Complete topological descriptions of certain Morse boundaries

We study direct limits of embedded Cantor sets and embedded Sierpi\'nski curves. We show that under appropriate conditions on the embeddings, all limits of Cantor spaces give rise to homeomorphic spaces, called $\omega$-Cantor spaces, and similarly, all limits of Sierpi\'nski curves give homeomorphic spaces, called to $\omega$-Sierpi\'nski curves. We then show that the former occur naturally as Morse boundaries of right-angled Artin groups, while the latter occur as Morse boundaries of fundamental groups of finite-volume, cusped hyperbolic 3-manifolds.


Introduction
Many geodesic metric spaces which are not Gromov hyperbolic nonetheless display some hyperbolic-like behavior.The Morse boundary was introduced in [CS15] and [Cor17] with the goal of identifying and encoding this behavior in a useful way.It is defined for any proper geodesic metric space (though in some cases it may be empty), and a key property states that is invariant under quasi-isometries.In particular, the Morse boundary of a finitely generated group G is well-defined and provides a quasi-isometry invariant for the group.
Set theoretically, the Morse boundary of X consists of equivalence classes of geodesic rays satisfying the Morse property, namely, the property that any (λ, )-quasi-geodesics with endpoints on the ray stays bounded distance from the ray, where the bound, N = N (λ, ), depends only on λ and .If one restricts the Morse gauge N , then the corresponding boundary points can be given a natural topology.This space is denoted ∂ N M X.The Morse boundary ∂ M X is defined as the direct limit of these spaces.(See Section 2 below for formal definitions.) It is shown in [CH17] that the intermediate strata ∂ N M X are, in essence, boundaries of hyperbolic spaces and thus have many nice properties.Using the direct limit topology on ∂ M X has the advantage that many questions can be reduced to understanding these intermediate spaces.Indeed, the Morse boundary has been shown to have many properties analogous to boundaries of hyperbolic spaces (see eg., [CH17], [CCM19], [Mur19], [Liu19], [Zal18]).On the other hand, the limit space itself can be quite complicated.In particular, if X itself is not a hyperbolic space, then the topology on ∂ M X is neither compact nor metrizable.However, in this paper we provide the first complete topological descriptions of non-compact Morse boundaries, showing that, at least sometimes, Morse boundaries are more "accessible" as topological spaces than previously thought.
More specifically, we study the Morse boundaries of right-angled Artin groups and fundamental groups of finite-volume cusped hyperbolic 3-manifolds.Understanding these boundaries involves not only characterizing the strata ∂ N M X, but also understanding how they sit inside each other.As we will discuss below, the process is more subtle than it at first appears.
Direct limits of Cantor spaces and right-angled Artin groups.For the case of a right-angled Artin group we prove: Theorem 1.1.Let A Γ be a right-angled Artin group, for Γ a finite graph.Then ∂ M A Γ satisfies exactly one of the following: (1) ∂ M A Γ is empty, (2) ∂ M A Γ consists of two points, (3) ∂ M A Γ is a Cantor space, or (4) ∂ M A Γ is an ω-Cantor space.
In addition, we give precise conditions on the defining graph Γ for when each of these cases occur.Here, an ω-Cantor space is defined as a direct limit of a sequence of Cantor spaces X 1 ⊂ X 2 ⊂ X 3 . . .such that X i has empty interior in X i+1 for all i.We call these limit spaces ω-spaces because all σ-compact Morse boundaries are examples of k ω -spaces (for more information see [FT77]).Crucially, in Theorem 3.3, we show that any two ω-Cantor spaces are homeomorphic, so that (4) describes a well-defined homeomorphism type.
The result for right-angled Artin groups in fact follows from a more general theorem: Theorem 1.2.Let G be a finitely generated group.Suppose that ∂ M G is totally disconnected, σ-compact, and contains a Cantor subspace.Then ∂ M G is either a Cantor space or an ω-Cantor space.
We note that all hypotheses on ∂ M G are necessary, but for the last two there is no known counterexample, for more discussion see below.
At first sight, one might think that Theorem 1.2 should be straightforward to prove using the stratification of ∂ M G given by Morse gauges.By hypothesis, each stratum is a totally disconnected compact space, which should make it a Cantor space, and hence would make ∂ M G a direct limit of Cantor spaces, as required.However, the fact that strata are compact and totally disconnected does not mean that they are Cantor spaces; they could be any closed subspace of a Cantor space, of which there are uncountably many homeomorphism types.The solution is to perturb the stratification by Morse guages to ensure that each stratum is a Cantor space.To do this we need to, roughly speaking, "add" a sequence of Cantor spaces converging to each isolated point of the stratum.This sequence of Cantor spaces is obtained as translates of the Cantor space given in the hypothesis of the theorem.
The hypothesis that ∂ M G is σ-compact is necessary since the ω-Cantor space is σcompact by construction.On the other hand, to the best of our knowledge there is no known example of a group G so that ∂ M G is not σ-compact.However, we believe that such examples can be found among small-cancellation groups, and it is even possible that every infinitely presented C (1/6)-small-cancellation group has non-σ-compact Morse boundary.
The hypothesis that ∂ M G contains a Cantor subspace is also a necessary condition.However, there is no known example of a group with non-compact Morse boundary that does not contain a Cantor subspace.In fact, in most of the motivating examples, one has a stable free subgroup (e.g., in acylindrically hyperbolic groups).On the other hand, there are examples by E. Fink of torsion groups with Morse rays [Fin17], but even those probably contain a Cantor subspace.In fact, Fink finds a tree in the Cayley graph where all rays are Morse.We believe that this tree is most likely stable, thereby exhibiting a Cantor subspace in the Morse boundary.This naturally raises a question: Does there exist a group whose Morse boundary is non-compact and does not contain a Cantor subspace?Direct limits of Sierpiński curves and cusped hyperbolic 3-manifolds.In the case of fundamental groups of cusped hyperbolic 3-manifolds, we show that their Morse boundaries are direct limits of a sequence S 1 ⊂ S 2 ⊂ S 3 . . . of Sierpiński curves.Moreover, these curves are nicely embedded, namely the peripheral curves of S i are disjoint from those of S i+1 .We call such a limit an ω-Sierpiński curve and prove that any two such are homeomorphic.
Theorem 1.3.Let M be a finite-volume hyperbolic 3-manifold with at least one cusp, and The difficulty here is similar to the problems that arise in the proof of Theorem 1.2.Roughly, each stratum is obtained from S 2 = ∂H 3 by removing "shadows" of horoballs, so one might at first expect that strata are Sierpiński curves.However, the shadows are not disjoint, so the situation is more complicated, and once again we have to perturb the stratification.In this case, more technology is needed to do so, namely technology from [Mac08,MS19], which allows us to carefully detour all the boundary circles of the shadows in order to make them disjoint.This is a rather delicate procedure, and in fact the technology to make something like this work in higher dimensions seems to not be available.It would be interesting to know whether Morse boundaries of fundamental groups of non-compact, finite-volume hyperbolic n-manifolds are all homeomorphic to a certain limit of the (n − 1)-dimensional analogues of Sierpiński curves.
Outline.Section 2 contains definitions and background on Morse boundaries.In Sections 3 and 4 we investigate limits of Cantor sets and apply this to characterize Morse boundaries of right-angled Artin groups.In Sections 5 and 6, we investigate limits of Sierpiński curves and prove Theorem 1.3.

Background
2.1.Morse boundary.We will assume throughout that X is a proper geodesic metric space.We begin with a definition of the Morse boundary of X and some properties that will be needed for the arguments below.We refer the reader to [Cor17] for more details.
Definition 2.1.A geodesic α in X is Morse if there exists a function N : R + × R + → R + such that any (λ, )-quasi-geodesic with endpoints on α, lies in the N (λ, )-neighborhood of α.The function N is called a Morse gauge for α.
Lemma 2.2 (Lemma 3.1 [Liu19]).For each Morse gauge N , there exists an N depending only on N , such that if α is an N -Morse geodesic in X and β is a subgeodesic of α, then β is N -Morse.
Remark 2.3.Taking Lemma 2.2 into consideration, we call a geodesic N -Morse if the geodesic and all subgeodesics have Morse gauge N .
It is shown in [Cor17] that if two sides of a triangle in X are N -Morse, then the third side is N 1 -Morse where N 1 depends only on N .This also holds for ideal triangles by [CCM19], Lemma 2.3.We record this property here as it will be used repeatedly below.
Lemma 2.4 ([CCM19]).Let X be a proper geodesic metric space.Let ∆(x, y, z) be a geodesic triangle with vertices x, y, z ∈ X ∪ ∂ M X and suppose that two sides of ∆ are N -Morse.Then the third side is N -Morse where N depends only on N.
For two Morse rays α, β in X, say α ∼ β if they have bounded Hausdorff distance.It is shown in [Cor17] that this bound depends only on the Morse gauge N , that is, there exists C N such that two N -Morse rays α, β are equivalent if and only if d(α(t), γ(t)) < C N for all t.The Morse boundary of X consists of the set of equivalence classes of Morse rays.To topologize this set, first choose a basepoint e ∈ X and let N be a Morse gauge.Set that is an N -Morse geodesic ray with β(0) = e} with the compact-open topology.These spaces are compact.This topology is equivalent to one defined by a system of neighborhoods, {V n (α) | n ∈ N}, defined as follows: M X e such that d(α(t), γ(t)) < C N for all t < n.Let M be the set of all Morse gauges.Put a partial ordering on M so that for two Morse gauges N, N ∈ M, we say N ≤ N if and only if N (λ, ) ≤ N (λ, ) for all λ, ∈ N. Define the Morse boundary of X to be M X e with the induced direct limit topology, i.e., a set U is open in ∂ M X if and only if U ∩ ∂ N M X e is open for all N .A change in basepoint results in a homeomorphic boundary, justifying the omission of the basepoint from the notation.More generally, we will usually assume the basepoint is fixed and omit it from the notation ∂ N M X e as well.The reader is warned, however, that unlike ∂ M X, these subspaces do depend on a choice of basepoint.
An alternate construction of the Morse boundary is given by the second author and Hume in [CH17].Define X (N ) e to be the set of all y ∈ X such that there exists a N -Morse geodesic [e, y] in X.Then X (N ) e satisfies the Gromov 4-point definition of hyperbolicity, hence, we may consider its Gromov boundary, ∂X M X e that induce a homeomorphism on the direct limits.Thus the Morse boundary, ∂ M X, can also be thought of as the direct limit of the metric spaces ∂X (N ) e .We now establish some useful properties of these spaces.
Lemma 2.5.For any N, N with N < N , the inclusion i : Proof.This map is continuous by Corollary 3.2 of [Cor17].What is left to show is that the map is closed.Let K be a closed set in ∂ N M X.Let γ i be a sequence in i(K) converging to a point γ.Since the γ i are represented by N -Morse geodesics and N -Morse geodesics converge to N -Morse geodesics [Cor17, Lemma 2.10] it converges to an N -Morse geodesic α in ∂ N M X.But since i is simply the inclusion and K is closed in ∂ N M X, then α = γ and thus i(K) is closed.
The following is a general fact about σ-compact Morse boundaries.It says that one can choose a countable exhaustion of the Morse boundary by strata.
Lemma 2.6.If ∂ M X is σ-compact, then there exists a sequence of Morse gauges N 1 , N 2 , . . .[CD19] we know that for any compact subset K ⊂ ∂ M X there exists an N so that K ⊂ ∂ M X (N ) e .Thus for each K i we have a Morse gauge N i and the ascending condition on the K i guarantees that N i ≤ N i+1 and since the K i exhaust ∂ M X then it follows that ∂ M X = ∂X

Limits of Cantor spaces
In this section we describe particular direct limits of Cantor sets, and more precisely of sequences of Cantor spaces each having empty interior in the next one.Definition 3.1.Let C D be Cantor spaces.We say that C is entwined in D if C has empty interior in D. An ω-Cantor space is a topological space lim − →i∈N X i , where each X i is a Cantor space and each X i is entwined in X i+1 .
Example 3.2.Let C be the usual middle-third Cantor space.For each middle third interval I, glue in a copy C of C by identifying the boundary points of C with the boundary points of I.Call the resulting space D. Then D is again a Cantor space and C is entwined in D.
A similar example can be described in terms of boundaries.Let T be a (rooted) trivalent tree so that ∂T can be identified with C. Now glue a separate copy of T to each vertex v of T by identifying the root to v. The result is a larger tree T with T ⊂ T such that ∂T is entwined in ∂T .
The main result of this section is that there is only one ω-Cantor space up to homemorphism: Theorem 3.3.Any two ω-Cantor spaces are homeomorphic.
The rest of this section is devoted to the proof of this theorem.After the preliminary observation in the remark below, we show that, roughly speaking, when C ⊆ D are Cantor spaces, with C entwined in D, and C is subdivided in two clopen sets C 0 , C 1 , we can take small clopen neighborhoods of C 0 , C 1 in D.
Remark 3.4.A clopen set in a Cantor space is either empty or a Cantor space.
Lemma 3.5.Let C ⊆ D be Cantor spaces, with C entwined in D. Then if C = C 0 C 1 with C i clopen and non-empty, then there exist clopen sets D 0 , D 1 ⊆ D so that: (1) (4) If d is any metric on D compatible with its topology, then for any > 0, we can choose D i ⊆ N (C i ).
(5) C i and D i are Cantor spaces, and C i is entwined in D i .
Proof.Fix a metric d on D compatible with its topology and > 0. We can choose > 0 so that D i = N (C i ) satisfy the first 4 properties listed above (in fact, up to now we only need C to be properly contained in D).However, the D i are open, but they might be not closed; we now shrink them to make them clopen.Since every point in a Cantor space has a neighborhood basis of clopen sets, for every point which is open), and it is readily seen that the first 4 properties are satisfied.
The fact that C i and D i are Cantor spaces follows now from Remark 3.4.We now argue that In particular, the interior of C i as a subspace of D i coincides with the interior of C i as a subspace of D. Since C ⊇ C i has empty interior in D i , we have that C i has empty interior in D, whence in D i , as required.
Next, we show that we can extend partial homeomorphisms of Cantor spaces defined over entwined Cantor subspaces.Proof.Let Ω be the set of finite words in the alphabet {0, 1}, and denote by e the empty word.We denote the word obtained by appending 0 (resp.1) at the end of the word ω ∈ Ω simply by ω0 (resp.ω1).Fix metrics on D and D compatible with their respective topologies.We can choose clopen sets C ω ⊆ C with the following properties: (1) ) for any > 0, there are only finitely many C ω whose diameter is larger than .
Set C ω = φ(C ω ), and notice that analogous properties hold for the C ω as well.Also, set D e = D, and, inductively on the length of ω, use Lemma 3.5 to construct clopen sets ) is non-empty, (5) for any > 0, there are only finitely many D ω whose diameter is larger than .Also, construct D ω ⊆ D (and K ω ) with the analogous properties.Notice that D − C = ω∈Ω K ω and that each K ω (which is clopen) is a Cantor space by Remark 3.4.Similar observations hold for D − C and K ω .
We are now ready to define φ.Choose for each ω any homeomorphism φ ω : K ω → K ω , and set φ(x) = x if x ∈ C, and φ(x) = φ ω (x) if x ∈ K ω .Then φ is bijective.Hence, since D, D are compact and Hausdorff, to show that φ is a homeomorphism we are left to show that it is continuous.Notice that by construction, we have φ(D ω ) = D ω for each ω.It is readily checked that {D ω } is a basis for the topology of D .In fact, for any x ∈ D and > 0 there exists some D ω of diameter at most that contains x, since there are infinitely many D ω containing x and only finitely many of them have diameter larger than .Hence, the preimage under φ of any member of a neighborhood basis for D is open, and hence φ is continuous.
Finally, we are ready to prove Theorem 3.3, by iterative extensions.
Proof of Theorem 3.3.Let X = lim − →i∈N X i and X = lim − →i∈N X i be ω-Cantor spaces.Start with any homeomorphism φ 0 : X 0 → X 0 and inductively define homeomorphisms φ n+1 : X n+1 → X n+1 extending φ n ; those exist by Lemma 3.6.Then the φ n give a well-defined bijection X → X , which is a homeomorphism by definition of the direct limit topology.

ω-Cantor boundaries
We now study totally disconnected Morse boundaries of finitely generated groups.In this case, we write ∂ M G for the Morse boundary of some (hence any) Cayley group of G.
Following [CH17], we say that a quasi-convex subspace Y of a geodesic metric space X is N -stable if every pair of points in Y can be connected by a geodesic which is N -Morse in X.We say that a subgroup is stable if it is stable as a subspace.
The main theorem in this section is the following.Theorem 1.2 Let G be a finitely generated group.Suppose that ∂ M G is totally disconnected, σ-compact, and contains a Cantor subspace.Then it is either a Cantor space or an ω-Cantor space.
We will need the following lemma to construct sequences limiting to a specific boundary point, to show that such boundary point is not isolated.Lemma 4.1.Let G be a group, let γ, α 1 , α 2 be Morse rays, with [α 1 ] = [α 2 ], where all rays are based at e. Also, let g i = γ(i) ∈ G be the sequence of group elements traversed by γ.Then for some j ∈ {1, 2} and some subsequence i(n) we have Proof.By Lemma 2.4, if two legs of a geodesic triangle are N -Morse then the third leg is N -Morse where N depends only on N .In our case, this says that there exists N , depending on the Morse gauges of γ, α 1 , α 2 , so that there exist N -Morse geodesic rays Moreover, there exists C, again depending only on the various Morse gauges, so that all triangles γ([ Suppose by contradiction that for j = 1, 2 no subsequence of β j i converges to γ.This means that for all i, j the Gromov products (γ, β j i ) e are uniformly bounded by, say, R. By slimness, this says that there exists R = R (R, C) sush that each g i α j intersects N R (e).As a consequence, the Gromov products (g i α 1 , g i α 2 ) g i diverge.However, all said Gromov products are equal to (α 1 , α 2 ) e .This is impossible unless this product is infinite, and hence, [α 1 ] = [α 2 ], contradicting one of the hypotheses.i do not converge to γ, then the g i α j both backtrack a lot along γ.But then, this means that the α j fellow-travel for arbitrarily long times.
We now augment stable subsets of G to make their boundary a Cantor space.
Lemma 4.2.Let G be a group with totally disconnected Morse boundary and a Cantor space C 0 ⊂ ∂ M G. Then for every Morse gauges N there exists an N and a Cantor space M G for some Morse gauge N 0 .Since ∂ M G is totally disconnected, it follows that for any Morse gauge N , so is ∂ N M G.We will consider some N so that In fact, by Lemma 2.4, there exists N depending on N, N 0 only so that A ⊆ ∂ N M G. Let now C = A, and we claim that C is a Cantor space.To check this we need show that C is non-empty, perfect, compact, totally disconnected, and metrizable.It is metrizable, totally disconnected, and compact because it is a closed subspace of ∂ N M G.It is clearly non-empty so what is left to show is that it is perfect, i.e., C has no isolated points.Any point in C − A is clearly not isolated.Also, any point in some gC 0 is not isolated, since gC 0 is a Cantor space.Finally, any point in ∂ N M G is a limit of points in C (in fact, even in A) by Lemma 4.1.
We now show that the subsets Proof.Note that it suffices to find N so that Then the sequence of points (g i ) converges to p. Passing to a subsequence if necessary, the sequence (g −1 i ) also converges to some point q ∈ ∂ M G. Since ∂ M G is not compact, we can choose a point z ∈ ∂ M G with z = q and z / ∈ ∂ N 1 M G, where N 1 is the Morse gauge from Lemma 2.4.It follows from [Liu19], Lemmas 5.3 and 6.9, that for some N , g i z converges to p in ∂ N M G.It remains to show that g i z / ∈ ∂ N M G for all i.To see this, let β i be a geodesic ray based at e representing g i z and let γ i be the segment of γ from e to g i .Then γ i , β i , g i β 0 form a geodesic triangle with vertices e, g i , g i z.By Lemma 2.4, if γ i and β i are both N -Morse, then g i β 0 is N 1 -Morse, and hence its translate, β 0 , is also N 1 -Morse.But z = [β 0 ], so this contradicts our choice of z and we conclude that β i is not N -Morse.Thus we have We are ready to prove Theorem 1.2.
Since it is a non-empty, compact, perfect, totally disconnected, metrizable metric space, it follows that its boundary is a Cantor space by [Bro10].
Assume now that G can be chosen to be a countable limit over gauges N 1 , N 2 , . ... By Lemma 4.2 there is a Cantor space G for a sufficiently large j(2).In view of Lemma 4.3, we may increase j(2) to ensure that 4.1.Totally disconnected boundary versus totally disconnected levels.In this subsection we show that, when ∂ M X is σ-compact, then it is totally disconnected if and only if all its strata are.This will be needed in our analysis of Morse boundaries of right-angled Artin groups.
Proposition 4.4.Let X be a proper metric space, let e ∈ X, and suppose that ∂ M X is σ-compact.Then ∂ M X is totally disconnected if and only if ∂ N M X is totally disconnected for every Morse gauge N .
Proof.If ∂ M X is totally disconnected then any given stratum, ∂ N M X is totally disconnected as well since the stratum is topologically embedded in ∂ M X by Lemma 2.5.
Suppose that ∂ N M X is totally disconnected for every Morse gauge N .By Lemma 2.6, there exists a sequence of Morse gauges N 1 , N 2 , . . .
M X.We will prove that for each pair of distinct points x, y ∈ ∂ M X there exists a clopen subset A ⊆ ∂ M X so that x ∈ A and y / ∈ A, which suffices to show that ∂ M X is totally disconnected.We can assume x, y ∈ ∂ N 1 M X.In fact, we will construct a sequence of clopen subsets M X = A j for each j < i; we can then just set A = A i .Also, we can replace the last condition simply by , and the stronger condition with j < i will follow by a simple inductive argument.
Let us start with the remark that, for each i, since ∂ N i M X is a totally disconnected compact metrizable space, we have that each point has a neighborhood basis of clopen subsets.In particular, there exists a clopen set Suppose that we constructed A i−1 with the required properties, for some i ≥ 2. Set and notice that it is also clopen in M X, since they are compact (because they are closed in the compact space M X has a basis of clopen neighborhoods, a straightforward compactness argument gives that A i−1 is contained in a finite union of clopen sets of ∂ N i M X, each contained in U .The union A i of said clopen sets has the required properties, and we are done.4.2.Artin groups.We now show that ω-Cantor spaces arise naturally as Morse boundaries of right-angled Artin groups.Let A Γ denote the right-angled Artin group with finite defining graph Γ. Theorem 1.1 Let A Γ be a right-angled Artin group, for Γ a finite graph.Then ∂ M A Γ satisfies exactly one of the following: (1) Proof.If A Γ is a direct product (⇔ Γ is a join), then every geodesic is contained in a flat, so the Morse boundary of A Γ is empty.If A Γ is a free group (⇔ Γ is discrete) then the Morse boundary is equal to the Gromov boundary which is either a Cantor space (if A Γ is non-abelian) or just two points (if In all other cases, it is shown in [KMT17] that A Γ contains a stable non-abelian free group and thus has a Cantor space as a subspace of the Morse boundary.Moreover, ∂ M A Γ is σ-compact since A Γ is CAT(0) and thus its Morse boundary is equivalent to the contracting boundary defined in [CS15].The boundary of ∂ N M A Γ is totally disconnected for all Morse gauges N by Theorem 5.1 in [CH17] and thus by Proposition 4.4 the whole Morse boundary is disconnected.
Applying Theorem 1.2 gives the desired result.

Limits of Sierpiński curves
We now begin our study of limits of Sierpiński curves.First, we state the characterization of the Sierpiński curve that will be most useful for our purposes.We then define entwined Sierpiński curves, and study direct limits of entwined Sierpiński curves.
A Sierpiński curve S is a topological space homeomorphic to S 2 − i∈N D i , where D i is the interior of a closed disk in S 2 satisfying . All Sierpiński curves are homeomorphic to each other [Why58].The peripheral circles of a Sierpiński curve are the boundaries of the D i as above, and they can be characterised as the subspaces γ of S homeomorphic to S 1 so that S − γ is connected (this follows from the Jordan curve theorem, see the argument in [Kra69, page 256, case (ii)]).
Definition 5.1.Let S ⊆ T be Sierpiński curves.We say that S is entwined in T if no peripheral circle of S intersects any peripheral circle of T .
The following lemma says the following.Suppose that S ⊆ T are Sierpiński curves, with S entwined in T .Then T is obtained from S by attaching Sierpiński curves onto each peripheral circle.
Lemma 5.2.Let S ⊆ T be Sierpiński curves, with S entwined in T .Then there exist Sierpiński curves S i ⊆ T with the following properties: (1) T = S ∪ S i , (2) S i ∩ S is a peripheral circle of S and S i , and any peripheral circle of S arises in this way, (3) S i ∩ S j = ∅ if i = j, (4) in any metric on T compatible with its topology, we have diam(S i ) → 0.
Proof.We can assume T = S 2 − D i , where the D i are open disks as described above.
Let γ be a peripheral circle of S. Then S 2 − γ is a union of two open disks D γ , D γ whose boundary is γ, by the Jordan curve theorem.One of these two disks, say D γ , is disjoint from S, for otherwise S − γ would be disconnected.Since S is entwined in T , each D i is contained in either D γ or D γ .Thus, D γ ∩ T is S 2 minus a union of interiors of closed disks, namely, D γ and the D i contained in D γ .Notice that the diameters of the disks D i go to 0, and that their union is dense in S 2 , and hence D γ ∩ T is a Sierpiński curve, which we will denote by S γ .We claim that the Sierpiński curves {S γ } have the required properties.
1) We first observe that S 2 − S = D γ .In fact, the closure of any complementary components of S is a closed disk by [Moo25, Theorem 9] and the Jordan curve theorem, and the boundary of such disk is then a peripheral circle.Thus, S 2 = S ∪ D γ and hence 2) We have S γ ∩ S = γ by construction.
3) It suffices to show that for two distinct peripheral circles γ, α, we have D γ ∩ D α = ∅.If not, since the boundary circles of γ and α are disjoint, up to switching γ and α, we would have D γ ⊆ D α (given complementary components of γ, α, they are either disjoint or one of them, as well as its closure, is contained in the other).But then we would have γ ⊆ D α , in contradiction with the fact that D α ∩ S = ∅.
4) This statement does not depend on the choice metric, so for convenience we endow S 2 with the metric d as a subspace of Euclidean space, and in turn endow S and T with their metrics subspaces of (S 2 , d).Since each S γ is contained in the closure of the corresponding D γ , it suffices to show that, given any > 0 there are only finitely many γ with diam(D γ ) ≥ .
Fix > 0, and we can further assume < 1.Notice that there exist only finitely many γ of diameter at least /2 (for example because S is homemorphic to the standard Sierpiński carpet, and this is true of the peripheral circles in that case), and from now on we only consider those with diameter at most /2.The issue we have to deal with now is that D γ can have diameter much larger than that of its boundary γ (in fact, it is always the case that one of the two components of the complement of γ is has "large" diameter).However, what we know, thanks to the fact that we are working with the restriction of the Euclidean metric, is that the component D γ of the complement of γ either has diameter at most , or it contains a ball in (S 2 , d) of radius 1.In fact, γ is contained in a closed Euclidean ball B ⊆ R 3 of radius /2, and B ∩ S 2 is connected.Hence, either D γ is contained in B, yielding the first case, or D γ contains S 2 − B, which is easily seen to contain a half-sphere (since is sufficiently small), yielding the second case.
The second case can occur at most finitely many times since, as we argued above, the D γ are pairwise disjoint, and we can only fit finitely many disjoint balls of radius 1 on S 2 .Hence, there are only finitely many γ with diam(D γ ) ≥ , as required.
The following point-set topology lemma will allow us to check continuity of certain maps, namely the ones that we will encounter when constructing homeomorphisms between limits of entwined Sierpiński curves.
Lemma 5.3.Let f : X → Y be a map between two metric spaces and let C 0 , C 1 , . . .be closed subsets of X. Suppose that (1) Proof.Let (x n ) be a sequence in X converging to x.It suffices to show that a subsequence of (f (x n )) converges to f (x).
We consider two cases, at least one of which applies, by property 1. 1) There is a subsequence (x n k ) contained in some C j .Since C j is closed, we also have x ∈ C j .Then (f (x n k )) converges to f (x) because f | C j is continuous by property 5.
