Last time, we had started to prove the following theorem [3, Theorem 2.2.20]:

**Theorem.** The projective limit of a projective system (*p _{ij }*:

*X*→

_{j}*X*)

_{i}*of compact sober spaces is compact and sober. It is non-empty if every*

_{i≤j ∈ I}*X*is non-empty.

_{i}My goal this time is to finish its proof. That is unfortunately pretty technical. If there were anything to remember from the proof, that would be that the following Proposition, which we had proved last time, is the cornerstone of the proof.

**Proposition. ** Let (*p’ _{jk }*:

*C*→

_{k}*C*)

_{j}*be a projective system of compact sober spaces. If every*

_{j≤k∈J}*C*is non-empty, then its projective limit is non-empty as well.

_{j}The Proposition of course only deals with the final part of the Theorem. Showing that the projective limit is sober is easy, and we deal with that point next. Showing that it is compact is much more technical, and we will repeatedly use the Proposition to prove that. The general idea will be to find non-empty closed (hence compact and sober) subsets *C _{i}* of each

*X*, in such a way as to define a new projective limit (

_{i}*p*:

_{ij }*C*→

_{j}*C*)

_{i}*of compact sober subspaces, and to use the fact that this projective limit is non-empty in order to make progress.*

_{i≤j ∈ I}## Sobriety

Not all subspaces of a sober space are sober: take any non-sober space *X*, and realize that it occurs (up to homeomorphism) as a subspace of its sobrification. However, any subspace [*g*_{1}=*g*_{2}] of *X* that is built as an equalizer of two continuous maps *g*_{1}, *g*_{2} : *X* → *Y* with *X* sober (Lemma 8.4.12 of the book; recall that, in an explicit form, [*g*_{1}=*g*_{2}] is the subspace of all *x* in *X* such that *g*_{1}(*x*)=*g*_{2}(*x*).)

This shows immediately that:

**Lemma 1.** A projective limit *X* of a projective system (*p _{ij }*:

*X*→

_{j}*X*)

_{i}*of sober spaces is sober.*

_{i≤j ∈ I}This is because Π_{i∈ I}*X _{i}* is sober (Theorem 8.4.8 in the book) and

*X*occurs as the equalizer [

*g*

_{1}=

*g*

_{2}] where

*g*

_{1}: Π

_{i∈ I}*X*→ Π

_{i}

_{i≤j∈ I}*X*maps (

_{i}*x*)

_{i}*to (*

_{i∈ I}*p*(

_{ij}*x*))

_{j}*and*

_{i≤j∈ I}*g*

_{2}maps (

*x*)

_{i}*to (*

_{i∈ I}*x*)

_{i}*. (Please pay attention to the fact that the target space of*

_{i≤j∈ I}*g*

_{1 }and

*g*

_{2}is a product indexed, not by

*i*, but by all pairs of indices

*i*,

*j*such that

*i*≤

*j*.)

This also shows, for example, that every open subspace *U* of a sober space *X* is sober, being the equalizer of the characteristic map of *U* (with values in Sierpiński space **S**) with the constant 1 map.

Every closed subspace *C* is also sober, as the equalizer of the characteristic map of its complement with the constant 0 map.

## A compactness lemma

Let *X* be the projective limit of a projective system of compact sober spaces (*p _{ij }*:

*X*→

_{j}*X*)

_{i}*. Let also*

_{i≤j ∈ I}*p*:

_{i}*X*→

*X*be the projection maps. We have the following, which one can think as a sort of compactness lemma for the set of indices

_{i }*I*.

We will not have any use of it in this post, but this is a generally useful lemma, and it can be taken as a gentle example of how we can make good use of the Proposition we mentioned at the beginning of the post (the projective limit of a projective system of non-empty compact sober spaces is non-empty).

**Lemma 2.** For every *i* in *I*, and every open neighborhood *V* of the image of *X* by *p _{i}*, there is an index

*j*≥

*i*in

*I*such that that

*V*already contains the image of

*X*by

_{j}*p*.

