I have devoted the last few posts in dealing with arcs and arc chains. The whole purpose of that was to give an introduction to some results that are needed to understand a few recent results of Jimmie Lawson [1]. I will only mention about Theorem 4.7 there, which is a pretty funny compactness theorem. There are additional results in Section 5 of this paper, which I will not talk about.
Let me recall that an arc chain is a compact, connected topological space of cardinality at least 2, with a semiclosed total ordering ≤. A semiclosed (partial) ordering is a partial ordering such that the downward closure and the upward closure of any point is closed. As last time, we will be looking at arc chains embedded in a pospace, and more generally in a topological space with a semiclosed partial ordering.
A chain in a space X is a totally ordered subset; we will also see such chains as subspaces of X. Hence chains in X are implicitly given the subspace topology from X.
The compact interval theorem
One of the important theorems of [1] is the following. This theorem would be easy if X were compact, but we only assume it to be locally compact. The compactness of the interval [a, b]C will arise from local compactness plus the inclusion in a closed connected chain C.
Theorem A [1, Theorem 4.7]. Let X be a locally compact space with a semiclosed partial ordering ≤, and let C be a closed connected chain in X. For every pair of points a, b ∈ C with a<b, the interval [a, b]C ≝ {x ∈ C | a≤x≤b} is compact, and therefore is an arc chain.
Proof. Just like C, [a, b]C is a closed connected chain. This is proved as follows.
- [a, b]C is equal to C ⋂ ↑a ⋂ ↓b, which is closed since ≤ is semiclosed. We will write ↓C for downward closure in C and ↑C for upward closure in C.
- [a, b]C is a chain, because ≤ restricts to a total ordering on any subset.
- If [a, b]C were not connected, then we could write it as the disjoint union of two non-empty closed subsets C1 and C2. Those are closed subsets of [a, b]C, hence also of C since [a, b]C is closed. If a ∈ C1 and b ∈ C2, then C1 ∪ ↓C a and C2 ∪ ↑C b would form a partition of C in two disjoint non-empty closed subsets, which is impossible since C is connected; similarly if a ∈ C2 and b ∈ C1. If both a and b are in C1, then C1 ∪ ↓C a ∪ ↑C b and C2 would form a partition of C in two disjoint non-empty closed subsets, which is impossible since C is connected; and similarly if a and b are in C2.
Hence, replacing C by [a, b]C if necessary, we will simply assume that C = [a, b]C, namely that C is a connected chain with a least element a and a largest element b. We will concentrate on showing that such a connected chain C is compact. It will almost immediately follow that C is an arc chain: we recall that C = [a, b]C is connected, as we have just seen, it has cardinality at least 2 since a<b, and we will see in item 2 below that its ordering is semiclosed.
Here are a few additional observations we will need in the following.
- As a closed subspace of X, C is locally compact. Indeed, let x ∈ C, let U ∩ C be an open neighborhood of x in C (where U is open in X). Since X is locally compact, there is a compact subset Q of X and an open subset V of X such that x ∈ V ⊆ Q ⊆ U. Then V ∩ C is an open neighborhood of x in C, it is included in Q ∩ C, which is compact in X since intersections of closed sets and compact sets are compact (by Corollary 4.4.10 in the book). Since Q ∩ C is compact in X and included in C, Q ∩ C is compact in C (exercice! just apply the definition). And Q ∩ C is included in U ∩ C.
- The restriction of ≤ to the subspace C is semiclosed. Indeed, for every x ∈ C, ↓C x = ↓x ∩ C and ↑C x = ↑x ∩ C are closed in C.
- C is a pospace, hence is Hausdorff. This is because any chain with a semiclosed ordering is a pospace, hence is Hausdorff, as we have seen in Proposition C of the January 2025 post.
- In particular, local compactness takes a new guise: for every x ∈ C, for every open neighborhood U of x in C, there is an open neighborhood V of x such that the closure clC(V) of V in C is compact, and is included in U. That would be wrong in a non-Hausdorff, locally compact space.
