# Projective limits of topological space II: Steenrod’s theorem

Last time, I explained some of the strange things that happen with projective limits of topological spaces: they can be empty, even if all the spaces in the given projective system are non-empty and all bonding maps are surjective, and they can fail to be compact, even if all the spaces in the projective system are compact.

Steenrod [1, Theorem 2.1] claimed that if all the spaces in the projective system are compact and T1, then the projective limit is non-empty and compact T1.  (The T1 condition is not mentioned explicitly, because what Steenrod includes the T1 separation axiom in his definition of topological spaces.  Also, note that he calls bicompact what we call compact.)

As I said last time, this is wrong.  But Fujiwara and Kato prove [3, Theorem 2.2.20] the following—and call it Steenrod’s theorem:

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.

Note that we do not need the bonding maps pij to be surjective for that to hold.

My goal is to explain the proof (credited to O. Gabber by Fujiwara and Kato).  We will do it in (at least) two passes.  This month, we will prove the following apparently very special case.  We will explain the rest next month

Proposition.  Let  (pij : Xj → Xi)i≤j ∈ I be a projective system of compact sober spaces.  If every Xi is non-empty, then its projective limit X is non-empty as well.

Recall that X consists of the tuples (xi)i∈ I where each xi is in Xi, and where xi = pij (xj) for all ij in I.  X is topologized as a subspace of the product space Πi∈ I Xi.  The projection map pi : X → Xmaps every such tuple to xi.

The proof is pretty subtle, and will occupy the rest of this post.  Because of sobriety, instead of picking one element xi from each Xi, it is enough to find an irreducible closed subset Zi in each Xi—subject to some conditions—instead.  That Zi will be the closure of a unique point by sobriety, and that will be xi.

## The dcpo Φ

Let Φ be the collection of all families (Ci)i∈ I where each Ci is closed and non-empty in Xi.  and such that the image of Cj by pij is included in Ci for all ij in I.  We order Φ by pointwise reverse inclusion, namely (Ci)i∈ I≤(C’i)i∈ I if and only if Ci contains C’i for every i in I.  The idea is to find a maximal element of Φ, and to show that it fits.

In order to show that Φ has a maximal element, we plan to use Zorn’s Lemma.  We first check that Φ is non-empty: (Xi)i∈ I is an element of Φ.

It is easy to see that given any directed family ((Cji)i∈ I)j∈ J of elements of Φ (namely, for each i in J, the collection of sets (Cji)j∈ J is filtered, since we are working with reverse inclusion), its pointwise intersection, (Ci)i∈ I, where each Ci is defined as ∩j∈ J Cji, is again an element of Φ. The key point is that Ci is non-empty, because every filtered intersection of non-empty closed subsets of a compact space is non-empty (Exercise 4.4.11 in the book.)

By Zorn’s Lemma, Φ has a maximal element.  Let us call it (Zi)i∈ I in the sequel.

## The key lemma

Given any property P of indices i in I, we will say “for cofinally many jP(j)” or “P(j) holds for cofinally many j” to say that for every k in I, there is a jk in I such that P(j) holds.  (In other words, the set of indices j such that P(j) holds is cofinal in I.)

If the index set were N instead of I, that would be equivalent to saying that P(j) holds for infinitely many values of j, if that can be of some help.

We will also say “P(j) holds for cofinally many j≥i” (or the obvious variants) to say that for every k in I with ki, there is a jk in I such that P(j) holds.  If the index set were N, that would mean that P(j) holds for infinitely many j≥i—and the `≥i‘ part would be useless.

Lemma. For every i in I, and every closed subset C of Zi, if pij[Zj] intersects C for cofinally many ji, then C=Zi.

Remark.  If pij[Zj] intersects C for cofinally many ji, then pij[Zj] intersects C for every ji.  (Indeed, take any ji.  Then there is a further kj such that pik[Zk] intersects C, by cofinality.  Hence there is a point x in Zk such that pik(xk) is in C, and then pjk(xk) is a point of Zj whose image by pij is in C.)  But we will really need the “cofinally many” part later.

Proof.  We build a new family of closed subset Z’k of each Xk, as follows.  By the Remark, for every ji,kpij[Zj] intersects C.  The set pij[Zj] is included in Xj, but the fact that it intersects C can be expressed by saying that, equivalently, pij-1(C) ∩ Zj is non-empty.  Note that the set pij-1(C) ∩ Zj is included in Xi instead.