_{ij}*Proof.* We reason by contradiction, and assume that *V* does not contain the image of *X _{j}* by

*p*for any

_{ij}*j*≥

*i*. Let

*J*be the subset of those indices of

*I*above

*i*. This is a directed set. For each

*j*in

*J*,

*C*=

_{j}*X*–

_{j}*p*

_{ij}^{-1}(

*V*) is then non-empty. It is closed in a compact space, hence is itself compact. As a compact subset, it is also a compact sub

*space*(Exercise 4.9.11 of the book). As a closed subspace of a sober space, it is sober.

For all *j*≤*k* in *J*, the restriction *p’ _{jk}* of

*p*to

_{jk}*C*maps every element to an element of

_{k}*C*. Indeed, for every element

_{j}*x*of

*C*, if

_{k}*p*(

_{jk}*x*) were in

*p*

_{ij}^{-1}(

*V*), then

*p*(

_{ij}*p*(

_{jk}*x*))=

*p*(

_{ik}*x*) would be in

*V*, and

*x*would be in

*p*

_{ik}^{-1}(

*V*): contradiction.

Therefore (*p’ _{jk }*:

*C*→

_{k}*C*)

_{j}*is a projective system of non-empty compact sober spaces. The Proposition we mentioned at the beginning of the post shows that it has a non-empty projective limit. We pick a tuple (*

_{j≤k∈J}*x*)

_{j}*in that projective limit. We can complete it to a tuple indexed by*

_{j∈ J}*I*instead, defining

*x*for

_{j}*j*not in

*J*as

*p*(

_{jk}*x*) for some arbitrary

_{k}*k*in

*I*above both

*i*and

*j*. This defines an element

*x*of

*X*such that

*p*(

_{j}*x*)=

*x*for every

_{j}*j*in

*J*, in particular

*p*(

_{i}*x*)=

*x*. It follows that

_{i}*x*is in the image of

_{i}*p*, hence in

_{i}*V*, and that is impossible since

*x*is in

_{i}*C*. ☐

_{i}## Compactness

In order to show the Theorem, we need to show that the projective limit *X* of a projective system of compact sober spaces (*p _{ij }*:

*X*→

_{j}*X*)

_{i}*is compact. Let also*

_{i≤j ∈ I}*p*:

_{i}*X*→

*X*be the projection maps, as usual. That is the hard part, and I hope you will not be lost in indices. There will be quite a lot of them. In the sequel,

_{i }*i*and

*j*will always denote indices from the index set

*I*, while

*k*will index the open subsets from a directed open cover of

*X*.

In order to show that *X* is compact, we consider a directed family (*U _{k}*)

*of open subsets of*

_{k ∈ K}*X*, whose union is equal to

*X*, and we wish to show that some

*U*is already equal to the whole of

_{k}*X*. (This is Proposition 4.4.7 of the book. Being able to consider the family as directed will simplify the argument.)

For each *k* in *K*, and every *i* in *I*, there is a largest open subset *U _{ki}* of

*X*such that

_{i }*p*

_{i}^{-1}(

*U*) ⊆

_{ki}*U*: we just take the union of all open subsets

_{k}*U*of

*X*such that

_{i }*p*

_{i}^{-1}(

*U*) ⊆

*U*.

_{k}We notice that as *i* grows, *p _{i}*

^{-1}(

*U*) becomes larger. (Formal argument: If

_{ki}*i*≤

*j*, then

*p*=

_{i}*p*o

_{ij}*p*,

_{j}*p*

_{i}^{-1}(

*U*) is equal to

_{ki}*p*

_{j}^{-1}(

*p*

_{ij}^{-1}(

*U*)). This shows that

_{ki}*p*

_{ij}^{-1}(

*U*) is an open subset

_{ki}*U*of

*X*such that

_{j }*p*

_{j}^{-1}(

*U*) ⊆

*U*. It must then be included in the largest one,

_{k}*U*. So

_{kj}*p*

_{i}^{-1}(

*U*) ⊆

_{ki}*p*

_{j}^{-1}(

*U*).) Hence the family of sets

_{kj}*p*

_{i}^{-1}(

*U*), when

_{ki}*i*varies, is directed. Its union is included in

*U*, and we claim that:

_{k}**Lemma 3.** The directed union of the sets *p _{i}*

^{-1}(

*U*),

_{ki}*i*∈

*I*, is equal to

*U*.