Here is how we justify this claim. Since C is locally compact, there is an open neighborhood V of x and a compact subset Q of C such that x ∈ V ⊆ Q ⊆ U. Then clC(V) contains V, and is the smallest closed subset of C containing V. But, since C is Hausdorff, the compact set Q is closed (Proposition 4.4.15 in the book); so clC(V) ⊆ Q. Additionally, since clC(V) is closed and included in the compact set Q, it is equal to the intersection clC(V) ⋂ Q, which is closed (by Corollary 4.4.10 in the book). - No element of C except a is a local minimum (in C, with respect to the restriction of ≤ to C). A local minimum is a point x of C such that there is an open neighborghood O of x in C such that O ∩ ↓C x = {x}; we have introduced the notion in the post on Koch’s arc theorem
In order to see this, we reason by contradiction. Let x be a local minimum of C, and let us assume that x is different from the least element a of C. There is an open neighborghood O of x in C such that O ∩ ↓C x = {x}. The two open sets C–↑C x and O ∪ (C–↓C x) are disjoint: any point in C–↑C x is strictly below x (since ≤ is total on C), hence cannot be in O, and cannot be in C–↓C x either. Their union is the whole of C: the complement of that union is the intersection of ↑C x, ↓C x, and the complement of O, but the only point in ↑C x ⋂ ↓C x is x, and it is in O. C–↑C x is non-empty, because otherwise x would be least in C, and O ∪ (C–↓C x) is non-empty since it contains x. To sum up, we have found a partition of C in two disjoint non-empty open sets, and that is impossible since C is connected.
The key observation that J. Lawson makes is the following.
Claim I: For every x ∈ C,
(i) if a<x, then there is a point y in C such that y<x and [y, x]C is compact;
(ii) if x<b, then there is a point z in C such that x<z and [x, z]C is compact.
Proof of Claim I. (i) Since C is Hausdorff, there are disjoint open neighborhoods U of x and W of a in C. By local compactness in its new guise (item 4 above), there is an open neighborhood V of x such that clC(V) is compact and included in U. V does not contain any local minimum of C, because the only possible local minimum would be a (by item 5), but all the elements of V are in U, and a is not in U. By item 3 above, C is a pospace. All of that allows us to use Koch’s arc theorem: there is an arc chain A from some point y of the boundary ∂V (in C) to x.
By definition of an arc chain, y<x, and A ⊆ [y, x]C. In fact, A = [y, x]C: otherwise, there would be a point y’ in [y, x]C that is not in A, and then ↓C y’ ∩ A and ↑C y’ ∩ A would form a partition of A in two disjoint non-empty closed subsets (they are non-empty since the first one contains y and the second one contains x), contradicting the fact that A is connected. Hence A = [y, x]C is an arc chain, and is in particular compact.
(ii) This is a consequence of (i), applied to C ordered with the reverse ordering ≥, exchanging a and b. (End of proof of Claim I.)
Let K be the collection of points x ∈ C such that [x, b]C is compact in the subspace topology from X. K contains b itself, since [b, b]C = {b}, and every finite set is compact. The strategy is to show that K contains smaller and smaller points, until we finally show that it contains a. This is a standard way of showing that closed bounded intervals in R are compact.
Claim II: for every x ∈ K such that a<x, there is a point y in K such that y<x.
Proof of Claim II. Let x ∈ K be such that a<x. By Claim I (i), there is a point y in C such that y<x and [y, x]C is compact. By definition of K, [x, b]C is compact. Since ≤ is total on C, x is comparable to any element of C, so [y, b]C = [y, x]C ∪ [x, b]C, which is therefore compact. Hence y is in K. (End of proof of Claim II.)
Claim III. K has an infimum.
Proof of Claim III. The intersection L of the sets ↓C x, where x ranges over K, is closed because the ordering on C is semiclosed (item 2), and contains a because a is the least element of C. L is the set of lower bounds of K, and we wish to show that it has a largest element.
If not, then the set UL of upper bounds of L would be disjoint from L. Since UL is the intersection of the sets ↑C y where y ranges over L, UL is closed (because the ordering on C is semiclosed, by item 2) and non-empty (because it contains b, since b is the largest element of C). Additionally, the union of L and of UL is the whole of C. In order to see this, we consider any element x of C that is not an upper bound of L and we will show that x is in L. Since x is not an upper bound of L, there is an element y of L such that y ≰ x, namely such that x<y (since ≤ is total on C), and then x is in L, since L is downwards closed. Hence L and UL form a partition of C in two disjoint non-empty closed subsets. This is impossible since C is connected. (End of proof of Claim III.)
Let x be the infimum of K, which exists by Claim III. We have x<b, otherwise b would be the sole element of K, which is impossible by Claim II. Hence Claim I (ii) applies, and there is a point z in C such that x<z and [x, z]C is compact.
Since x<z and x is the infimum of K, there must be a point y ∈ K such that y<z. Otherwise, since ≤ is total on C, all the points y ∈ K would be such that z≤y, and then the infimum x of K would be larger than or equal to z. Since x is the infimum of K and y ∈ K, additionally, x≤y.
Then [x, y]C is a closed subset of C, being equal to ↑C x ∩ ↓C y, and is included in the compact set [x, z]C. Therefore [x, y]C is itself compact. Since y ∈ K, [y, b]C is compact, so [x, b]C = [x, y]C ∪ [y, b]C is compact, and therefore x ∈ K.
We have obtained that x is not just the infimum of K, but its least element. By Claim II, if a<x, then there is a point y in K such that y<x, which would contradict the minimality of x in K. Hence x=a. We have obtained that a is in K, namely that [a, b]C is compact. ☐
Theorem 4.7 in [1] also states that [a, b]C has the order topology. By definition, the order topology in a poset is the topology generated by open intervals ]x, y[, the set of points strictly larger than x and strictly smaller than y; plus the half-open intervals [⊥, y[ with y≠⊥ if the poset has a least element ⊥, the half-open intervals ]x, ⊤] with x≠⊤, if the poset has a largest element ⊤. We do not need to prove it: this is a consequence of Proposition D of the January 2025 post, which says that every arc chain has the order topology, and even says that the open subsets are in fact disjoint unions of what I called o-intervals there.
Local compactness is important, through a counterexample
The fact that [a, b]C is compact in Theorem A is something I didn’t know initially whether it should be considered obvious or difficult.
My first thought was that it was obvious, but this is because my mind is trained to think of closed intervals as compact, just like on the real line. Hence it is “obvious for lack of imagination” (namely, lacking the imagination for a counterexample).
My second thought was that I had absolutely no idea how to prove it. My favorite method of showing that a set is compact is by using the definition, or its refined form, Alexander’s subbase lemma. As you can see, J. Lawson’s proof is a lot more sophisticated, and relies on Koch’s arc theorem. I don’t think we need the full force of Koch’s arc theorem to prove the theorem, but that is convenient, and something non-trivial is needed, at any rate.
In order to settle the debate, one might observe that Theorem A would be wrong without local compactness. Jimmie Lawson gives an example of a non-locally compact space in which the conclusion of Theorem A fails [1, Example 3.8]. This is a variant of Countexample 64 (page 96) in [2], the real line with Smirnov’s deleted sequence topology.
This space is obtained as follows. We fix a strictly decreasing sequence of points an in [0, 1] that converges to 0, and we will take an ≝ 1/2n, n ∈ N (another classic choice is an ≝ 1/n, n≥1). Let A ≝ {an | n ∈ N}. A basic Smirnov open subset is a subset of the form ]a, b[ or ]a, 1] or [0, c[–A, where 0<a<b<1, 0<c<1. A Smirnov open subset is a union of basic Smirnov open subsets.
Lemma B. An intersection of basic Smirnov open subsets is a finite union of basic Smirnov open subsets, hence the Smirnov open subsets form a topology, with a base given by the basic Smirnov open subsets.
Proof. This is a case analysis with 6 cases. The only interesting cases are when one of the basic Smirnov open subsets is of the form [0, c[–A and the other one is ]a, b[ or ]a, 1] with a>0. Since a>0, the interval ]a, b[ (resp., ]a, 1]) contains only finitely points from A, and then we can cut up the intersection ]a, c[–A in finitely many open intervals. ☐
Definition C. We write [0, 1]Sm for [0, 1] with its topology consisting of the Smirnov open subsets.
Fact D. The topology of [0, 1]Sm is finer than the usual metric topology on [0, 1].
