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An ind-scheme over a base scheme $S$ can be defined in several ways. For simplicity, lets assume that $S$ is the spectrum of an algebraically closed field $k$. We can define a $k$-ind-scheme as a functor from $k$-algebras to sets, such that there exists a sequence of $k$-schemes with closed embeddings

$X_1 \to X_2 \to ...\to X_n \to ...$

and a natural isomorphism of $X$ with the functor $R \mapsto \lim_{\rightarrow}(X_n(R))$. Such a functor is a sheaf in the Zariski topology and can be uniquely extended as a sheaf to operate on all $k$-schemes.

Is there a sensible way to define the "underlying topological space" of an ind-scheme $X$? In other words, is it true that the direct limit of the topological spaces underlying the $X_n$ (an increasing union really) is independent of the presentation?

If we can take all $X_n$ to be quasi-compact, it seems to me that the answer is positive. since for different (quasi-compact) presentations, we will have closed embeddings $f_n:X_n\to X'_{m(n)}$. Am I right?

Edit:

Even though I wrote "any sensible way to define the underlying topological space", I now realize that I am actually only interested in the "direct limit" topology. The main question is the independence of presentation in the general or some special cases.

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The underlying set of a functor $X : \mathsf{Alg}(k) \to \mathsf{Set}$ is given by $\mathrm{colim}_{K/k} X(K)$, where $K/k$ runs through all field extensions. The topology is defined using open subfunctors. A reference is the book by Demazure and Gabriel on algebraic groups. I have learned it from a script by Marc Nieper-Wißkirchen. I'm not sure what happens for Ind-schemes, though.

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    $\begingroup$ Nice answer. It gives the topology introduced by Kambayashi, which is not the same as the one given by the inductive limit (see the differences here arxiv.org/abs/1109.4088 ) $\endgroup$ – Jérémy Blanc Sep 18 '13 at 14:00

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