_{k}*Proof.* Consider any point (*x _{i}*)

*in*

_{i∈ I}*U*. Since

_{k}*U*is open, and remembering that the topology on

_{k}*X*is the subspace topology from a product topology, there is a finite subset

*J*of

*I*and open subsets

*V*of

_{i }*X*,

_{i}*i*∈

*J*, such that

*x*∈

_{i}*V*for every

_{i}*i*∈

*J*, and such that every tuple whose

*i*th coordinate is in

*V*for every

_{i }*i*∈

*J*, is in

*U*. Since

_{k}*I*is directed, there is a an index

*j*in

*I*above every element of

*J*. For every

*i*∈

*J*,

*p*(

_{ij}*x*)=

_{j}*x*is in

_{i}*V*, so

_{i}*x*is in ∩

_{j}

_{i∈J}*p*

_{ij}^{-1}(

*V*). The open set

_{i}*p*

_{j}^{-1}(∩

_{i∈J}*p*

_{ij}^{-1}(

*V*)) consists of tuples (

_{i}*y*)

_{i}*such that*

_{i∈ I}*y*is in ∩

_{j}

_{i∈J}*p*

_{ij}^{-1}(

*V*), namely such that

_{i}*y*is in

_{i }*V*for every

_{i}*i*∈

*J*, and is therefore included in

*U*. By the maximality property of

_{k}*U*, ∩

_{kj}

_{i∈J}*p*

_{ij}^{-1}(

*V*) is included in

_{i}*U*. Therefore

_{kj}*x*is in

_{j}*U*. It follows that (

_{kj}*x*)

_{i}*is in*

_{i∈ I}*p*

_{j}^{-1}(

*U*), showing the claim. ☐

_{kj}Since *X* is the directed union of the sets *U _{k}*,

*k*∈

*K*,

*X*is also the directed union over

*k*of the directed union over

*i*of the sets

*p*

_{i}^{-1}(

*U*). Switching the two unions,

_{ki}*X*is also the directed union over

*i*of the open sets

*V*, where

_{i}*V*is defined as the (directed) union of (

_{i}*U*)

_{ki}*.*

_{k∈K}Note that *V _{i}* is an open subset of the compact set

*X*, and is defined as a directed union of open subsets. This sounds good, but the latter is not (yet known to be) an open

_{i}*cover*, so we cannot conclude (yet).

Instead, we define *C _{i}* as the complement of

*V*in

_{i}*X*. This is a closed subset of

_{i}*X*, and as in the proof of Lemma 2,

_{i}*C*is a compact sober subspace of

_{i}*X*.

_{i}For all *i*≤*j* in *I*, *p _{ij}*

^{-1}(

*U*) is included in

_{ki}*U*. (Formal argument: It suffices to check that

_{kj}*p*

_{j}^{-1}(

*p*

_{ij}^{-1}(

*U*)), which is equal to

_{ki}*p*

_{i}^{-1}(

*U*), is included in

_{ki}*U*, and to invoke the maximality of

_{k}*U*.) Taking unions over

_{ki}*k*in

*K*,

*p*

_{ij}^{-1}(

*V*) is included in

_{i}*V*. Hence the image of

_{j}*C*by

_{j}*p*is included in

_{ij}*C*: if there were a point of

_{i}*C*whose image by

_{j }*p*were not in

_{ij}*C*, hence in

_{i}*V*, it would be in

_{i}*p*

_{ij}^{-1}(

*V*) and therefore in

_{i}*V*, contradiction.

_{j}All this means that (*p’ _{ij }*:

*C*→

_{j}*C*)

_{i}*is a(nother) projective system of compact sober spaces, where*

_{i≤j ∈ I}*p’*is the restriction of

_{ij }*p*to

_{ij}*C*.

_{j}But that projective system has an empty projective limit! Let us check that.

Imagine there were an element its its projective limit. That element would be in *p _{i}*

^{-1}(

*C*) for every

_{i}*i*in

*I*, hence not in

*p*

_{i}^{-1}(

*V*) for any

_{i}*i*in

*I*, but that is impossible since

*X*is the union over all

*i*in

*I*over

*p*

_{i}^{-1}(

*V*).

