(k_s)$ with a quotient of $U(k_s)$ as a set. Give it the quotient topology of the natural topology on $U(k_s)$ arising from the topology on $k_s$, and give $X(k)$ the subspace topology from $X(k_s)$. This does give the right answer (same as via the rigid-analytic method), but the only way I see this is locally compact when $k$ is
(noting that $X(k_s)$ is essentially never locally compact) and functorial is to invoke comparison with the rigid-analytic method. (Otherwise I get bogged down in rising chains of finite Galois extensions and it feels like it becomes a mess. Doing the $k$-analytic manifold structure for smooth $X$ in this way also seems to become quite unpleasant. Note, by the way, that $k_s$ is not complete, so can't speak of $k_s$-analytic manifolds.) So this idea doesn't seem to provide an answer. Or maybe I am missing a simple trick?
I am happy to restrict to the case of locally compact $k$ (though allowing completed algebraic closures is perfectly interesting), but do not want to restrict to characteristic 0 (though ideas in that case are appreciated too).
Best Answer
Assume $k$ is a field and $F$ is a contravariant functor from $k$-schemes of finite type to sets. For each $X/k$ and $x\in F(X)$ we get a "characteristic map" $X(k)\to F(k)$ by pulling back $x$. Now if $k$ is a topological field, we can define a topology on $F(k)$, which is the finest making all these maps (for all pairs $(X,x)$) continuous. Some facts are easy to check, for instance:
With this generality, of course, one doesn't get very far without making some assumptions on $F$ and/or $k$. For instance, compatibility with products is not a formal consequence of the definition and therefore, probably, does not hold in general (example, please?). The nasty thing about this topology is that it is a quotient topology ($F(k)$ is a quotient space of the disjoint sum of all $X(k)$ indexed by pairs $(X,x)$) and general quotient topologies don't behave nicely.
Assume now that $k$ satisfies the following "weak henselian" property:
(WH) Every étale morphism $Y\to X$ gives rise to an open map $Y(k)\to X(k)$.
This is satisfied by local fields and more generally by henselian valued fields. It is of course particularly relevant if $F$ is an algebraic space of finite type over $k$. In this case, it is easy to check that if $U\to F$ is an étale presentation, then, assuming (WH), $U(k)\to F(k)$ is an open map. This (together with the existence of pointed étale neighborhoods) is enough to imply, for instance, that the topology on $F(k)$ is compatible with products of algebraic spaces. (The point is that we now have a quotient by an open equivalence relation, and these quotients do behave nicely). I have not checked the other conditions in detail but I am confident that it works.
Some remarks: