Lorentzian metric spaces and their Gromov-Hausdorff convergence

We present an abstract approach to Lorentzian Gromov-Hausdorff distance and convergence, and an alternative approach to Lorentzian length spaces that does not use auxiliary ``positive signature'' metrics or other unobserved fields. We begin by defining a notion of (abstract) bounded Lorentzian-metric space which is sufficiently general to comprise compact causally convex subsets of globally hyperbolic spacetimes and causets. We define the Gromov-Hausdorff distance and show that two bounded Lorentzian-metric spaces at zero GH distance are indeed both isometric and homeomorphic. Then we show how to define from the Lorentzian distance, beside topology, the causal relation and the causal curves for these spaces, obtaining useful limit curve theorems. Next, we define Lorentzian (length) prelength spaces via suitable (maximal) chronal connectedness properties. These definitions are proved to be stable under GH limits. Furthermore, we define bounds on sectional curvature for our Lorentzian length spaces and prove that they are also stable under GH limits. We conclude with a (pre)compactness theorem.


Introduction
This paper is devoted to the development of a Lorentzian-signature version of the by now classical theory of metric geometry, as developed, for instance, in the book by Burago et al. [5]. This theory comprises results on sectional curvature bounds via comparison geometry, as obtained by the A.D. Alexandrov school, and results on Ricci bounds from below as obtained by Gromov. Our work fits well in the current trend in low regularity spacetime (i.e. Lorentzian) geometry. Although there were contributions in non-regular Lorentzian geometry in the last two decades, it is only in the last few years that mathematicians have started getting some systematic results.
It must be said that the investigation of Lorentzian analogs to the Gromov-Hausdorff convergence has likely been one of the least explored. Among the most relevant papers we find those by Noldus who introduced and studied a Gromov-Hausdorff distance for compact globally hyperbolic manifolds with spacelike boundary [16,17] and later went on, in a joint work with Bombelli [4], to explore the non-manifold case.
Müller recently reconsidered some constructions by Noldus, including his strong metric, in a broad categorical analysis of the Gromov-Hausdorff convergence problem 1 [12,13].
Sormani and Wenger [19] followed a rather different route introducing the notion of intrinsic flat distance which is a kind of analog to the Gromov-Hausdorff distance when convergence of sets in the Hausdorff-distance sense is replaced 1 Actually our work is just a first version that we wish to present in order to show the stage of our research in the Lorentzian Gromov-Hausdorff convergence problem. We were prompted to post it due to a recent paper by Müller [14], which appeared just a few days ago, and which included some results that seem comparable to our own. A more polished version with more accurate references to previous work will follow.
by Federer-Fleming's flat convergence. This type of convergence is used to study sequences of Lorentzian manifolds foliated by spacelike manifolds with non-negative scalar curvature. Following a similar 'positive signature' approach Sormani and Vega [18] show that in a cosmological spacetime it is possible to introduce a distance by taking advantage of Andersson et al. cosmological time function [3]. Their canonical distance is well adapted to express convergence of cosmological spacetimes, a point of view taken up by Allen and Burtscher [2] where they show that in the class of warped product (smooth) spacetimes, uniform convergence of warping functions can be related to Gromov-Hausdorff and Sormani-Wenger intrinsic flat convergence.
Our approach is closer in spirit to Noldus' work, but starts with the identification of a convenient notion of (bounded) Lorentzian-metric space. This notion is more general than the subsequently introduced notion of (pre)length space. In this sense our work sets the stage for a complete Lorentzian analog of metric geometry. It is possibly this general framework that sets apart our work with respect to previous approaches. Also we define a notion of Gromov-Hausdorff convergence and show that all our relevant definitions, from the mentioned (pre)length spaces, to the sectional curvature bounds are stable under GH limits. In fact, it is this stability criteria that helped us in selecting what we believe to be a most convenient notion of (pre)length space.
Although this work presents results for just the timelike-diameter bounded spaces, we also explored the non-bounded case, obtaining some results that will likely appear in a next version.
Since we shall be introducing new definitions for Lorentzian (pre)length spaces, we wish to comment already the differences with the approach by Kunzinger and Sämann [8]. There a prelength space is a 5-tuple (X, d, , ≤, ρ), where X is the set, d the Lorentzian distance, the chronological relation, ≤ the causal relation, and ρ a metric, all these elements satisfying some compatibility conditions. Causal curves are defined by imposing the locally Lipschitz property with respect to ρ, see [8,Def. 2.18].
In our work we derive the topology, the chronological relation and the causal relation from just (X, d), which is what makes the GH stability property possible, Moreover, our causal curves are not defined imposing the locally Lipschitz property with respect to some metric. Actually, there is a metric that can be defined from d, this is the distinction or Noldus metric, but in simple smooth examples this choice shows to be untenable as the usual causal curves are nonrectifiable with respect to it [16,Thm. 6] [12,Thm. 3].
In fact in our work we do not really feel the need to restrict the class of causal curves with some kind of rectifiability property with respect to some metric. Indeed, we are able to obtain standard limit curve theorems by parametrizing causal curves (or better isocausal curves) via time functions constructed from the distance function, as these type of function has within it information that behaves well under GH limits.
In conclusion, our definition of Lorentzian length space seems to be different from that by Kunzinger and Sämann. More work will be required to establish if the two notions really coincide in some cases.
Finally, we mention that our work contains a (pre)compactness result for certain families of bounded Lorentzian-metric spaces with uniform bounds on the timelike diameter and on the number of certain -nets. Unfortunately, we could not establish a clear connection with Ricci bounds introduced via comparison geometry. It seems that the notion of -net is too much adapted to the notion of distinction-metric ball, which is, unfortunately, a non-local concept. More work and new ideas will be needed to produce a precompactness theorem involving bounds on the Ricci curvature.

Basic definitions
Definition 1.1. A bounded Lorentzian-metric space (X, d), is a set X endowed with a map d : X × X → [0, ∞) such that (i) For every x, y, z ∈ X such that d(x, y) > 0, d(y, z) > 0 we have d(x, z) ≥ d(x, y) + d(y, z).
(ii) Let T be the coarsest topology on X for which d is continuous. For every > 0 the sets {(p, q) : d(p, q) ≥ } are compact with respect to the product topology T × T .
(iii) d distinguishes points, i.e. for every pair x, y, x = y there is z such that d(x, z) = d(y, z) or d(z, x) = d(z, y).
(iv) There is a countable subset S ⊂ X such that in (iii) the point z can be found in S.
We write q ∈ I + (p), (p, q) ∈ I, or q p if d(p, q) > 0, and say that q is in the chronological future of p, and dually in the past case. By the reverse triangle inequality p q and q r implies p r.
1. It follows from the finiteness of d and the reverse triangle inequality in (i) that d(x, x) = 0 for all x ∈ X. This property reads p p, and we refer to it as chronology.
2. There is always a topology on X for which d is continuous, namely the discrete topology. Thus the coarsest topology in point (ii) exists. 3. By (iii) there is just one point, if it exists, denoted i 0 and called spacelike boundary, such that d(i 0 , p) = d(p, i 0 ) = 0 for every p ∈ X. As a consequence, we can always remove one point from S so that i 0 / ∈ S. Any point q different from i 0 has another point z in its chronological past or future. By (iv) the point z can be found in S. 4. By 1. if |X| = 1 then the only point in X is i 0 . If |X| > 1 then X contains at least two points p, q with p q, (X has at least three points if there is i 0 ∈ X) thus the set {d ≥ } ⊂ X × X, is not only compact, as by assumption (ii), but also non-empty, thus d reaches a positive maximum on X × X. In any case we have a bound on d which justifies the adjective bounded in the definition. 5. We let the set consisting of only one point be included into the definition. The purpose is to have a well defined Gromov-Hausdorff limit for "collapsing" spaces.  It is understood that the boundary of the gray region belongs to X, save for the point i 0 that might or might not belong to X.