2) There is a subsequence (x n k ) is so that no two elements of the subsequence are contained in the same C j .Suppose which exists by property 2).Then (x n k ) converges to x in view of property 3, which implies that d X (x n k , x n k ) tends to 0. By case 1 (with j = 0), we have that (f (x n k )) converges to f (x).Finally, by property 4, d Y (f (x n k ), f (x n k )) also tends to 0, so (f (x n k )) also converges to f (x).
The following proposition is, essentially, the inductive step in the definition of homeomorphisms between limits of entwined Sierpiński curves.
Proposition 5.4.Let S ⊆ T, S ⊆ T be Sierpiński curves, with S entwined in T and S entwined in T .Then any homeomorphism φ : S → S can be extended to a homeomorphism ψ : T → T .
Proof.Let S i ⊆ T, S i ⊆ T be the Sierpiński curves that we obtain from Lemma 5.2.Since {S i ∩ S} is the set of peripheral circles of S and {S i ∩ S } the set of peripheral circles of S , we can choose the indices in such a way that φ(S i ∩ S) = S i ∩ S.
The arguments from [Why58, show that any homeomorphism between peripheral circles of Sierpiński curves can be extended to a homeomorphism between the Sierpiński curves.In particular, there are homeomorphisms φ i : S i → S i that extend φ| S i ∩S .Hence, we can define ψ : T → T by requiring ψ| S = φ, ψ| S i = φ i .This is a well-defined map in view of Lemma 5.2-(3).Then ψ is clearly bijective.By Lemma 5.2-(1)-( 2)-(4)), the hypotheses of Lemma 5.3 hold, so ψ is continuous.Hence, it is a homeomorphism because T, T are compact and Hausdorff.Thus, ψ is the desired extension of φ.
Finally, we are ready for the main theorem of this section.Definition 5.5.An ω-Sierpiński curve is a topological space lim − →i∈N X i , where each X i is a Sierpiński curve and each X i is entwined in X i+1 .
Proof.The proof is identical to the proof of Theorem 3.3, using Proposition 5.4 instead of Lemma 3.6.

ω-Sierpiński boundaries
We now study Morse boundaries of fundamental groups of finite-volume hyperbolic 3-manifolds.The main theorem of this section is: Theorem 1.3 Let M be a finite-volume hyperbolic 3-manifold with at least one cusp, and let G = π 1 (M ).Then ∂ M G is an ω-Sierpiński curve.
6.1.Setup and notation.Let M be a finite-volume hyperbolic 3-manifold with at least one cusp, and let G = π 1 (M ), regarded as a subgroup of Isom(H 3 ).Lifting disjoint cuspidal neighborhoods to H 3 , we obtain an equivariant family of disjoint horoballs {H p } p∈P .Regarding S 2 with the standard metric as the boundary of H 3 , we have the collection P ⊆ S 2 of parabolic points for G.For each p, we denote r p = e −d H 3 (x 0 ,Hp) , where x 0 is the origin of the Poincaré disk model of H 3 .
For a given 0 < λ ≤ 1, we denote V λ = S 2 \ p∈P B(p, λr p ). Set X = H 3 − H p , a neutered space for M .Denote by O p ⊆ X the horosphere that bounds H p .
We will use the fact that X is a CAT(0) space and that each O p is a flat in X, meaning that it is convex and isometric to R 2 , see e.g.[BH99,Chapter II.11].In addition, the inclusion map ι : X → H 3 is 1-Lipschitz and proper.
We will use this notation throughout this section.
6.2.From the Morse boundary to ∂H 3 .In this section we relate the Morse boundary of X with ∂H 3 ; the main result is Proposition 6.4.Roughly speaking, we have to show that Morse geodesic rays in X correspond to geodesic rays in H 3 that do not spend too much time in horoballs, and also relate the latter rays with points in the subspace V λ ⊂ ∂H 3 defined above.This can be done in greater generality than our case (groups hyperbolic relative to subgroups with empty Morse boundaries, replacing H 3 with the cusped space), but we decided to use the extra structure available to us to make the proofs simpler.See [Tra18] for related work.First of all, we show that Morse geodesic rays in X give quasi-geodesics in H 3 .Lemma 6.1.Let γ be an N -Morse geodesic ray in X.Then ι(γ) is a (K, L)-quasi-geodesic ray where K, L depend only on N .
product (γ, γ p ) x 0 will be at least d(x 0 , H p ) − ln(λ), and hence the distance between x and p will be at most λr p .
The proof of the following lemma is a variation on the usual arguments to show that contracting properties of various kinds imply the Morse property.We will need it to show that certain rays in X, constructed from rays in H 3 , are Morse.Lemma 6.3.For every diverging function f : R + → R + there exists a Morse gauge N with the following property.Let Z be a geodesic metric space and let γ ⊆ Z be a geodesic.Suppose that for each R > 0 any path β that intersects N R (γ) only at its endpoints x, y has the property that l(β We will show that any subpath β of α that intersects N R (γ) only at its endpoints x = α(t) and y = α(u) has length bounded by a constant D depending only on K, C, f (R), showing that α is contained in the (R + D)-neighborhood of γ.
We have, Finally, we compare the spaces ∂ N M X with the spaces V λ and the describe how, in order to study ∂ M X, one can study the V λ instead.Proposition 6.4.There exists a continuous injective map Ψ : ∂ M X → S 2 with the following properties.
(1) For each Morse gauge N there exist λ = λ(N ) so that Ψ ∂ N M X ⊆ V λ .(2) For every 0 < λ ≤ 1, there exists a Morse gauge N λ so that for each N ≥ N λ we have V λ ⊆ Ψ ∂ N M X .Proof.Throughout the proof, we fix basepoints x 0 = ι(x 0 ) of X and H 3 , and when discussing (quasi-)geodesic rays, we will assume that they are based at x 0 .
