# Projective limits of topological spaces III: finishing the proof of Steenrod’s theorem

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

Theorem. The projective limit of a projective system (pij : Xj → Xi)i≤j ∈ I of compact sober spaces is compact and sober. It is non-empty if every Xi is non-empty.

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 : Ck → Cj)j≤k∈J be a projective system of compact sober spaces.  If every Cj is non-empty, then its projective limit is non-empty as well.

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 Ci of each Xi, in such a way as to define a new projective limit (pij : Cj → Ci)i≤j ∈ I of compact sober subspaces, and to use the fact that this projective limit is non-empty in order to make progress.

## 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 [g1=g2] of X that is built as an equalizer of two continuous maps g1, g2 : X → Y with X sober (Lemma 8.4.12 of the book; recall that, in an explicit form, [g1=g2] is the subspace of all x in X such that g1(x)=g2(x).)

This shows immediately that:

Lemma 1.  A projective limit X of a projective system (pij : Xj → Xi)i≤j ∈ I of sober spaces is sober.

This is because Πi∈ I Xi is sober (Theorem 8.4.8 in the book) and X occurs as the equalizer [g1=g2] where g1 : Πi∈ I Xi → Πi≤j∈ I Xi maps (xi)i∈ I to (pij(xj))i≤j∈ I and g2 maps (xi)i∈ I to (xi)i≤j∈ I.  (Please pay attention to the fact that the target space of gand g2 is a product indexed, not by i, but by all pairs of indices ij such that ij.)

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 (pij : Xj → Xi)i≤j ∈ I.   Let also pi : X → Xbe the projection maps.  We have the following, which one can think as a sort of compactness lemma for the set of indices 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 pi, there is an index ji in I such that that V already contains the image of Xj by pij.

Proof.  We reason by contradiction, and assume that V does not contain the image of Xj by pij for any ji.  Let J be the subset of those indices of I above i.  This is a directed set.  For each j in JCj=Xjpij-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 subspace (Exercise 4.9.11 of the book).  As a closed subspace of a sober space, it is sober.

For all jk in J, the restriction p’jk of pjk to Ck maps every element to an element of Cj.  Indeed, for every element x of Ck, if pjk(x) were in pij-1(V), then pij(pjk(x))=pik(x) would be in V, and x would be in pik-1(V): contradiction.

Therefore  (p’jk : Ck → Cj)j≤k∈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 (xj)j∈ J in that projective limit.  We can complete it to a tuple indexed by I instead, defining xj for j not in J as pjk(xk) for some arbitrary k in I above both i and j.  This defines an element x of X such that pj(x)=xj for every in J, in particular pi(x)=xi.  It follows that xi is in the image of pi, hence in V, and that is impossible since xi is in Ci. ☐

## Compactness

In order to show the Theorem, we need to show that the projective limit X of a projective system of compact sober spaces (pij : Xj → Xi)i≤j ∈ I is compact.   Let also pi : X → Xbe 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 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 (Uk)k ∈ K of open subsets of X, whose union is equal to X, and we wish to show that some Uk is already equal to the whole of 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 Uki of Xsuch that pi-1(Uki) ⊆ Uk: we just take the union of all open subsets U of Xsuch that pi-1(U) ⊆ Uk.

We notice that as i grows, pi-1(Uki) becomes larger.  (Formal argument: If ij, then pi= pij o pj, pi-1(Uki) is equal to pj-1(pij-1(Uki)).  This shows that pij-1(Uki) is an open subset U of Xsuch that pj-1(U) ⊆ Uk.  It must then be included in the largest one, Ukj.  So pi-1(Uki) ⊆ pj-1(Ukj).)  Hence the family of sets pi-1(Uki), when i varies, is directed.  Its union is included in Uk, and we claim that:

Lemma 3.  The directed union of the sets pi-1(Uki), iI, is equal to Uk.

Proof.  Consider any point (xi)i∈ I in Uk.  Since Uk is open, and remembering that the topology on X is the subspace topology from a product topology, there is a finite subset J of I and open subsets Vof Xii ∈ J, such that xi ∈ Vi for every i ∈ J, and such that every tuple whose ith coordinate is in Vfor every i ∈ J, is in Uk.  Since I is directed, there is a an index j in I above every element of J.  For every i ∈ J, pij(xj)=xi is in Vi, so xj is in ∩i∈J pij-1(Vi). The open set pj-1(∩i∈J pij-1(Vi)) consists of tuples (yi)i∈ I such that yj is in ∩i∈J pij-1(Vi), namely such that yis in Vi for every i ∈ J, and is therefore included in Uk.  By the maximality property of Ukj, ∩i∈J pij-1(Vi) is included in Ukj.  Therefore xj is in Ukj.  It follows that (xi)i∈ I is in pj-1(Ukj), showing the claim.  ☐

Since X is the directed union of the sets UkkKX is also the directed union over k of the directed union over i of the sets pi-1(Uki). Switching the two unions, X is also the directed union over i of the open sets Vi, where Vi is defined as the (directed) union of (Uki)k∈K.

Note that Vi is an open subset of the compact set Xi, and is defined as a directed union of open subsets.  This sounds good, but the latter is not (yet known to be) an open cover, so we cannot conclude (yet).

Instead, we define Ci as the complement of Vi in Xi.  This is a closed subset of Xi, and as in the proof of Lemma 2, Ci is a compact sober subspace of Xi.

For all ij in Ipij-1(Uki) is included in Ukj.  (Formal argument: It suffices to check that pj-1(pij-1(Uki)), which is equal to pi-1(Uki), is included in Uk, and to invoke the maximality of Uki.)  Taking unions over k in Kpij-1(Vi) is included in Vj.  Hence the image of Cj by pij is included in Ci: if there were a point of Cwhose image by pij were not in Ci, hence in Vi, it would be in pij-1(Vi) and therefore in Vj, contradiction.

All this means that (p’ij : Cj → Ci)i≤j ∈ I is a(nother) projective system of compact sober spaces, where p’ij is the restriction of pij to Cj.

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 pi-1(Ci) for every i in I, hence not in pi-1(Vi) for any i in I, but that is impossible since X is the union over all i in I over  pi-1(Vi).

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 Ci is non-empty.  Hence Ci is empty for some i in I, and this means that Vi=Xi.

Since Vi is the directed union of (Uki)k∈K, and Xi is compact, Xi=Uki for some k in K.  Finally, every element of the projective limit X is in pi-1(Xi) (meaning that it ith coordinate must be in Xi), namely in pi-1(Uki), which is included in Uk.  Hence X=Uk.  ☐

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 nN, we define Xn 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 pmn : XnXm (mn) are identity maps.

We had seen that every Xn is compact, in fact Noetherian, and T1.  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 Xn.  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 Xn.

It follows that the sobrification S(Xn) of Xn is obtained by adding a fresh element ω(corresponding to ↑n), above nn+1, …, but incomparable with 0, 1, …, n–1.

Since sobrification is a functor, the projective system (pmn : Xn → Xm)m≤n ∈ N gives rise to a new projective system (S(pmn): S(Xn) → S(Xm))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.

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

Jean Goubault-Larrecq 