Topology
Observe that for every p the map q → (p, q) is continuous from (X, T ) to (X × X, T × T ). Therefore, the map In what follows we denoteX = X\{i 0 }, and withT the topology induced onX. Of course, if i 0 / ∈ X thenX = X andT = T .
is the projection of the intersection of a closed set and a compact set, hence compact, and similarly for {d r ≥ d}). The open neighborhood system of i 0 in T contains at least the family F of finite intersections of sets of the above form.
If T had other neighborhoods of i 0 that do not contain elements of F then it is possible to reduce the neighborhoods of i 0 to those of the above system while preserving continuity of d. This is obvious for any pair (p, q) with p, q = i 0 , since we are not altering the neighborhood system of points different from i 0 .
Let > 0. The compact set B = π 1 ({d ≥ }) consists of points p that admit some r such that d(p, r) ≥ , thus p is distinguished from i 0 and there is somẽ r ∈ S and δ ∈ Q, δ > 0, such that p ∈ {dr > δ} or p ∈ {dr > δ}. This means that B can be covered with sets of this form. Pass to a finite subcovering Similarly, we can find V ∈ F such that for every x ∈ V , d(·, x) < . This shows that the neighborhood system F of i 0 is sufficient to guarantee continuity of d on π −1 1 (i 0 )∪π −1 2 (i 0 ) and hence that T , being the coarser topology for which d is continuous, induces this neighborhood system of i 0 .
Let us prove that (X, T ) is compact.
Corollary 1.7. T andT are the initial topologies of the family of functions {d p , d p : p ∈ S}. More precisely, the sets of the form {q : a < d(p, q) < b} ∩ {q : c < d(q, r) < d} (2) with p, r ∈ S and a, b, c, d, ∈ Q ∪ {−∞, ∞} form a subbasis for the topologies. Thus both T andT are second countable and hence separable.
Proof. Let us first give the proof forT . Let A be the family of the form (2). We already know that they separate points (Remark 1.4). As already observed if o is a point that admits a point p ∈ S in its chronological past (the other case is analogous), {q : d(p, q) ≥ } = π 2 ({d ≥ } ∩ π −1 1 (p)), is a compact neighborhood of it for some rational > 0 as it contains {q : d(p, q) > } which is open. But the latter sets belongs to A, thus A satisfies both (a) and (b) in Prop. 1.6. Then the subbasis is countable and so the basis given by the finite intersections is also countable (a countable set has countably many finite subsets).
For T the proof is identical but easier since local compactness follows from compactness.
Proposition 1.8. The topologies T andT are normal, paracompact and metrizable.
Proof. A topology which is Hausdorff and locally compact is regular. And every regular σ-compact space is normal. Thus T andT are normal.
The topologies are also paracompact because every locally compact σ-compact topological space is paracompact (notice that Hausdorff paracompact spaces are normal, this gives another proof of normality).
The next result clarifies that (v), though it does not mention the topology, can be really reinterpreted as a separability property for T . Proposition 1.9. In the definition of bounded Lorentzian-metric space we can replace (iv) for (iv') T is separable.
Moreover, in a bounded Lorentzian-metric space any countable dense subset S satisfies the property of Definition 1.1(iv).
Proof. We already know that under (i)-(iv) T is second countable hence separable.
Suppose that (i), (ii), (iii) and (iv') hold. We have previously shown, without making use of (iv), that the sets in (1) are open and that the functions d r , d r , r ∈ X, are continuous.
By assumption there is a countable set S such that every non-empty open set of T contains some element of S.
In (iii) suppose that the first case applies (the other case being analogous), d(x, z) = d(y, z), with z ∈ X, then we can find some a ∈ R such that d(x, z) < a < d(y, z) (or similarly with > in place of <), which implies that z ∈ d −1 x ((−∞, a)) ∩ d −1 y ((a, +∞)) where the right-hand side is open. Thus we can find s ∈ S, such that s ∈ d −1 r ((−∞, a)) ∩ d y ((a, +∞)), which implies d(x, s) < a < d(y, s) and hence d(x, s) = d(y, s). Remark 1.10 (One point compactification). By the previous results if X does not include the spacelike boundary i 0 , we can always enlarge X so as to include it. That means adding an abstract element i 0 whose distance to all the other elements in X is defined to be zero and whose neighborhood system is defined as in Prop. 1.5.
The fact that d remains continuous after the extension is proved with the argument contained in the proof of Prop. 1.5, as the fact that the new topology still satisfies (ii). This means that the so defined one-point compactification of a bounded Lorentzian-metric space not including i 0 returns a bounded Lorentzianmetric space including i 0 , hence compact. Remark 1.11. Using the preceding results we can give a characterization of bounded Lorentzian-metric space which can also be used as a definition.
A space (X, d) consisting of a set X, a map d : X × X → [0, ∞), and for which there is a point i 0 at zero distance from any other point, is a bounded Lorentzian-metric space if and only if (i) For every x, y, z ∈ X such that d(x, y) > 0, d(y, z) > 0 we have (ii') The coarsest topology T on X for which d is continuous is separable and (X, T ) is compact.
(iii) d distinguishes points, i.e. for every pair x, y, x = y there is z such that Proof of the equivalence. We already proved that (i)-(iv) implies (i), (ii') and (iii). The converse is also clear from Prop. 1.9 and from the compactness of X ×X in T ×T which implies that of its closed subsets of the form {d ≥ }.
for all x, x ∈ X. A bijective and distance preserving map is an isometry.
We have Theorem 1.13. Let X, Y be bounded Lorentzian-metric spaces. If f : X → Y is an isometry then it is a homeomorphism.
Proof. Indeed, a subbasis of T Y is provided by the sets in (1), whether i 0 Y ∈ Y or not, which implies that their preimage have the same form with d Y replaced by d X and so are elements of a subbasis for T X . This proves that f is continuous (using the surjectivity of f ). The proof that f −1 is continuous is analogous (using the surjectivity of f −1 ). Corollary 1.14. Let f : X → Y be an isometry. If X and Y do not contain the spacelike boundary then their one-point compactifications are isometric and hence homeomorphic. Just extend f as follows f (i 0 X ) := i 0 Y . If X and Y do contain the spacelike boundary then we get an isometry and hence an homeomorphism ofX andY , through the restriction f |X .
These results clarify that it is not restrictive to work with bounded Lorentzianmetric spaces that contain i 0 , or with those that do not contain i 0 .

Distance quotient
Let (X, T ) be a topological space endowed with a continuous function d : We define an equivalence relation "∼" on X by p ∼ q if d(r, p) = d(r, q) and d(p, r) = d(q, r) for every r ∈ X. It is easy to check that this is indeed an equivalence relation. LetX = X/ ∼ be the quotient, letT = T / ∼ be the quotient topology, and let π : X →X be the quotient projection. The function d passes to the quotient. Indeed if p, p are two representatives of a class [p] and q, q are two representative of a class [q] we have d(p, q) = d(p , q) = d(p , q ), so that we can define a functiond([p], [q]) := d(p, q). It is easy to check thatd is continuous.
Thus we have obtained a quotient topological space (X,T ) endowed with a continuous functiond :X ×X → [0, ∞).
Let us observe how the properties for d and T on X are related to the analogous properties onX.
• d satisfies the reverse triangle inequality, as in (i), iffd does.
•d distinguishes points (i.e. satisfies (iii)) even if d does not.
• There is a point of X (possibly non unique) at vanishing distance from all the other points iffX contains the spacelike boundary point.
• If X satisfies (iv'), that is the topology is separable, the same holds forX.
The next result is not obvious and requires a proof Proposition 1.15. If X satisfies (ii) thenX satisfies (ii), where the coarsest topology for whichd is continuous is the quotient topology.
Proof. Clearly every set of the form d −1 ((a, b)) ⊂ X×X with a, b in the extended real line is projectable toX×X. Thus, d is continuous for some product topology T × T iffd is continuous for the product of the quotient topologyT ×T (noncalligraphic font intended). Thusd is continuous forT ×T . But there cannot be any coarser topologyT T onX such thatd is continuous inT ×T . Otherwise consider the coarser topology T of X that projects on it (namely the open sets of T are the preimages of the open sets ofT under the quotient map). We conclude that d is continuous with respect to T × T . Since T is coarser than T , we arrive at a contradiction. We conclude that the quotient topology is the coarsest topology for whichd is continuous. Since the quotient projection is continuous, the setd ≥ is compact, as it is the π × π-projection of a compact set. So point (ii) is satisfied for (X,d) with the quotient topology.  Proof. (iv) is obvious with S := S, so we need only to prove (ii). It is sufficient to prove that T is the discrete topology . Also the set of pairs d ≥ will be finite by the cardinality of S × S hence compact. Let p ∈ S, we want to show that {p} is open for T . We know from Section 1 that the functions d r , d r are Tcontinuous. Let r ∈ X and let D r = {d(r, s), s ∈ S} and D r = {d(s, r), s ∈ V }. For each q, since D q is finite, the number d(p, q) is isolated from the other different numbers in D q . Thus we can find a q , b q ∈ R such that (a q , b q ) contains only d(p, q) and no other element of D q . Similarly, we can find c q , d q ∈ R such that (c q , d q ) contains only d(q, p) among the numbers in D q . Now the open set contains p by construction, but it cannot contain any other point p = p, for otherwise d would not distinguish p from p , contradicting point (iii). Thus {p} is an open set.
We recall that a directed graph is an ordered pair G = (S, A) where S is a set whose elements of S are called vertices and A is a set of ordered pairs of vertices, so A ⊂ S × S, each pair being called an arrow. Given an arrow (p, q), we call p the starting point and q the ending point.
Causets can be regarded as weighted directed graphs, where we say that (p, q) ∈ A if d(p, q) > 0. The distance d(p, q) is then the weight of the arrow (p, q). The weight cannot be arbitrary as they have to satisfy the reverse triangle inequality.
We know that causets, being bounded Lorentzian-metric space, satisfy chronology which means that the directed graph is acyclic. Observe that every point of the directed graph is connected to some other point by an arrow, save for one point i 0 , if present, that would be isolated. Proof. By Prop. 2.2 we know that every causet is a finite bounded Lorentzianmetric space. Conversely, a finite bounded Lorentzian-metric space satisfies (i) and (iii), so it is a causet.
It is convenient to define causets via Definition 2.1 as it does not mention a topology.
Proposition 2.4. Let S be a finite subset of a bounded Lorentzian-metric space endowed with the induced topology and distance. Then S/ ∼ is a causet.
Proof. It is clear that S inherits property (i) and that it satisfies (iii) as it is a distance quotient.

Subsets of globally hyperbolic spacetimes
LetX be a compact, causally convex subset of a globally hyperbolic spacetime (M, g). Denote with ∂ 0X := {q ∈X : ∃p ∈X, p q and ∃r ∈X, q r}.
Let d X := d| X×X where d is the Lorentzian distance of M , and similarly let dX := d|X ×X . By causal convexity dX is the supremum of the Lorentzian lengths of timelike connecting curves inX, but none of these curves can pass through ∂ 0X , thus d X = dX | X×X = d| X×X which is then the supremum of the Lorentzian lengths of the causal connecting curves in X.
is a bounded Lorentzian-metric space that does not contain the spacelike boundary point i 0 , and T is the topology induced on X from the manifold topology.
A posteriori we can identify the abstract point i 0 with ∂ 0X .
Proof. Let us verify the properties of bounded Lorentzian-metric space.
(i) This is obvious by taking the restriction of analogous equations in M .
(ii) We want to show that T is the induced topology on X. Since d is continuous we know that d X is continuous in the induced topology, so by definition of T , T is coarser that the induced topology. There will be p ∈ X, d(p, q) > 0, (or the other way). Any timelike curve from p to q is contained in X (by causal convexity ofX), so we can assume, without loss of generality that p ∈ O and hence p ∈ O (otherwise follow a timelike curve from p to q and choose a point sufficiently close to q). Actually, in the same way we can find a second point p ∈ O sufficiently close to q, p p q and constants a, b > 0 such that ) ⊂ X is contained in O and contains q, thus T is finer than the induced topology (remember that for every r, (d X ) r and (d X ) r are continuous in T ).
We know that {(p, q) ∈X ×X : d(p, q) ≥ } is compact in the product of the induced topologies because it is the intersection of a closed set with the compact setX ×X. But the previous set coincides with {(p, q) ∈ X × X : d(p, q) ≥ }, which therefore it is compact.
(iii) Let x, y ∈ X with x = y. We can find z x, z ∈ X (or viceversa). By the past distinction property of M there is p ∈ M such that p x but p y. We know that I + (z) ∩ I − (x) ⊂ X. Since x ∈ I + (p) and the last set is open in M , there will be some point p ∈ I + (p) ∩ [I + (z) ∩ I − (x)] ⊂ X. Now, p y, otherwise p y, a contradiction. Thus d(p , x) > 0 but d(p , y) = 0.
(iv') A topology induced from a second countable space is second countable and hence separable.

Distance preserving maps
In this section we assume that the bounded Lorentzian-metric spaces do not include the spacelike boundary. Generalizations to the other cases are straightforward. We recall that a bijection that preserves the distance is an isometry.
be bounded Lorentzian-metric spaces, and let f : X → Y be distance preserving, then f is injective.
Proof. Consider two distinct points x, x ∈ X. Choose z ∈ X such that d X (x, z) = d X (x , z) (the other case being analogous). Then Lemma 3.2. Let (X, d) be a bounded Lorentzian-metric space and let f : X → X be distance preserving, then f is T -continuous.
The proof uses the existence of the countable subset S. We do not assume that X is compact.
Proof. Since T is first-countable it is sufficient to prove the sequential continuity of f . Let x ∈ X and let {x n } be a sequence with x n → x. First note that the sequence {f (x n )} n is contained in a compact subset of X. Indeed there exist z 0 ∈ X and a > 0, with d(z 0 , x) > a > 0 (the case with d(x, z 0 ) > a > 0 is treated similarly) so for any sufficiently large n, Suppose that the claim is false, namely that the sequence of images {f (x n )} does not converge to f (x), then we can assume, by passing to a subsequence, that . We consider just the former possibility, the treatment of the latter case being completely analogous. Choose It is convenient to observe that for each q ∈ X we can find p ∈ X such that d(p, q) > δ > 0, (or r ∈ X such that d(q, r) > δ > 0 in which case we argue analogously), which implies that for any increasing sequence k m , we have d(f km (p), f km (q)) > δ > 0. By the compactness of {d ≥ δ} we can conclude that f km (q) admits a converging subsequence. Now, for each such n as in the last but one paragraph choose a strictly increasing sequence {m n k } k ⊂ N such that the sequences converge to points a, b, c n respectively. Then by the continuity of d Again by the continuity of d these inequalities must hold in suitable neighborhoods of a, b, c n and hence for sufficiently large k, k we have Choose for every n an exponent m(n) := m n k − m n are chosen sufficiently large so as to satisfy If there are infinitely many former cases, the compactness of {d ≥ ε} tells us that there is a subsequence of {(f m(n) (z), x)} n that converges and hence a subsequence of {f m(n) (z)} n that converges. Actually, the same conclusion can be drawn if there are infinitely many latter cases, for the compactness of {d ≥ ε} tells us that there is a subsequence of {(f m(n) (z), x n )} n that converges and hence a subsequence of {f m(n) (z)} n that converges. In either case, denoting with z ∞ the point to which a subsequence of {f m(n) (z)} n converges we get, taking the limit of the previous equation in display The contradiction proves that Theorem 3.3. Let (X, d) be a bounded Lorentzian-metric space and let f : X → X be distance preserving. Then f is surjective.
Proof. Suppose not. We claim that under the assumption there exists an nonempty open set O ⊂ X disjoint from f (X). Let p ∈ X\f (X). If there exists a neighborhood of p disjoint from f (X) there is nothing to prove. Otherwise every neighborhood U p intersects f (X), hence we can find r n ∈ X, such that f (r n ) → p. If there is a converging subsequence r n k → r ∈ X, then as f is continuous (by Lemma 3.2) f (r n ) → f (r) which implies p = f (r), hence p ∈ f (X), a contradiction. Thus no subsequence of r n converges. By (iii) I + (p) or I − (p) is non-empty. We assume the former possibility, the latter case being analogous. The nonempty set I + (p) does not intersect f (X), for if it had some point f (q) ∈ I + (p), d(p, f (q)) > > 0, q ∈ X, then d(f (r n ), f (q)) > for sufficiently large n, which would imply d(r n , q) > and hence by the compactness of {d ≥ } there would be some converging subsequence of r n , a contradiction. As I + (p) is a non-empty open set not intersecting f (X) we conclude that claim is true.
and hence There is an increasing sequence k n such that f kn (r i ) →r i , f kn (p) →p, f kn (s i ) →s i , for every i (the proof was given in the 4th paragraph of the proof of Lemma 3.2). By preservation of distance the sequence where in the last step we used the continuity of d. Thus a i < d(r i ,p) < b i . This implies that for n big enough a i < d(f kn (r i ),p) < b i (and similarly for the other inequalities involving c i and d i ), which implies that for m > n big enough Thus considering all inequalities we have that for m > n big enough taking into account that k s is increasing, k m −k n ≥ 1, which contradicts (4).
be bounded Lorentzian-metric spaces, and let f : X → Y be distance preserving and surjective, then f is continuous (with respect to the topologies T X and T Y ).
Proof. Since the topologies are first countable it suffices to prove sequential continuity of f . Suppose that f is not continuous at x. We can find a sequence x n → x such that {f (x n )} n does not converge to f (x). We know that there exists a point in the chronological past or in the chronological future of x. Let us assume the former possibility, the latter being analogous, and let x x. For sufficiently large n we have x x n and This proves that f (x n ) belongs to a compact set. Without loss of generality we can assume that f (

or the other analogous equation). Thus we can find
By surjectivity there is w ∈ X such that z = f (w). Thus for sufficiently large n we obtain |d X (x n , w) − d X (x, w)| > , which contradicts the continuity of d X (·, w).
Two metric spaces that are in bijection under a distance preserving map are said to be isometric, and so they are also homeomorphic as the topology is induced from the metric. In the Lorentzian signature the homeomorphic nature of the bijection does not follow from the distance preserving property in an obvious way because topology and metric are less tightly linked than in positive signature.
be bounded Lorentzian-metric spaces, and let f : X → Y , g : Y → X be distance preserving. Then f and g are continuous isometries (with continuous inverses).
Proof. By Theorem 3.1 f and g are injective. By Theorem 3.3 g • f : X → X is surjective. Thus by the injectivity of g, f is surjective. Similarly g is surjective. By Theorem 3.4 they are continuous bijections. The inverses are also surjective and distance preserving hence continuous.
Remark 3.6. The statement of the previous theorem does not change if one (or both) X, Y , contain the spacelike boundary (as a consequence, one contains the spacelike boundary if so does the other). Indeed, the theorem applies toX andY and the restrictions f |X , g|X (note that by distance preservation they cannot have the spacelike boundary in their image) so they are indeed maps f |X :X →Y and g|Y :Y →X hence bijective by the previous theorem. But is in the chronological future or past of some point necessarily inY and hence of the form f (q). By distance preservation we would get that i 0 X is in the chronological future or past of q, a contradiction. Thus necessarily f (i 0 X ) = i 0 Y since all the other points inY are already in the image of f |X . From here it is immediate that f is an isometry and hence a homeomorphism (Theorem 1.13).

Lorentzian Gromov-Hausdorff distances 4.1 Abstract Gromov-Hausdorff semi-distance
A pair (X, d X ) consisting of a nonempty set X and a bounded function Proposition 4.5. Let R ⊂ X × Y be a correspondence between two bounded Lorentzian metric spaces. ThenR is a closed correspondence with the same distortion.
Proof. For every x ∈ X we can find y ∈ Y , such that (x, y) ∈ R ⊂R, and analogously, given y ∈ Y we can find x ∈ X such that (x, y) ∈ R ⊂R.
Let (x, y), (x , y ) ∈R, then we can find x n → x, y n → y, (x n , y n ) ∈ R, and x n → x , y n → y , (x n , y n ) ∈ R. Thus for every n, thus taking the limit and using the continuity of d X and d Y , which implies that disR ≤ disR, the other direction being obvious.
where the infimum is taken over all correspondences R ⊂ X × Y .
Notice that for every bounded space X, d GH (X, X) = 0, as the correspondence given by the diagonal ∆ ⊂ X × X has vanishing distortion.
Proof. By definition we have Note that for every y, y ∈ Y and in particular for y, y such that (x, y), Proposition 4.9. The Gromov-Hausdorff semi-distance satisfies the triangle inequality. Namely, for any (X, Proof. The family of correspondences R X,Z between X and Z contains the family of those correspondences that are obtained through composition of correspondences R X,Y and R Y,Z . With Lemma 4.8 follows In summary we have Proposition 4.10. The Gromov-Hausdorff semi-distance between bounded spaces has the following properties: Proof. Point (a) is obvious from the definition. Point (b) follows from Remark 4.4. Point (c) is Proposition 4.9.
Remark 4.11. The addition of the spacelike boundary to two bounded Lorentzianmetric spaces not including it does not alter their Gromov-Hausdorff semidistance. The reason is that any correspondence can be enlarged including the element (i 0 X , i 0 Y ) without altering its distortion. Similarly, if X, Y include the spacelike boundary then we can remove it from both without altering their Gromov-Hausdorff semi-distance (because any correspondence can see its distortion reduced letting i 0 X correspond just to i 0 Y and conversely, and so both can be really removed without altering the distortion). This does not work if the spacelike boundary is removed on just one of the two spaces.
The next result clarifies the behavior of the Gromov-Hausdorff semi-distance with reference to the distance quotient (see Section 1.3).
Proposition 4.12. Let (X, T ) be a topological space endowed with a continuous function d : X × X → [0, ∞), hence a bounded space. Let S be a subset of X, endowed with the induced distance. Then d GH (S, S/ ∼) = 0. Moreover, there is a subsetŠ ⊂ S, isometric to S/ ∼, such that the distinguishing property of Definition 1.1(iii) holds forŠ, and d GH (S,Š) = 0.
By Proposition 2.4 if X is a bounded Lorentzian-metric space and S is a finite subset then, by first taking the distance quotient and then the representatives, we can findŠ ⊂ S, d GH (Š, S) = 0, which is a causet.
by the definition of quotient distanced. Thus dis R = 0 and hence Let us defineŠ by picking an element in each class [s] ∈ S/ ∼. This establishes a bijection between the two sets (hence a natural correspondence). By the same calculation of the previous paragraph,Š and S/ ∼ are isometric, thus d GH (S/ ∼,Š) = 0. The desired equality d GH (S,Š) = 0 follows from the triangle inequality for d GH . Further, we know that S/ ∼ satisfies the distinguishing property of Definition 1.1(iii) so, by the isometry, the same holds forŠ.

Lorentzian Gromov-Hausdorff distance
The following result will be proved through several propositions in what follows.
Theorem 4.13. The Gromov-Hausdorff semi-distance between bounded Lorentzianmetric spaces that contain i 0 has the following properties: (a) The Gromov-Hausdorff semi-distance is non-negative. Further, the spaces (X, d X ) and (Y, d Y ) are isometric and homeomorphic, that is, there exists an isometry (which then is also a homeomorphism) f : The same result holds with "contain" replaced by "do not contain".
The next result is then an immediate consequence of the previous one [5, Corollary 4.14. The Gromov-Hausdorff semi-distance d GH descends to the set of isometry classes of bounded Lorentzian-metric spaces that contain i 0 (equivalently, that do not contain i 0 ). The resulting space is a metric space.
be bounded Lorentzian-metric spaces that either both contain the spacelike boundary or that do not. If d GH (X, Y ) = 0 then the spaces (X, d X ) and (Y, d Y ) are isometric (and the isometry is a homeomorphism).
The fact that the isometry is a homeomorphism follows from Theorem 3.5 and Remark 3.6. Observe that removing i 0 from a bounded Lorentzian-metric space gives a bounded Lorentzian-metric space which, however, is not in general at zero Gromov-Hausdorff distance from the original space.
Proof. We give the proof for X, Y not containing the spacelike boundary. The other case follows from Remark 4.11. By Proposition 4.17 we know that there are maps f n : X → Y which are n -isometries on the image for any sequence n → 0. We observe that for every x ∈ X we can find a subsequence {f ns (x)} s such that f ns (x) converges in Y . Indeed, we can find z x (or the other inequality), hence for n sufficiently large such that n < d(z, x)/2 we have which proves that the sequence f n (x) is contained in a compact set for n large enough.
Let S = {x k } be a countable dense subset of X. For every k we can find a subsequence {f n k m (x k )} m that converges to some point of Y which we denote f (x k ). Notice that the sequence {n k+1 m } m can be chosen to be a subsequence of {n k m } m . Then by the Cantor diagonal trick, {f n m m (x)} m is a subsequence that converges to f (x) for every x ∈ S. Thus without loss of generality we can assume that f n | S converges to some function f : S → Y . Since f n is an thus f preserves d. Now we extend f to X as follows. If x ∈ X choose a sequence x n → x, x n ∈ S, such that lim n f (x n ) exists (this is possible by the usual compactness argument. Namely take z x, or the other inequality, and use that d(f (z), f (x n )) = d(z, x n ) > d(x, z)/2 > 0 for large n) and set f (x) := lim n f (x n ). Let x, x ∈ X, and let x n , x n the just mentioned sequences defining f (x) and f (x ). We have where in the last step we used the fact that f preserves d on S. We conclude that f preserves d on X. We can construct an analogous distance preserving map g : Y → X. By Theorem 3.5 f and g are isometries.
We are ready to prove Theorem 4.13 Proof. The non-trivial direction of (a) is

The distinction metric
Let (X, d) be a bounded Lorentzian-metric space. We assume that i 0 ∈ X, i.e. (X, T ) is compact by Proposition 1.5.
We define the following distinction metric on X which measures how much a point x ∈ X can be distinguished from a point y ∈ X by using the function d. The property γ(x, y) = 0 ⇒ x = y follows from Definition 1.1(iii). The triangle inequality is straightforward. so it is a metric is the classical sense. Note that for each z ∈ X the functions d z := d(z, ·) and d z = d(·, z), are 1-Lipschitz continuous with respect to γ. The Noldus metric is defined as It was originally introduced in [17, Definition 3] for smooth manifolds and called strong metric. It was also very much related to an earlier metric construction by Meyer [9]. For the next result see also [ Proof. We consider the case x = y otherwise the identity D(x, y) = 0 = γ(x, y) is clear. Since d is continuous and X is compact, there is some w ∈ X such that Let us assume the former case, the latter being analogous (one is obtained from the other exchanging x ↔ y). Only one among d(w, x) and d(x, w) can be positive, and similarly only one among d(w, y) and d(y, w) can be positive. If d(w, x) > 0 and d(y, w) > 0 we have y w x and therefore contradicting the choice of w. Similar the case d(x, w) > 0 and d(w, y) > 0 leads to a contradiction. As a consequence we have The same conclusion is reached in the latter case (observe that the last expression is invariant under exchanges x ↔ y), and hence For the converse, since d is continuous and X is compact, there is some w ∈ X such that (a) γ(x, y) = d(x, w)−d(y, w), or (b) γ(x, y) = d(y, w)−d(x, w), or (c) γ(x, y) = d(w, x) − d(w, y), or (d) γ(x, y) = d(w, y) − d(w, x). It is sufficient to consider case (a), the other cases following by time duality or by the symmetry x ↔ y.
We know that γ(x, y) > 0 thus d(x, w) > 0 and hence d(w, x) = 0 by chronology. If d(y, w) = 0 we must have d(w, y) = 0 for if d(w, y) > 0 then x w y, and then the choice z = y would contradict the fact that w gives the maximal value. If d(y, w) > 0 then again d(w, y) = 0 by chronology.
This shows that in case (a) Since the penultimate expression is invariant by time duality (d(u, v) → d(v, u)) and by exchanges x ↔ y, we conclude that it also holds in cases (b), (c) and (d).
Proposition 4.20. The Lorentzian distance d is 1-Lipschitz with respect to γ. Proof.
The topology induced by γ is called γ-topology. The next result is stated in [17,Thm. 8 (c)] without proof.
Proposition 4.21. Let (X, d) be a bounded Lorentzian-metric space. The γtopology coincides with the topology T of (X, d). Thus (X, γ) is a compact metric space.
Proof. By Prop. 4.20 d is continuous in the γ-topology and so are the functions d p , d p for p ∈ X. By Prop. 1.7 the γ-topology is finer than T .
For the other direction, suppose by contradiction that there is a γ-ball {z : γ(x, z) < r}, x ∈ X, r > 0, such that no T -open neighborhood of x is contained in it. As T is first-countable we can find x n → x such that γ(x, x n ) ≥ . By the T -compactness of X there is w n such that γ(x, x n ) = |d(w n , x) − d(w n , x n )| or γ(x, x n ) = |d(x, w n ) − d(x n , w n )|. We can pass to a subsequence so that only one of the equalities holds, say the former, and w n → w, for some w ∈ X. By taking the limit, we have as 0 < ≤ |d(w, x) − d(w, x)| = 0, a contradiction.
Proposition 4.23. Let (X, d) be a bounded Lorentzian metric space. The diameter diam γ X for the distinction metric coincides with the diameter diamX for the Lorentzian metric.
Elsewhere the Lorentzian diameter of a bounded Lorentzian metric space X is also denoted diamX.
Proof. We can assume that i 0 ∈ X, as adding it does not change both the Lorentzian and the distiction-distance diameter. As X is compact and d is continuous there are x, y ∈ X such that diamX = d(x, y). Now, by the expression for γ(x, y) we have for every z ∈ X γ(x, y) ≥ max{|d(x, z) − d(y, z)|, |d(z, x) − d(z, x)|}, thus choosing z = y we get γ(x, y) ≥ d(x, y) − d(y, y) = d(x, y) = diamX which proves diam γ X ≥ diamX.
For the other inequality, observe that for every x, y, z ∈ X, |d(x, z) − d(y, z)| ≤ diamX and |d(z, The associated metric is