We first define Ψ : ∂ M X → S 2 .By Lemma 6.1, given ∈ ∂ N M X and an N -Morse geodesic ray γ representing , we can define Ψ( ) as the limit point of ι(γ).The fact that Ψ is well-defined and continuous follows from the fact that if two N -Morse geodesic rays γ, γ have initial subgeodesics of length L that stay within distance C of each other, then ι(γ), ι(γ ) have the same property (since ι is 1-Lipschitz).Injectivity follows from the fact that if the distance between γ(t), γ (t) diverges as t goes to infinity, then the same is true for ι(γ), ι(γ ) since ι is a proper map.
In view of Lemma 6.2, in order to prove item (1), we have to show that given an N -Morse geodesic ray γ in X, the geodesic ray α in H 3 asymptotic to ι(γ) intersects each horoball H p in a set of diameter bounded by a constant depending only N .Since the Hausdorff distance between α and ι(γ) is bounded in terms of N , it suffices to bound the diameter of the intersection of ι(γ) with a suitable neighborhood of H p .In turn, since ι is a proper map, it suffices to do the same in X.That is, we need to show that the diameter of the intersection of γ with the R-neighborhood of O p is bounded in terms of R and N .To see this, note that O p is a flat, and as the diameter of this intersection increases, we can find quasi-geodesic in the flat straying farther and farther away from γ. Since γ is N -Morse, the intersection must have bounded diameter.
In view of Lemma 6.2, in order to prove item (2), given a geodesic ray α in H 3 that intersects any horoball H p in a set of diameter at most C, we have to find an N -Morse geodesic ray γ in X so that ι(γ) lies within finite Hausdorff distance of α, where N = N (C).
We first do the case C = 0, and then show how to reduce to this case.If C = 0, then α is contained in X, and moreover it is a geodesic ray in X since ι is 1-Lipschitz.We only have to argue that α is Morse in X, with controlled Morse gauge.By Lemma 6.3, it suffices to prove that whenever β is a path (in X) intersecting the R-neighborhood of α only at its endpoints x, y, then l(β) ≥ f (R)d X (x, y) − f (R), where f is some fixed diverging function.Consider such a path β, and regard it now as a path in H 3 .Since ι is proper, β lies outside the ρ(R)-neighborhood of α (in H 3 ), where ρ is some diverging function.Moreover, ι is 1-Lipschitz and hence the endpoints of β are R-close to α.Using the hyperbolicity of H 3 (or even just that α is strongly contracting), one can then show that the length of β is at least g(ρ(R))d H 3 (x, y) − 2R, for some diverging function g.Notice that |d H 3 (x, y) − d X (x, y)| ≤ 10R, since x, y lie R-close to α both in X and in H 3 , and α is a geodesic in both X and H 3 .Hence, we get the required inequality.
We are only left to reduce the case of a general C to the case C = 0, and we will do so by changing the neutered space.Given a geodesic ray α in H 3 that intersects any horoball H p in a set of diameter at most C, we can regard it as a geodesic ray in a neutered space X containing X, in which we shrunk all the horoballs a uniform amount.The previous argument yields that α is a Morse geodesic ray in X .There is a quasi-isometry φ : X → X, whose constants depend only on C, that moves each point a bounded amount.Hence, φ(α) is an N -Morse (K, C)−quasi-geodesic, with N, K, C depending only on C. It is readily seen that a geodesic ray γ within bounded Hausdorff distance from φ(α) has the property that ι(γ) lies within finite Hausdorff distance of α, and we are done.6.3.Approximating strata with Sierpiński curves.As in the case of totally disconnected boundaries, we have to wiggle the strata in order for them to be actual Sierpiński curves.The following proposition, which we prove later, is what will allow us to do this: Recall that P ⊂ ∂H 3 denotes the set of parabolic points for the group G = π 1 (M ).Proposition 6.5.For each sufficiently small 0 < λ ≤ 1, and each p ∈ P , there exists a circle γ so that (1) γ ⊆ B(p, 3λr p /4) − B(p, λr p /4) = A and γ is homotopically non-trivial in A, (2) for each p ∈ P − {p} with r p ≤ r p we have γ ∩ B(p , 3λr p /4) = ∅.Corollary 6.6.For each sufficiently small 0 < λ ≤ 1 there exists a Sierpiński curve S with V λ ⊆ S ⊆ V λ/4 , with each peripheral circle of S contained in some B(p, 3λr p /4).
Proof.Consider for each p the open disk containing p bounded by the circle γ as in the previous lemma.Such disks are either disjoint or nested, so that we see that the complement of all the disks is the required Sierpiński curve.
We can now prove Theorem 1.3.Theorem 1.3.Let M be a finite-volume hyperbolic 3-manifold with at least one cusp, and let G = π 1 (M ).Then ∂ M G is an ω-Sierpiński curve.
Proof.Since G is quasi-isometric to the neutered space X, we can work with ∂ M X instead.First of all, ∂ M X is σ-compact since X is CAT(0) by [CS15, Main Theorem].Hence, by Lemma 2.6 we have that ∂ M X = lim − → ∂ N M X can be chosen to be a countable limit over gauges N 1 , N 2 , . ... By Proposition 6.4 and Corollary 6.6, and provided that N 1 is large enough, there is a Sierpiński curve S 1 so that ∂ X for a sufficiently large j(2).In fact, we can further require that λ(N j(2) ) < λ(N 1 )/4, for λ(N ) as in Proposition 6.4.Applying Proposition 6.4 and Corollary 6.6 again, we find j(3) and another Sierpiński curve S 2 so that ∂ X, again with λ(N j(3) ) < λ(N j(2) )/4.The condition λ(N j(2) ) < λ(N 1 )/4 ensures that S 1 is entwined in S 2 .Proceeding inductively, we see that ∂ M X is a limit of entwined Sierpiński curves.Thus, ∂ M X is an ω-Sierpiński curve, as required.