In order to see this, we only need to observe that half-open intervals [0, b[ can be written as the union of [0, b[–A with ]0, 1], and that ]0, 1] is itself the union of the intervals ]1/2n, 1], n ∈ N.
Corollary E. [0, 1]Sm is Hausdorff.
It will be useful to refine the basis of [0, 1]Sm as follows.
Lemma F. For every ε such that 0<ε≤1, a base of open subsets of [0, 1]Sm is given by the sets ]a, b[ or ]a, 1] or [0, c[–A, where 0<a<b<1, and 0<c<ε. (The only difference with basic Smirnov open subsets is that we only allow c to be strictless less than ε.)
Proof. For ε≤c<1, we can write the missing basic Smirnov open subset [0, c[–A as (([0, 1/2n[–A) ∪ ]1/2n, 1/2n–1[ ∪ … ∪ ]1/2, 1[) ⋂ ]1/2n, c[. ☐
Proposition G. [0, 1]Sm is connected.
Proof. Let us assume that it is not, and that we can write it as a union of two disjoint non-empty open sets U and V. Exactly one of them contains the point 0, say U. Let v≠0 be some point of V.
Since 0 ∈ U, U contains a basic open subset [0, c[–A, for some c with 0<c<1. We pick a point u≠0 in [0, c[–A, and ε≝min(u, v)/2. Using the alternate base of Lemma F, U is a union of subsets ]a, b[ or ]a, 1] or [0, d[–A, where 0<a<b<1, and 0<d<ε. The intersection of [0, d[–A with ]ε, 1] is empty, so U ⋂ ]ε, 1] is a union of intervals ]a, b[ or ]a, 1], hence is open in the usual metric topology on ]ε, 1]. Since V does not contain 0, V does not contain any basic open subset of the form [0, d[–A, hence is also open in the metric topology; in particular, V ⋂ ]ε, 1] is open in the usual metric topology on ]ε, 1].
U ⋂ ]ε, 1] and V ⋂ ]ε, 1] are disjoint, their union is ]ε, 1], and they are both non-empty, since we have chosen ε < u, v. This entails that ]ε, 1] is not connected, which is absurd. ☐
An altnerative description of the Smirnov deleted sequence topology is the following.
Lemma H. The open subsets of [0, 1]Sm are the sets U–B where U is open in [0, 1] and B ⊆ A.
Proof. We first show that U–B is open in [0, 1]Sm. For every point x ∈ U–B, it suffices to find a basic Smirnov open subset of U–B containing x.
- If x≠0 and x ∉ A (in particular, x≠1), then there is a unique natural number n such that 1/2n+1 < x < 1/2n, and there is an open interval ]a, b[ included in U with 0 < a < b < 1 that contains x; then ]a, b[ ⋂ ]1/2n+1, 1/2n[ = ]max (a, 1/2n+1), min (b, 1/2n)[ is a basic Smirnov open set included in U–B that contains x.
- If x=1, then U contains a half-open interval ]a, 1] with 0 < a < 1; our desired Smirnov open set is ]max (a, 1/2), 1].
- If x ∈ A and x≠1, say x = 1/2n with n≥1, then 1/2n+1 < x < 1/2n–1, and there is an open interval ]a, b[ included in U with 0 < a < b < 1 that contains x; then ]a, b[ ⋂ ]1/2n+1, 1/2n–1[ = ]max (a, 1/2n+1), min (b, 1/2n–1)[ is a basic Smirnov open set included in U–B that contains x.
- If x=0, then U contains a half-open interval [0, c[ with 0 < c < 1. Then U–B contains the basic Smirnov open set [0, c[–A, which contains x.
Conversely, we need to show that every basic Smirnov open set ]a, b[ or ]a, 1] or [0, c[–A can be written as U–B with U open in [0, 1] and B ⊆ A, and this is immediate. ☐
Proposition I. [0, 1]Sm is not compact. More generally, no subset of [0, 1]Sm containing infinitely many numbers 1/2n is compact.
