Here’s a nice little theorem in rigid analytic geometry.
Theorem. Let be an affinoid rigid space, with a closed analytic subset and an admissible open subset containing . Then for some , the “tube” around is contained in .
To quote from Conrad’s notes on nonarchimedean geometry: “This can be proved by the methods of rigid geometry, but the only proof along such lines which I know is long and complicated. A short proof was given by Kisin via Raynaud’s formal models…” Conrad then goes on to give a proof using Berkovich spaces. Here we give a very short adic proof, with spectrality of qcqs adic spaces as the key ingredient.
Proof. WLOG is qc (for any covering of by affinoids, some finite subcover will contain all of ). Passing to adic spaces, it clearly suffices to show that for some . Since , we clearly have . But is a closed subset of the spectral space , so is spectral itself, and hence compact and Hausdorff in the constructible topology. Therefore (putting the constructible topology on everything) is an empty inverse limit of compact Hausdorff spaces, and thus one of the terms in the limit must be empty. (A reference for this last implication is e.g. Proposition 1.1.4 in this book.)
People often say things like “Berkovich spaces are great because they’re locally compact Hausdorff and so you can make nice topological arguments on them”; the moral here is that adic spaces have the same flexibility once one passes to the constructible topology.
(I don’t know Kisin’s proof, but it must be related somehow, since the topological spaces underlying the adic spaces come from the inverse limit of all formal models.)