Let be a spectral space. Following Bhatt-Scholze, say is *w-local *if the subset of closed points is closed and if every connected component of has a unique closed point. This implies that the natural composite map is a homeomorphism (cf. Lemma 2.1.4 of BS).

For the purposes of this post, a *w-local adic space* is a qcqs analytic adic space whose underlying spectral topological space is w-local. These are very clean sorts of spaces: in particular, each connected component of such a space is of the form , where is a nonarchimedean field and is an open and bounded valuation subring of , and therefore has a unique closed point and a unique generic point.

I’ve been slowly internalizing the philosophy that w-local affinoid perfectoid spaces have a lot of amazing properties. Here I want to record an example of this sort of thing.

Given a perfectoid space together with a subset , let’s say is perfectoid (resp. affinoid perfectoid) if there is a pair where is a perfectoid space (resp. affinoid perfectoid space) and is a map of adic spaces identifying homeomorphically with and which is universal for maps of perfectoid spaces which factor through on topological spaces. Note that if the pair exists, it’s unique up to unique isomorphism.

**Theorem. ***Let be a w-local affinoid perfectoid space. Then any subset of which is closed and generalizing, or which is quasicompact open, is affinoid perfectoid.*

*Proof when is closed and generalizing. *The key point here is that the map defines a bijection between closed generalizing subsets of and closed subsets of the (profinite) space , by taking preimages of the latter or images of the former. To check that this is true, note that if is closed and generalizing, then its intersection with a connected component of being nonempty implies (since is generalizing) that contains the unique rank one point of . But then contains all specializations of that point (since is closed in ), so , so any given connected component of is either disjoint from or contained entirely in . This implies that can be read off from which closed points of it contains. Finally, one easily checks that is closed in , since is profinite and is quasicompact. Therefore .

Returning to the matter at hand, write as a cofiltered intersection of qc opens , . But qc opens in are the same as open-closed subsets, so each pulls back to an open-closed subset , and its easy to check that any such comes from a unique rational subset . Then is the perfectoid space we seek.

*Proof when is quasicompact open. *

First we prove the result when is connected, i.e. when as above. We claim that in fact is a rational subset of . When is empty, this is true in many stupid ways, so we can assume is nonempty. Since is a qc open, we can find finitely many nonempty rational subsets such that . But the ‘s are totally ordered, since any finite set of open bounded valuation subrings of is totally ordered by inclusion (in the opposite direction), so where is the largest .

Now we turn to the general case. For each point , we’ve proved that is a rational subset (possibly empty) of the fiber . Since and each is the topological space of a rational subset of , it’s now easy to check* that for every and for some small as above, there is a rational subset such that . Choose such a for each point in . Since , we can choose finitely many ‘s such that the ‘s give a covering of . Since each of these subsets is open-closed in , we can refine this covering to a covering of by finitely many *pairwise-disjoint* open-closed subsets where for all and for some (choice of) . Then again comes from a rational subset of , so the intersection comes from the rational subset of , and since by design, we (finally) have that is affinoid perfectoid. Whew!

*Here we’re using the “standard” facts that if is a cofiltered inverse system of affinoid perfectoid spaces with limit , then , and any rational subset is the preimage of some rational subset , and moreover if we have two such pairs and with the ‘s both pulling back to then they pull back to the same rational subset of for some large .

Let be a subset of a spectral space ; according to the incredible Lemma recorded in Tag 0A31 in the Stacks Project, the following are equivalent:

- is generalizing and pro-constructible;
- is generalizing and quasicompact;
- is an intersection of quasicompact open subsets of .

Moreover, if has one of these equivalent properties, is spectral. (Johan tells me this lemma is “basically due to Gabber”.) Combining this result with the Theorem above, and using the fact that the category of affinoid perfectoid spaces has all small limits, we get the following disgustingly general statement.

**Theorem. ***Let be a w-local affinoid perfectoid space. Then any generalizing quasicompact subset is affinoid perfectoid.*

By an easy gluing argument, this implies even more generally (!) that if is a subset such that every point has a qc open neighborhood in such that is quasicompact and generalizing, then is perfectoid (not necessarily affinoid perfectoid). This condition* holds, for example, if is locally closed and generalizing; in that situation, I’d managed to prove that is perfectoid back in May (by a somewhat clumsy argument, cf. Section 2.7 of this thing if you’re curious) after Peter told me it was so. But the argument here gives a lot more.

*Johan’s opinion of this condition: “I have no words for how nasty this is.”