Kuratowski-type embeddings and -nets
The product B×B can be regarded as a Banach space with norm (b 1 , b 2 ) ∞ = max( b 1 ∞ , b 2 ∞ ). Of course, it is also a metric space once given the associated metric The completeness of this metric space, which is inherited from that of the Banach space B, will be particularly important for us. Let is an embedding, i.e. I is a homeomorphism of X with its (compact, hence closed) image I(X).
In the following we will call I the Kuratowski embedding of (X, d) with respect to S . Observe that γ does not depend on S .
Proof. We know that X is compact while (B × B, dist ∞ ), being a metric space, is Hausdorff. It is well known that a continuous injective map from a compact space X to a Hausdorff space is actually a homeomorphism of X with its image [20,Theorem 17.14] (the image is necessarily closed as the continuous image of a compact set is compact, hence closed). Thus it is sufficient to prove that I is injective and continuous.
If x, y ∈ X are mapped to the same element of B × B then for every s ∈ S, d(x, s) = d(y, s) and d(s, x) = d(s, y) which implies x = y by the defining property (iii)-(iv) of bounded Lorentzian-metric space. Thus I is injective.
Let us prove continuity of I. Let y n ∈ X converge to y ∈ X. We will show that d yn converges uniformly to d y . The other assertion that d yn converges uniformly to d y follows analogously. The claim then follows readily from the construction of I.
Assume to the contrary that d yn does not converge uniformly to d y . Then, passing to a subsequence if necessary, we can assume that there exists > 0 such that for every n ∈ N there is r n ∈ X with |d yn (r n ) − d y (r n )| ≥ ε. We thus have |d(y n , r n ) − d(y, r n )| ≥ ε for all n. Since (X, T ) is compact we can choose a converging subsequence {r n k } k converging to r ∈ X. Then d(y n k , r n k ) and d(y, r n k ) both converge to d(y, r), which contradicts (7). We know that I is a bijection between X and I(X). Moreover, we have from the definitions that Since S is dense and d is continuous we have that the right-hand side coincides with γ(x, y), namely I is an isometry.   for the metric space (X, γ).
2. If N ⊆ X is an ε-net and Y ⊆ X is a bounded Lorentzian metric space such that N ⊆ Y , then N is an ε-net of Y . This follows from Remark 4.22.
3. For every ε > 0 every bounded Lorentzian-metric space contains a finite ε-net. This follows directly from the compactness of X. Just cover X with γ-open balls of radius and pass to a finite covering {B i }. The set N consisting of the centers of the balls provides a finite ε-net.

4.
For an ε-net N ⊆ X we have d GH (N , X) ≤ 2ε. This can be seen as follows: For every x ∈ X choose η x ∈ N with γ(x, η x ) ≤ with the additional convention η ν = ν for ν ∈ N . Define the correspondence R ⊂ X × N as For all x, y ∈ X we have It follows that the distortion of R is bounded by 2 .
5. Every ε-net contains a finite 2ε-net. This follows from the compactness of (X, T ).
6. Every finite ε-net N ⊆ X contains a 3ε-netŇ which is additionally a causet and such that d GH (N ,Ň ) = 0. We are going to constructŇ as in Proposition 4.12, via the quotient distance and then by choice of representatives, thus the fact thatŇ is a causet and d GH (N ,Ň ) = 0 follows from that proposition.
A consequence of the previous observations is that every bounded Lorentzianmetric space contains a causet N which is an -net and such that d GH (N , X) ≤ .  Proof. The last statement follows from Remark 4.27, 4 & 6. Just apply the first statement to a sequence of /6-nets.
We start by assuming that X m GH −−→ X. Let > 0 be given. Consider m sufficiently large such that d GH (X m , X) < /3. Choose for every such m a correspondence R m ⊂ X m × X with δ m := disR m < /3 and δ m → 0 for m → ∞.
Next choose an /3-net N ⊂ X. For m ∈ N and η ∈ N selectř m (η) ∈ X m with (r m (η), η) ∈ R m and define We claim that N m is an -net in X m . Let x m , y m ∈ X m . Choose x, y ∈ X with (x m , x), (y m , y) ∈ R m . We can find η ∈ N such that γ(η, x) ≤ 3 , thus we have By the arbitrariness of y m , γ m (r m (η), x) ≤ , that is, N m is an -net.
Further the restriction of R m to N m × N defines a correspondence with distortion bounded by δ m ≤ /3, hence d GH (N m , N ) → 0.
For the converse implication let us assume that for every δ > 0 we have finite This proves that X m GH −−→ X.
Finally, we relate the convergence of bounded Lorentzian metric spaces (X, d) to the Gromov-Hausdorff convergence of the associated metric (X, γ). The following is another instructive proof of one direction that does not use the notion of -net. Proof. By assumption we can find a correspondence R n ⊂ X × X n such that n := dis d R n → 0. We want to show that δ n := dis γ R n → 0. Let x, y ∈ X, and let x n , y n ∈ X n , be any pair such that (x, x n ) ∈ R n , (y, y n ) ∈ R n . We want to prove that |γ(x, y) − γ n (x n , y n )| ≤ 3 n , from which dis γ R n ≤ 3 n follows and hence the thesis.
Let w n ∈ X n be such that (w, w n ) ∈ R n , then |d n (x n , w n ) − d n (y n , w n )| ≥ γ(x, y) − 3 which implies Observe that from the definition of γ n we get that there is z n ∈ X n such that |d n (x n , z n )−d n (y n , z n )| ≥ γ n (x n , y n )− (or the other analogous inequality with z n in the first entry) so we can find z ∈ X, (z, z n ) ∈ R n so that hence |γ(x, y) − γ n (x n , y n )| ≤ 3 as desired.
Theorem 4.32. Let X, Y be bounded Lorentzian metric spaces. We have the inequality where the infimum is over all possible isometric injections of X, Y in a bounded Lorentzian metric space Z. Here d γ H is the Hausdorff distance for (Z, γ), with γ the distinction metric on Z.
Proof. Suppose we one such injection. Let r > d γ H (X, Y ), we need to find a correspondence R ⊂ X × Y , with disR ≤ 2r, from which the desired inequality would follow. Let R be given by all those pairs (x, y) for which γ(x, y) < r. It is easy to check that this is a correspondence by the definition of the Hausdorff distance.
By Prop. 4.20 if (x 1 , y 1 ) ∈ R and (x 2 , y 2 ) ∈ R, as all these points can be seen as belonging to Z |d(x 1 , x 2 ) − d(y 1 , y 2 )| ≤ γ(x 1 , y 1 ) + γ(x 2 , y 2 )| < 2r, which implies that disR ≤ 2r. Now, let Z = X Y be the disjoint union (but identify the i 0 s if both spaces have), and define d so that d| X×X = d X , d| Y ×Y = d Y , and d| X×Y = d| Y ×X = 0. Then Z is a bounded Lorentzian-metric space. Observe that by Prop. 4.23 This result is important as it allows us to construct several examples of Lorentzian Gromov-Hausdorff convergence. Consider for instance a causal diamond in Minkowski spacetime, so as to get a bounded Lorentzian metric space. Two timelike hypersurfaces on it converging to each other would converge in a Gromov-Hausdorff sense, as the Hausdorff distance for the distinction metric goes to zero (a precise determination of the distinction distance is not required for this conclusion).  (8) is called the (extended) causal relation. We write x ≤ y for (x, y) ∈ J and x < y for x ≤ y and x = y. We also write J + (x) := {y : x ≤ y}, J − (x) := {y : y ≤ x}.
Remark 5.2. This causal relation will be very useful though it is intuitively a bit too large, particularly at the boundary points. Let X + ⊂ X be the subset of points that do not admit a point in the chornological future, and analogously for X − , so that i 0 = X + ∩ X − provided i 0 ∈ X. Observe that if (x, y) ∈ X − × X + then (x, y) ∈ J. Therefore, as examples obtained cutting out subsets of Minkowski spacetime show, the relation J is in general strictly larger than the closure of I, and it is also different from the smallest closed, reflexive and transitive relation containing I.
On the other hand if y / ∈ X + and assuming every neighborhood of y intersects I + (y) every pair (x, y) ∈ J lies in the closure of I. The reason for this is the reverse triangle inequality proven in Theorem 5.7 below. The time reversed case holds as well.
One should not expect the just mentioned inclusion to hold without any assumption. Indeed, this is a "no bubbling" property which, already for proper cone structures makes use of the 'proper' condition [11,Theorem 2.8]. The assumption on the non-isolateness of y in terms of chronology is a kind of metric replacement for the proper condition [11,Theorem 2.2]. Observe that proper cone structures can collapse to closed cone structures, and in the process we expect that the no bubbling property can be lost. In other words we do not expect the the mentioned inclusion to be preserved under Gromov-Hausdorff limits. Examples of spaces with this property will be given below in the context of chronally connected spaces, see Definition 5.13. Remark 5.3. If (x, z) are such that d(x, z) = 0 and y z then (x, y) / ∈ J (and dually). In particular, in a causet this shows that J is not much bigger than I, and that, roughly speaking, away from the boundary J 'wraps' I. Proof. We can assume that X contains i 0 as the inclusion does not alter γ, so that X can be assumed to be compact. We prove the version with the equality sign. Suppose that γ(x, p) ≤ r, γ(x, q) ≤ r, and let y ∈ J + (p) ∩ J − (q). By compactness the supremum in the definition of γ(x, y) is realized by a point w, and we have four possible cases.
The strict inequality sign case follows from the previous case, as there is r < r such that p, q ∈ {z : γ(x, z) ≤ r }.
By considering balls of radius γ(x, y) centered in x and y we get Corollary 5.5. For any pair of points x, y ∈ X we have γ(x, z) and γ(z, y) ≤ γ(x, y) We recall that for two relations A, B ⊂ X × X the composition A • B is defined as in Eq. (6).
Definition 5.6. Let (X, d) be a bounded Lorentzian metric space. A continuous function τ : X → R with τ (x) < τ (y) whenever x < y is called a time function. It is well known that if J is closed then J + (p) and J − (p) are closed for every p ∈ X (a closed preordered space is a semiclosed preordered space, see [15]). If we assume that i 0 ∈ X then by the compactness of X, J + (p)∩J − (q) is compact for every p, q ∈ X.
Proof. Reflexivity is immediate. Closedness follows from the continuity of d. As for transitivity, let (x, y), (y, z) ∈ J, and let p ∈ X, then d(p, y) − d(p, x) ≥ 0 and d(p, z) − d(p, y) ≥ 0, which summed give d(p, z) − d(p, x) ≥ 0, and similarly for the other type of inequality, i.e. (x, z) ∈ J.
Let us prove the inclusion I ⊂ J. Let (x, y) ∈ I and let p ∈ X. If p is such that d(p, x) > 0 then d(p, y) − d(p, x) ≥ d(x, y) > 0 follows from the reverse triangle inequality for I. If d(p, x) = 0, then necessarily d(p, y) − d(p, x) ≥ 0 as d is non-negative. This shows the validity of one type of inequality, the other type being similarly proved.
Let (x, y), (y, z) ∈ J. If d(x, y) = d(y, z) = 0 then as d(x, z) ≥ 0, the reverse triangle inequality is satisfied. If the two distances are both positive, the reverse triangle inequality follows from that for I. If one distance is zero and the other positive, say d(x, y) = 0, d(y, z) > 0, then we need only show d(x, z) − d(y, z) ≥ 0 which follows from (x, y) ∈ J. The other case is analogous. The formula I • J ∪ J • I ⊂ I is a consequence of the reverse triangle inequality.
Let us come to the last statement. We know that the Lorentzian distance d is bounded by some constant M < ∞ (cf. Remark 1.2). Recall that by Definition 1.1(iv) (X, T ) admits a dense denumerable subset S = {s n } n∈N that distinguishes points. Let us consider the following function It is continuous in the topology T as it is a uniform limit of continuous functions. Let two points x, y ∈ X, (x, y) ∈ J be given. Then we have d(s n , x) ≤ d(s n , y) and d(x, s n ) ≥ d(y, s n ) for each n ∈ N. As a consequence, τ (x) ≤ τ (y). But x = y implies that they are distinguished by some s k , which means that some of the previous inequalities were strict, which implies τ (x) < τ (y). For the Lipschitz property note that ≤ γ(x, y).

Lorentzian length spaces
In a bounded Lorentzian metric space (X, d) we call a continuous curve σ : Definition 5.9. An isocausal curve σ : for every t, t , t ∈ [a, b], t < t < t .
Clearly, it is equivalent to the same property for t ≤ t ≤ t . Note that it might be the case that d(σ(t), σ(t )) = 0 for some t < t ∈ [a, b] though d(σ(a), σ(b)) > 0. Any restriction σ| [c,d] , [c, d] ⊂ [a, b], c < d, is also maximal by the fact that any restriction of an isocausal curve is isocausal.
In order to formulate the limit curve theorem, it is convenient to extend the interval of definition of the curves (note that the extension is not isocausal) mostly due to slight complications caused by the fact that each curve has its own domain of definition. Alternatively, we could proceed using the notion of uniform convergence introduced in [10, Definition 2.1].
For an isocausal curve σ : [a, b] → X we set Definition 5.10. Let σ n : [a n , b n ] → X, σ : [a, b] → X be continuous isocausal curves. We say that σ n converges uniformly to σ with respect to the metric (distance) h if a n → a, b n → b, andσ n converges uniformly toσ with respect to the metric h. In particular, σ n converges to σ pointwisely.
Remark 5.11. In a bounded Lorentzian metric space that contains i 0 , (X, T ) is a compact metrizable space (the topology T being induced by γ) and hence admits a unique uniform structure [20,Theorem 36.19]. Thus, although we shall prove the following theorem by using the uniformity induced by γ, there is really no need to mention it in the statement of the theorem.
Theorem 5.12 (Limit Curve Theorem). Let (X, d) be a bounded Lorentzian metric space that includes i 0 . Let σ n : [a n , b n ] → X be a sequence of continuous isocausal curves parametrized with respect to a given time function τ , τ (σ(t)) = t. Suppose that {a n } has no accumulation point in common with {b n }. Then there exists a continuous isocausal curve σ : [a, b] → X and a subsequence {σ n k } k that converges uniformly to σ. If the curves σ n are maximizing (i.e. Equation (10) holds) then so is σ.
The last result is really just a consequence of pointwise convergence.
Proof. Set x n := σ n (a n ), y n := (σ n (b n )). We can pass to a subsequence, denoted in the same way, so that they converge, x n → x, y n → y. But τ (x n ) = τ (σ n (a n )) = a n thus in the limit a n → a := τ (x), and similarly b n → b := τ (y). Set σ(a) := x and σ(b) := y. Observe that a < b due to a n ≤ b n and the accumulation condition on {a n } and {b n }. For q ∈ Q consider the sequence {σ n (q)} n with n sufficiently large. Since (X, T ) is compact the sequence admits a converging subsequence. Via a Cantor diagonal argument we can pass to a subsequence so that the sequence {σ n (q)} n converges to some point that we denoteσ(q) for every q ∈ Q.
We shall prove in a moment that, by a similar reasoning,σ : Q → X is Cauchy continuous, i.e. it maps Cauchy sequences to Cauchy sequences. Then, by a standard result in topology (recall that by Theorem 4.24, (X, γ) is a compact metric space, hence complete),σ has a unique continuous extension to R, that we denote in the same way. Observe that by continuity of the extension σ, τ (σ(r)) = r for every r ∈ [a, b]. Similarly, by continuity ofσ and the closure of J,σ(r) ≤σ(r ), for every r, r ∈ [a, b]. If r, r ∈ [a, b] are such that r < r then as τ takes different values onσ(r) andσ(r ), we getσ(r) <σ(r ), namely σ :=σ| [a,b] is a continuous isocausal curve.
Observe that for q ∈ Q, q < a, we have for sufficiently large n, q < a n , hencê σ n (q) = x n , which implies taking the limit,σ(q) = x. Any continuous extension σ satisfiesσ(r) = x for every r ∈ R, r ≤ a, and analogously for r ≥ b. This proves thatσ is indeed the extension of σ in the sense of Eq. (11).
In order to prove the Cauchy continuity ofσ : Q → X, let us proceed by contradiction. Suppose that there exists a Cauchy sequence {q n } ⊂ Q that does not map to a γ-Cauchy sequence. Let r ∈ R be the limit of q n . There is ε > 0 and two subsequences {q Without loss of generality we can assume that q m k < q n k for infintely many k and that both sequences converge withσ(q m k ) → w andσ(q n k ) → z, respectively. We have γ(w, z) ≥ ε. Observe thatσ(q m k ) ≤σ(q n k ), hence w ≤ z, and finally w < z.
For r ∈ (a, b) we have for sufficiently large k, q m k , q n k ∈ (a, b). Since τ (σ(q m k )) = q m k → r, τ (σ(q n k )) = q n k → r we obtain τ (w) = τ (z) = r. This contradicts the properties of the time function τ .
If r ≤ a it must be q n k > a for every sufficiently large k, otherwise q m k , q n k ≤ a which would implyσ(q m k ) =σ(q n k ) = x and hence w = z = x, γ(x, x) ≥ > 0, a contradiction. However, if q n k > a for every sufficiently large k, we have τ (σ(q n k )) = q n k > a and hence τ (z) = r = a. If q m k ≤ a for an infinite number of k, thenσ(q m k ) = x for those values of k and hence w = x, which implies τ (w) = a. If q m k > a for sufficiently large k, then as proved above for q n k we get τ (w) = a. In any cae τ (w) = τ (z) = a that contradicts the properties of the time function τ . The case r ≥ b is analogous.
Since for all n sufficiently large all curves are constant on the complement of [a − 1, b + 1] we see that the convergence is uniform on all of R. The equations d(σ n (t), σ n (t )) + d(σ n (t ), σ n (t )) = d(σ n (t), σ n (t )) clearly pass in the limit to Equation (10) for σ.