6.4.Proof of Proposition 6.5.We are only left to prove Proposition 6.5.We will follow arguments from [MS19], and the construction is roughly as follows.We start with a circle that possibly does not avoid all required balls around parabolic points, and we make detours to avoid balls of a certain size.We iterate the procedure for balls of smaller and smaller size, and then we take a limit.For technical reasons, we will work mostly with arcs rather than circles.
As in [MS19], in order to make this work we need machinery from [Mac08]; we now state all relevant facts and definitions.First of all, after detouring, we would like the new circle to be close to the previous one, and this is captured by the following definition Definition 6.7.For any x and y in an embedded arc A, let A[x, y] be the closed, possibly trivial, subarc of A that lies between them.
An arc B ι-follows an arc A if there exists a (not necessarily continuous) map p : B → A, sending endpoints to endpoints, such that for all x, y ∈ B, B[x, y] is in the ι-neighborhood of A[p(x), p(y)]; in particular, p displaces points at most ι.
In [Mac08], Mackay defines a space X to be N -doubling if every ball can be covered by at most N balls of half the radius and L-linearly connected if for all x, y ∈ X there exists a compact, connected subset Y containing x, y of diameter less than or equal to Ld(x, y).
The following proposition will be used to remove unwanted detours from our circles (which in the limit might create unwanted topology).Proposition 6.8 ([Mac08, Proposition 2.1]).Given a complete metric space X that is L-linearly connected and N -doubling, there exist constants s = s(L, N ) > 0 and S = S(L, N ) > 0 with the following property: for each ι > 0 and each arc A ⊂ X, there exists an arc J that ι-follows A, has the same endpoints as A, and satisfies (1) ∀x, y ∈ J, d(x, y) < sι =⇒ diam(J[x, y]) < Sι.
The following lemma will allow us to take limits.Lemma 6.9 ([Mac08, Lemma 2.2]).Suppose (X, d) is an L-linearly connected, N -doubling, complete metric space, and let s, S, ε and δ be fixed positive constants satisfying δ ≤ min{ s 4+2S , 1 10 }.If we have a sequence of arcs J 1 , J 2 , . . ., J n , . . . in X, such that for every n ≥ 1 • J n+1 εδ n -follows J n , and • J n+1 satisfies (1) with ι = εδ n and s, S as fixed above, then the Hausdorff limit J = lim H J n exists, and is an arc.Moreover, the endpoints of J n converge to the endpoints of J.
Proof.In view of Lemma 6.2, a point in B(p, λ 1 r p ) ∩ B(p , λ 2 r p ) is represented by a ray from the basepoint that spends a long time in both horoballs H p and H p .If the geodesic ray goes through H p first (the other case is symmetric), then H p is much further away from the basepoint than H p .Choosing an appropriate value for λ 0 , we can make the segment inside H p as long as we want, and thus control the distance to H p .In particular, we can guarantee that r p ≤ r p /100.
We are now ready to prove Proposition 6.5.
Start with the circle γ 1 of radius λ/2 around p. We think of γ 1 as the union of two arcs J 1 , J 1 joining diametrically opposite points.
We will construct a sequence of arcs J n , J n with the same endpoints that have the following properties.
Assume we have constructed J i , J i for i ≤ n.We now construct J n+1 and J n+1 .If J n and J n do not intersect any B p = B(p , 2 × 50 −n λr p ) with 50 −n−1 r p ≤ r p ≤ r p 50 −n , we can take J n+1 = J n and J n+1 = J n and we are done (notice the "2" for later purposes).Otherwise, the B p they intersect are disjoint by Lemma 6.10.We can then replace each maximal segment of J n or J n contained in some B p with an arc along the boundary of B p .If the segment is initial or terminal, we ensure that we move the corresponding endpoints of J n and J n to the same point.In this way we obtain immersed paths, which are not necessarily arcs, but we can restrict to arcs contained in the image of such paths.By applying this procedure, we obtain arcs K n , K n that (4 × 50 −n λr p )-follow J n and J n (the constant is the upper bound on the diameters of the B p ).
We now apply Proposition 6.8 with ι = 50 −n λr p , obtaining arcs J n+1 and J n+1 , satisfying (1), that (50 −n λr p )-follow K n and K n , and hence (5 × 50 −n λr p )-follow J n and J n .It is straightforward to check that J n+1 , J n+1 have the required properties.
According to Lemma 6.9, there are limit arcs J and J , that clearly share the same endpoints.While these may intersect in multiple points, there exist a pair of subarcs sharing only endpoints, that form a circle γ around p. We now check that γ satisfies the required properties.
A similar statement holds for J , and since J 1 ∪ J 1 is the circle of radius λr p /2 around p, we are done.
Moreover, the same computation as above also yields that J (λr p /9)-follows J 1 , and similarly for J .One can use this to show that the concatenation γ of J and J is homotopic to γ 1 in A = B(p, 3λr p /4) − B(p, λr p /4): Subdivide each into small arcs and retract arcs in γ to corresponding arcs in γ 1 along geodesics in ∂H 3 .Also, γ and γ are homotopic since they only differ in the 2λr p /9-neighborhood of the endpoints of J, J .
the associated visual metric d (N ) .The collection of boundaries ∂X (N ) e , d (N ) is called the metric Morse boundary of X.It is shown in [CH17] that there is are natural homeomorphisms between ∂X (N ) e and ∂ N

Lemma 3. 6 .
Let C ⊆ D and C ⊆ D be Cantor spaces, with C entwined in D and C entwined in D .Let φ : C → C be a homeomorphism.Then there exists a homeomorphism φ : D → D that extends φ.

Figure 1 .
Figure 1.If the β ji do not converge to γ, then the g i α j both backtrack a lot along γ.But then, this means that the α j fellow-travel for arbitrarily long times.