_{i}The Proposition mentioned at the beginning of this post (projective limits of non-empty compact sober spaces are non-empty) tells us that this cannot happen if every *C _{i}* is non-empty. Hence

*C*is empty for some

_{i}*i*in

*I*, and this means that

*V*=

_{i}*X*.

_{i}Since *V _{i}* is the directed union of (

*U*)

_{ki}*, and*

_{k∈K}*X*is compact,

_{i}*X*=

_{i}*U*for some

_{ki}*k*in

*K*. Finally, every element of the projective limit

*X*is in

*p*

_{i}^{-1}(

*X*) (meaning that it

_{i}*i*th coordinate must be in

*X*), namely in

_{i}*p*

_{i}^{-1}(

*U*), which is included in

_{ki}*U*. Hence

_{k}*X*=

*U*. ☐

_{k}That finishes the proof of the Theorem.

Instead of ending this post here, let me conclude with an easy observation on sobrifications of projective limits.

## Sobrifications of projective limits

Remember that the sobrification functor **S** preserves products (Theorem 8.4.8 in the book), so we may think that it would perhaps preserve all limits in **Top**. That is wrong, since it does not preserve equalizers: see the warning comments after Lemma 8.4.12 in the book.

We will now argue that **S** does not preserve projective limits in general either.

Recall the following counterexample, due to A. H. Stone, and mentioned as Example 3 in my first post on projective limits. For every *n* ∈ **N**, we define *X _{n}* as

**N**with the following topology, akin to the cofinite topology: its closed subsets

*C*are those subsets such that

*C*∩ ↑

*n*is finite or equal to the whole of ↑

*n*. The bonding maps

*p*:

_{mn }*X*→

_{n}*X*(

_{m}*m*≤

*n*) are identity maps.

We had seen that every *X _{n}* is compact, in fact Noetherian, and T

_{1}. The projective limit

*X*is

**N**with the

*discrete*topology, and we had observed that this is not compact.

Let us look at the irreducible closed subsets *C* of *X _{n}*. The only ones that are finite are the one-element sets. It remains to look at those such that

*C*∩ ↑

*n*=↑

*n*. Those are obtained as ↑

*n*union finitely many elements below

*n–*1. Irreducibility implies there cannot be any element below

*n–*1, hence

*C*=↑

*n*. Conversely, it is easy to see that ↑

*n*is irreducible closed in

*X*.

_{n}It follows that the sobrification **S**(*X _{n}*) of

*X*is obtained by adding a fresh element ω

_{n}*(corresponding to ↑*

_{n }*n*), above

*n*,

*n*+1, …, but incomparable with 0, 1, …,

*n*–1.

Since sobrification is a functor, the projective system (*p _{mn }*:

*X*→

_{n}*X*)

_{m}

_{m≤n ∈ }_{N}gives rise to a new projective system (

**S**(

*p*):

_{mn}**S**(

*X*) →

_{n}**S**(

*X*))

_{m}

_{m≤n ∈ }_{N}. The sobrification of a compact space is compact, since it has an isomorphic lattice of open subsets. Hence we have obtained a projective system of compact sober spaces, and Steenrod’s theorem tells us that its projective limit

*X’*is compact and sober.

It follows that *X’* cannot be the sobrification of *X*, which is not compact. So the sobrification functor does not preserve projective limits.

Explicitly, *X* is **N** with the discrete topology, hence is already sober. *X’*, instead, is **N** plus an extra element ω, incomparable with all natural numbers, and is indeed different from *X.*

- Steenrod, Norman E. 1936. Universal Homology Groups. American Journal of Mathematics, 58(4), 661–701.
- Stone, Arthur Harold. 1979. Inverse Limits of Compact Spaces. General Topology and its Applications, 10, 203–211.
- Fujiwara, Kazuhiro, and Kato, Fumiharu. 2017 (Feb.). Foundations of Rigid Geometry I. arXiv 1308.4734, v5.

— Jean Goubault-Larrecq (October 23rd, 2018)