Definition 5.13. A bounded Lorentzian metric space (X, d) is a (bounded)
Lorentzian prelength space if for every pair x, y ∈ X with x y there exists an isocausal curve σ : [0, 1] → X from x to y.
We refer to the above property as chronal connecteness by isocausal curves.
Proposition 5.14. Let (X, d) be a bounded Lorentzian prelength space that includes i 0 . Then for every (x, y) ∈Ī, x = y, there exists a continuous isocausal curve σ : [a, b] → X connecting x to y.
Proof. Let τ be a time function and let (x n , y n ) → (x, y), (x n , y n ) ∈ I. Since J is closed,Ī ⊂ J, thus x < y, which implies τ (x) < τ (y). Let σ n : [a n , b n ] → X be continuous isocausal curves connecting x n to y n and parametrized with the time function. The assumptions of the limit curve theorem is satisfied as a n → τ (x), b n → τ (y), which differ.  The relationJ is meant to cure the fact that, as previously observed, J is too large, particularly at boundary points.
Definition 5.17. A bounded Lorentzian prelength space (X, d) is a bounded Lorentzian length space if for every pair x, y ∈ X with x y there exists a maximal curve connecting x and y.
We refer to the above property as maximal chronal connectedness by isocausal curves.
Observe that if an isocausal curve σ : [0, 1] → X connects two points x, y such that d(x, y) = 0, then d(σ(t), σ(t )) = 0 for every t, t ∈ [0, 1]. In particular it is maximizing in the sense of Equation (10). Thus any two distinctJ-related points in a bounded Lorentzian length space are connected by a maximizing isocausal curve. Proof. We proceed by proving the former statement, the latter being obtained dropping the maximization condition on the curves σ n introduced below.
Let x, y ∈ X with x y be given. In order to construct a maximal isocausal curve ζ : [0, 1] → X between x and y we first define its values for rational parameter values a ∈ [0, 1] and extend this later to all real numbers τ ∈ [0, 1]. The resulting map will be shown to be maximizing in the sense of Definition 5.17.
Set n := d GH (X, X n ) and fix a correspondence R n ⊂ X × X n with disR n ≤ 2 n (we can assume that n > 0 otherwise the conclusion follows by isometry). Choose x n , y n ∈ X n such that (x, x n ), (y, y n ) ∈ R n . Since d(x, y) > 0 we can assume that d n (x n , y n ) > 0, namely x n y n (otherwise pass to a subsequence). Let S = {s k } k∈N be a dense subset of (X, d). It distinguishes points as in Definition 1.1(iv) (X, T ) by Proposition 1.9. Choose s n k ∈ X n with (s k , s n k ) ∈ R n and set S n := {s n k } k∈N . Notice that in general S n might not distinguish points in X n .
Consider the function where constants α > 0 and β are chosen such that τ (x) = 0, τ (y) = 1. By replacing d s k with (d n ) s n k , and d s k with (d n ) s n k we obtain a continuous function τ n : X n → R such that τ n (p) ≤ τ n (q) for every (p, q) ∈ J n . Note, however, that it is not necessarily the case that p n , q n ∈ X n , p n < n q n , implies τ n (p n ) < τ n (q n ). For every p ∈ X, p n ∈ X n with (p, p n ) ∈ R n we have which implies τ n (x n ) → 0, τ n (y n ) → 1 as a special case.
Let σ n : [0, 1] → X n be a continuous maximizing isocausal curves connecting x n to y n . For each rational number a ∈ (0, 1), choose a point z n (a) on σ n with τ n (z n (a)) = a.
Let rational numbers a, b ∈ [0, 1] with a < b be given. Both z n (a) and z n (b) belong to the image of σ n , but it cannot be z n (b) ≤ z n (a), otherwise, as τ n is non-decreasing over isocausal curves, b = τ n (z n (b)) ≤ τ n (z n (a)) = a, which gives a contradiction. Thus a < b implies z n (a) < z n (b).
Taking the limit we obtain Let a, b ∈ [0, 1] be rational numbers with a < b. Next we prove that (ζ(a), ζ(b)) ∈ J. Let w ∈ X be given and choose w n ∈ X n such that (w, w n ) ∈ R n . Then as (z n (a), z n (b)) ∈ J n we have d n (w n , z n (a)) ≤ d n (w n , z n (b)), and d n (z n (a), w n ) ≥ d n (z n (b), w n ) which imply d(w, ζ n (a)) ≤ d(w, ζ n (b)) + 4 n , and d(ζ n (a), w) ≥ d(ζ n (b), w) − 4 n .
Next we extend ζ continuously to the whole interval [0, 1] as an isocausal curve. This follows again, like in the proof of Theorem 5.12, by the Cauchy continuity of the map ζ on rational numbers in [0, 1]. Observe that the argument uses just the following facts: that there is a time function τ with τ (ζ(a)) = a, for any rational number a ∈ [0, 1], and the property that for any pair of rational numbers a, b ∈ [0, 1] with a < b we have ζ(a) < ζ(b) (there is no need to distinguish the cases r = a, r = b and r ∈ (a, b) as done there, as we can work directly with the only case r ∈ [a, b]). The maximization property is obtained by continuity from Equation 13.
6 Length and Curvature

The length functional
This section can be skipped on first reading. Its goal is to introduce the length functional though most of the theory can be developed without this concept. Definition 6.1. Let (X, d) be a bounded Lorentzian metric space, and let σ : [0, 1] → X be a continuous isocausal curve. Its Lorentzian length is where the infimum is over the set of all partitions Clearly, by the reverse triangle inequality, if x = σ(0), y = σ(1), Proposition 6.2. Let σ : [0, 1] → X be a continuous isocausal curve of endpoints x and y. We have L(σ) = d(x, y) iff σ is maximal.
We observe that the following result does not depend on the existence of convex neighborhoods (for a similar observation in the smooth manifold context see [11]) Proof. We need to show that for every > 0, we have for sufficiently large n, L(σ n ) ≤ L(σ) + .