Proof. Let E be any subset of [0, 1]Sm consisting of infinitely many numbers 1/2n, say 1/2n0, 1/2n1, …, where n0 < n1 < … nk Since every sequence in a compact subset has a cluster point, it suffices to see that the sequence (1/2nk)k ∈ N has no cluster point in [0, 1]Sm. If x is a cluster point of (1/2nk)k ∈ N in [0, 1]Sm, then it is also a cluster point of (1/2nk)k ∈ N in [0, 1], because the topology of [0, 1]Sm is finer than that of [0, 1] (Fact D): every open neighborhood of x in [0, 1] is open in [0, 1]Sm, hence must contain infinitely many points 1/2nk. In [0, 1], the only cluster point of (1/2nk)k ∈ N is its unique limit 0. But 0 is not a cluster point of (1/2nk)k ∈ N in [0, 1]Sm, because the open subset [0, 3/4[–A contains 0 and simply no point 1/2nk at all.
Alternatively, we define 1/2nk as above, and we define Bk ≝ {1/2nk, 1/2nk+1, 1/2nk+2, …} for every k ∈ N. By Lemma H, the sets [0, 1]–Bk are all open in [0, 1]Sm. Those sets form a directed family of open subsets, since B0 ⊇ B1 ⊇ …, hence [0, 1]–B0 ⊆ [0, 1]–B1 ⊆ …, and their union is the whole of [0, 1]Sm, since the sets Bk have empty intersection. In particular, they form a directed open cover of E. If E were compact, then some set [0, 1]–Bk would already contain E (Proposition 4.4.7 in the book). But no set [0, 1]–Bk contains E, since 1/2nk is in E but not in [0, 1]–Bk. ☐
In particular, the conclusion of Theorem A fails, by taking C ≝ [0, 1]Sm itself, a ≝ 0, and b ≝ 1. Let us see what assumptions of Theorem A are wrong. C is a closed connected chain, and the ordering is semiclosed:
Lemma J. The usual ordering ≤ on [0, 1]Sm is semiclosed.
Proof. It is already semiclosed on [0, 1] with its usual metric topology, and the topology of [0, 1]Sm is finer (Fact D). ☐
Hence the only possible assumption of Theorem A that can fail is local compactness. We verify this by hand.
Proposition K. 0 has no compact neighborhood in [0, 1]Sm. In particular, [0, 1]Sm is not locally compact.
Proof. Let us imagine a compact neighborhood K of 0 in [0, 1]Sm. K must contain a basic Smirnov open subset containing 0, and that subset must be of the form [0, b[–A.
We claim that every element 1/2n of A that is in [0, b[ is also in K. In order to see this, we build the sequence (1/2n–1/2n+k+1)k ∈ N, and that is a sequence of elements of K: since 1/2n–1/2n+k+1 ≤ 1/2n ∈ [0, b[, 1/2n–1/2n+k+1 is also in [0, b[, and it is not in A, so it is in [0, b[–A, and therefore in K. That sequence has a unique limit in [0, 1], which is 1/2n. Since the topology of [0, 1]Sm is finer than that of [0, 1] (Fact D), K is also compact in [0, 1], hence closed, and therefore 1/2n must be in K.
Hence K must contain the whole of [0, b[.
The subsequence of those points 1/2n that are in [0, b[ must have a convergent subsequence in [0, 1]Sm, since K is assumed compact. The limit of that subsequence in [0, 1]Sm must be its unique limit in [0, 1], namely 0: indeed, every open neighborhood of 0 in [0, 1] is an open neighborhood of 0 in [0, 1]Sm, hence must contain all the terms of the subsequence from a certain point. But 0 is not a limit of any sequence of points of A, because [0, 3/4[–A is an open neighborhood of 0 that does not contain any point of the sequence.
This is a contradiction; hence K is not a compact neighborhood of 0 in [0, 1]Sm after all. ☐
While 0 has no compact neighborhood in [0, 1]Sm, every point x≠0 has a base of compact neighborhoods. One might say that [0, 1]Sm is locally compact around every point except 0. In order to see this, it is enough to show that every closed interval [x–ε, x+ε] included in ]0, 1] (namely, missing 0) is compact in [0, 1]Sm. To this end, we use Alexander’s subbase lemma (Theorem 4.4.29 in the book), and we consider any cover by subbasic open sets ]a, b[ or ]a, 1] or [0, c[–A, where 0<a<b<1, and 0<c<ε (see Lemma F). Since sets of the form [0, c[–A with 0<c<ε are disjoint from [x–ε, x+ε], we can remove them from the cover. What remains is an open cover of [x–ε, x+ε] is an open cover by open subsets of [0, 1], with its usual metric topology. Since [x–ε, x+ε] is compact in [0, 1] (see Example 4.4.3 in the book), we can extract a finite cover from that.