We can project back pij-1(C) ∩ Zj onto a closed subset of Xk, by taking the closure cl (pkj[pij-1(C) ∩ Zj]) of its image by pkj.  This is a familiar trick with directed families (here, I): to go from index i to index k, where k is possibly incomparable to i, we find an index j above both i and k, then we go from i to j, and then back from j to k.

One checks easily that the family F of closed sets cl (pkj[pij-1(C) ∩ Zj]) where j ranges over the indices that are above both i and k is filtered: as j grows, the sets cl (pkj[pij-1(C) ∩ Zj]) become smaller and smaller.  Since F is a filtered family of non-empty closed sets in a compact space, its intersection is closed and non-empty: this is what we choose to be Z’k.

It is also elementary to check that if kk’, then pkk’ maps every element of Z’k’ to Z’k.  (Exercise.  Think of using the fact that pkk’ is continuous, so the image of a closure is included in the closure of the image.)  Therefore (Z’k)k∈ I is an element of Φ.

By construction of Z’k, Z’is included in cl (pkj[pij-1(C) ∩ Zj]) for at least one j, hence in cl (pkj[Zj]) ⊆ cl (Zk) = Zk. By maximality of (Zk)k∈ I, Z’k=Zk for every k in I.

In particular, Z’i=Zi.  But   Z’⊆ cl (pij[pij-1(C) ∩ Zj]) for at least one j, and that is included in cl (pij[pij-1(C)]) ⊆ C.  Therefore Z’⊆ C.  Equality follows since C is a subset of Z’i.   ☐

## Zi is irreducible closed

Using the Lemma, we can now proceed and show that Zi is irreducible closed for every i.  Imagine that Zi is included in the union of two closed subsets C’ and C” of Xi.  We wish to show that Zi is included in C’ or in C”, namely that Zi ∩ C’ or Zi ∩ C” equals Zi.  To do so, we use the Lemma with C equal to Zi ∩ C’ or to Zi ∩ C”.

Explicitly, let J’ be the set of indices ji such that pij[Zj] intersects Zi ∩ C’, and let J” be the set of indices ji such that pij[Zj] intersects Zi ∩ C”: we must show that J’ or J” is cofinal in I.

To do so, we claim that J’ ∪ J” is cofinal in I.  In fact every ji is in J’ ∪ J”. Indeed, for every ji, Zj is non-empty, so there is a point x in Zj, and pij(x) is in Zi.  If pij(x) is in C’ then j is in J’, otherwise pij(x) is in C” and j is in J”.

Now, since J’ ∪ J” is cofinal in IJ’ or J” must be cofinal in I: otherwise after a certain rank no element of I would be in J’, and no element of I would be in J”.  This concludes the proof: summing up, if J’ is cofinal in I, then we apply the Lemma with C=Zi ∩ C’, so that Zi ⊆ C’, and if J” is cofinal in I, then we apply the Lemma with C=Zi ∩ C”, so that Zi ⊆ C”.  ☐

## Finishing the proof of the Proposition

Since every Xi is sober, the irreducible closed subset Zi is the closure ↓xi of a unique point xi.  By the definition of Φ, pij maps every point of Zj, in particular xj, to a point of Zi=↓xi, so pij(xj)≤xi.

In order to show the converse inequality, we recall from the proof of the Lemma that Z’i=Zi.  In particular xi is in Z’i, hence in cl (pij[pij-1(C) ∩ Zj]) ⊆ cl (pij[Zj]) = cl (pij[↓xj]).  Since pij is continuous, hence monotonic, pij[↓xj] ⊆ ↓pij(xj).  Therefore xi is in cl (↓pij(xj)) = ↓pij(xj).  This shows that xipij(xj), hence xi=pij(xj).

As this holds for all ij, the tuple (xi)i∈ I is in the projective limit X.  So X is non-empty.  This concludes the proof of the Proposition.  ☐

That is enough for this month.  We have only proved that the projective limit of non-empty compact sober spaces is non-empty.  We have done the most complicated part!  Next month, we will see that this entails that projective limits of compact sober spaces are compact.  This requires much less effort.

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 