Now, since the conclusion of Theorem A fails on [0, 1]Sm, something in the proof must fail. (We take C ≝ [0, 1], a ≝ 0, b ≝ 1.) No element of [0, 1]Sm except 0 is a local minimum, so that one (item 5 in the proof) is fine. Claim I, item (i) holds here: for every point x such that 0<x, there is a point y such that y<x and [y, x] us compact. Claim II (for every x ∈ K such that a<x, there is a point y in K such that y<x, where K is the set of points x such that [x, 1] is compact) holds, too. Claim III (K has an infimum) is true, too; in fact the infimum of K is 0. But 0 is not in K. The final part of the proof of Theorem A shows that the infimum of K is in K by relying on Claim I, item (ii), which says that for every point x<1, there is a point z such that x<z and [x, z] is compact. This is true for every x≠0, but fails when x=0, because of Proposition I.
Further properties of [0, 1]Sm
We know that [0, 1]Sm is not compact, not locally compact, but it is Hausdorff and connected.
As far as separation properties are concerned, [0, 1]Sm is not regular. Indeed, the set A itself is a closed subset of [0, 1]Sm, being the complement of the set [0, 1]–A, which is open by Lemma H. The point 0 is not in A, but there is no open neighborhood U of 0 and no open neighborhood V of A such that U and V would be disjoint. Indeed, U must contain a basic Smirnov open subset of the form [0, c[–A where 0<c<1. Then any element 1/2n of A such that 1/2n<c has an open neighborhood of the form ]1/2n–ε, 1/2n+ε[ included in V, for some ε>0 small enough, and ]1/2n–ε, 1/2n+ε[ intersects [0, c[–A, hence U, at uncountably many points.
Although [0, 1]Sm is connected, it is not path-connected. More precisely, there is no continuous path γ : [0, 1] → [0, 1]Sm from 0 to any other point, namely such that γ(0) = 0 and γ(1) ≠ 0. (There is a path from any point different of 0 to any point different from 0.) We argue by assuming γ exists, and by aiming for a contradiction. The image of the compact space [0, 1] under γ is compact in [0, 1]Sm. By Proposition I, it cannot contain infinitely many elements of A. Hence, for some k ∈ N, the points 1/2k, 1/2k+1, … must all lie outside the image of γ. Since the topology of [0, 1]Sm is finer than that of [0, 1] (Fact D), γ is also continuous from [0, 1] to [0, 1], so the image of γ is a compact connected subset of [0, 1] that contains 0, hence an interval of the form [0, b]. This interval contains no point 1/2k, 1/2k+1, …, so b=0. This is impossible since γ(1) ≠ 0.
Although [0, 1]Sm is connected, it is not locally connected. Specifically, 0 does not have any base of connected open neighborhoods. (All the others have one.) To this end, it suffices to observe that [0, 3/4[–A, which is a basic Smirnov open set, does not contain any open connected subset U that contains 0. Indeed, for every open subset U of [0, 3/4[–A that contains 0, U must contain a basic Smirnov open set [0, c[–A. We pick 1/2n with n large enough, so that 1/2n<c, and then we realize that U ⋂ [0, 1/2k[ is also a basic Smirnov open set, which contains 0, that U – [0, 1/2k[ = U ⋂ ]1/2k, 1] (since 1/2k is in A, hence not in U) is open and contains uncountably many points above and arbitrarily close to 1/2k. This produces a partition of U in two disjoint non-empty open subsets, showing that U is not connected.
- Jimmie Lawson. Chains in partially ordered spaces. Topology Proceedings 61:123–137, 2023.
- Lynn Arthur Steen and J. Arthur Seebach, Jr., Counterexamples in Topology. Springer-Verlag, New York, 1978. Reprinted by Dover Publications, New York, 1995. ISBN 0-486-68735-X
— Jean Goubault-Larrecq (April 20th, 2025)