Triangle comparison
Definition 6.5. Let (X, d) be a bounded Lorentzian length space. A timelike triangle is given by a triple (x, y, z), x y z.
A point p is said to be on the side xy, if d(x, p) + d(p, y) = d(x, y), in which case we define α ∈ [0, 1] such that d(x, p) = αd(x, y). Similarly, p is said to be on the side yz, if d(y, p) + d(p, z) = d(y, z), in which case we define β ∈ [0, 1] such that d(y, p) = βd(y, z). Finally, point p is said to be on the side xz, if d(x, p) + d(p, z) = d(x, z), in which case we define γ ∈ [0, 1] such that d(x, p) = γd(x, z). A point that belongs to one of the sides is said to belong to the perimenter of the timelike triangle.
Let us consider an abstract triangle ∆ of vertices A, B, C on a 2-dimensional affine space and parametrize each of the sides AB, BC, AC, with the parameters α, β, γ ∈ [0, 1]. Each choice of point on the triangle ∆ determines the values of one parameter, and hence, (recall that by the length space property (X, d) is maximally chronally connected and d is continuous) there is some point on the perimeter of the timelike triangle on X, with a corresponding value of parameter (e.g. if it belongs to the side xy, d(x, p) = αd(x, y)). Note that there could be more than one such point. Proof. Let us set n := 2d GH (X, X n ) > 0 (if one of them is zero the conclusion is immediate by isometry), so that n → 0, and let us fix a correspondence R n ⊂ X × X n with disR n ≤ n .
Let (x, y, z) be a timelike triangle in X such that (a, b, c) ∈ O, and let (d 1 , d 2 ) ∈ ∆. Let x n , y n , z n ∈ X n be such that (x, x n ), (y, y n ), (z, z n ) ∈ R n . Observe that a n := d n (x n , y n ) → d(x, y), b n := d n (y n , z n ) → d(y, z), c n := d n (x n , z n ) → d(x, z), so that for sufficiently large n the three distances are positive and (a n , b n , c n ) ∈ O.
Let p n , q n ∈ X n be in the perimeter of the timelike triangle of vertices (x n , y n , z n ), chosen so as to correspond to the parameters determined by the choice d 1 , d 2 ∈ ∆. By assumption there is one such choice such that (d 1 , d 2 , d(p, q)) ∈ F . Letp n ,q n ∈ X be chosen so that (p n , p n ) ∈ R n , and (q n , q n ) ∈ R n , and let us pass to a subsequence (denoted in the same way) such thatp n → p,q n → q. We want to show that p, q correspond to d 1 , d 2 ∈ ∆, respectively. For instance, if d 1 belongs to the side parametrized by α, we have |d(x,p n ) − αd n (x n , y n )| = |d(x,p n ) − d n (x n , p n )| ≤ n which implies d(x, p) = αd(x, y). Additionally, from d n (x n , p n ) + d n (p n , y n ) = d n (x n , y n ) |d(x,p n ) + d(p n , y) − d(x, y)| ≤ 3 n which in the limit gives d(x, p) + d(p, y) = d(x, y). The other cases are treated similarly.
Finally, (d 1 , d 2 , d n (p n , q n )) → (d 1 , d 2 , d(p, q)), and since F is closed, we have (d 1 , d 2 , d(p, q)) ∈ F . This shows that for each choice d 1 , d 2 we can find (p, q) as desired.
We define the sectional curvature on a smooth spacetime (M, g), where g has signature (−, +, . . . , +) and This definition differs by a sign with that adopted in [1,6]. It is called timelike (spacelike) sectional curvature whenever span(X, Y ) is a timelike (resp. spacelike) plane. It should be observed that if there is k ∈ R such that K ≥ k, over every timelike plane containing X, then Ric(X) ≥ nkg(X, X) where n + 1 is the spacetime dimension. Thus a lower bound on the timelike sectional curvature implies a lower bound for Ricci in the timelike direction (this is a type of strong energy condition which appears in singularity theorems), and similarly for upper bounds (with the convention in [1,6] a lower bounds leads to an upper bound and viceversa).
We say that the sectional curvature is lower bounded, and write K ≥ k, if the following inequality holds on any 2-plane regardless of whether span(X, Y ) is timelike or spacelike. This condition implies that the timelike sectional curvature is lower bounded by k.
We recall that a timelike triangle of given sides (a, b, c) is realized as a timelike triangle in a model space accordingly to the next realizability lemma [6] (where it is understood that 1/ √ k = ∞ for k ≤ 0). Then for every (a, b, c) ∈ O k there is a timelike triangle in M k with sides (a, b, c) (i.e. x, y, z ∈ M k , x y z, and a = d(x, y), b = d(y, z), c = d(x, z)).
Under the conditions of the theorem the realizing triangle on M k is actually unique up to isometries and the vertices are connected by unique maximizing timelike geodesics.
Let us consider a realizing triangle of sides a, b, c on M k . Let d 1 , d 2 ∈ ∆ and letp,q be corresponding points on the realizing triangle. On M k there is a continuous function H k : for every triangle. The expression will not be relevant here. Definition 6.9. Let k ∈ R. We say that the bounded Lorentzian length space (X, d) has sectional curvature bounded from below by k, if it is or, equivalently, for any timelike triangle and for any choice of parameters d 1 , d 2 ∈ ∆ we can find two points p, q on the timelike triangle with such parameters and such that d(p, q) ≤ d M k (p,q) wherep,q are corresponding points on the comparison triangle on M k . The upper bounded version is obtained replacing ≤ with ≥.
If we know that the vertices of the timelike triangle are connected by unique maximal isocausal curves, then after "equivalently" we can replace: for any timelike triangle and for any two points p, q on the timelike triangle we have d(p, q) ≤ d M k (p,q) wherep,q are corresponding points on the comparison triangle on M k . This will be possible within convex neighborhoods (see below).
Intuitively, under a lower bound on the sectional curvature the timelike triangle should be slender than the comparison triangle, while under an upper bound it should be fatter than the comparison triangle.
Of course, the bound on sectional curvature introduced here is meant to be the low regularity version of the bound on sectional curvature of smooth spacetimes, while at the same time be preserved under GH limits.
The fact that a lower bound on the sectional curvature leads indeed to the type of inequality used in the above definition was proved, at least for triangles contained in normal neighborhoods, by Alexander and Bishop [1] (actually they prove a stronger result because they used a signed distance that captures information on non-causally related events), see also Harris [6].
Our definition has similarities with that used in [8], but our geodesic triangle is just a triple of points, not of curves and we demand the existence of some pair p, q with specific properties while other pair choices with equal parameters (roughly, same distances from the vertices) might not satisfy those properties. Definition 6.10. We say that the bounded Lorentzian length space (X, d) admits convex neighborhoods, if every point p ∈ X admits some neighborhood U such that (U, d| U ) is a bounded Lorentzian length space and for every pair of chronologically related points in U there is just one maximal curve which, furthermore, is isochronal and entirely contained in U . Proposition 6.11. In a bounded Lorentzian length space that admits convex neighborhoods a maximal isocausal curve connecting two chronologically related points is actually isochronal (i.e. maximal curves have definite causal character).
Proof. By contradiction suppose that the maximal curve σ : [0, 1] → X, admits two points σ(t 1 ), σ(t 2 ), t 1 < t 2 , such that d(σ(t 1 ), σ(t 2 )) = 0. By continuity of d there is a largest non-empty closed interval [a, b] ⊂ [0, 1] such that d(σ(a), σ(b)) = 0. Since the inclusion [a, b] ⊂ [0, 1] is proper, we might assume, without loss of generality b < 1. Let q = σ(b), and let U q be a convex neighborhood. We can find s < b sufficiently close to b such that p := σ(s) ∈ U q , and t > b sufficiently close to b such that r := σ(t) ∈ U q . By the maximality of σ, d(p, q) = 0 and d(p, r) = d(q, r) is the length of σ| [s,t] . However, this curve is not isochronal, thus, by the property of convex neighborhoods, it is not actually maximal, which implies that d(p, q) > L(σ| [s,t] ) = d(p, q), a contradiction. Definition 6.12. We say that on (X, d) maximal isocausal curves branch in the future if we can find two maximal isocausal curves σ : [0, u] → X and γ : [0, v] → X whose images coincide over two non-empty closed intervals, i.e. σ([0, b]) = γ([0, c]), b < u, c < v, that cannot be extended to the right while preserving the same property. Then σ(b) (γ(c)) is a future branching point for σ (resp. γ). Similar definitions hold in the past case.
, then there will largest closed sets [0, b], [0, c], u < b, c < v, for which this identity holds and that cannot further extended to the right.
Note that our non-branching property is somewhat stronger than [8, Def. 4.10] (their non-branching spaces might admit curves that after being coincident on an interval are again coincident on a sequence of points approaching the edge of the interval).
The proof of the next result follows ideas in [8,Thm. 4.12], but adapted to our notions. Proposition 6.13. Let (X, d) be a bounded Lorentzian length space that admits convex neighborhoods and suppose that the sectional curvature is bounded from below by k ∈ R. Maximal isochronal curves do not branch.
One could just impose some form of local sectional curvature boundedness, the proof argument taking place within a convex neighborhood. Observe that these local conditions do not pass GH-limits as likely does not the existence of convex neighborhoods.
Proof. We prove that future branching is impossible, the other case being analogous. Let q = σ(b) and let U q be a convex neighborhood. We can find s < b sufficiently close to b that x := σ(s) ∈ U q and σ([s, b]) ⊂ U q . We can find t > b sufficiently close to b that z := σ(t) ∈ U q , σ([b, t]) ⊂ U q , and σ(t) / ∈ γ([0, 1]). Since σ(b) σ(t), and since for some c, γ(c) = σ(b), we have γ(c) σ(t). Let e > c be such that y := γ(e) σ(t), γ([c, e]) ⊂ U q and γ(e) / ∈ σ([0, 1]). We consider the triangle (x, y, z). Since σ is maximal L(σ| [s,t] ) = d(x, z). Similarly, since γ is maximal L(γ| [f,e] ) = d(x, y), where γ(f ) = σ(s). It cannot be d(y, z) = d(x, z) − d(x, y) otherwise the isochronal curve obtained concatenating γ| [f,e] with the maximal isochronal curve connecting y to z would give a maximal isochronal curve connecting x to z but different from σ| [s,t] as passing from y, contradicting the uniqueness of maximal isochronal curves in convex neighborhoods. Thus the triangle is associated with a triangle inequality with the strict sign, d(x, z) > d(x, y) + d(y, z). The realizing triangle in M k is therefore non-degenerate (sides are not aligned). Observe that the point σ(b) is the only point with distances d(x, σ(b)), d(σ(b), z) from x and z (by the uniqueness of maximal isochronal curves implied by the convex neighborhood). Letσ(b) the corresponding point on the sidexz on the realizing triangle on M k . Similarly, letγ(c) be the point that corresponds to γ(c) on the sidexȳ. Note that d M k (γ(c),z) < d M k (σ(b),z) otherwise one could go fromx toz with a curve of length d M k (x,γ(c)) + d M k (γ(c),z) ≥ d M k (x,σ(b)) + d M k (σ(b),z) = d M k (x,z), passing throughγ(c) and, due to the corner atγ(c), (due to the fact that the realizing triangle in non-degenerate) we would actually have that this timelike curve could be deformed to a longer timelike curve, a contradiction.
Finally, since the timelike curvature is bounded from below by k (used in the penultimate step) Remark 6.14. It can be observed that in the above proof we are comparing d(w, z) with d M k (w,z) where w is a point on the side xy andw the corresponding point on the sidexȳ. In the smooth case this type of comparison result follow just from the boundedness on the timelike sectional curvature due to Harris' results [6], with no need for conditions on spacelike plane as in Alexander and Bishop [1]. This means that we could have weakened the conditions in the definition of sectional curvature bound for Lorentzian length spaces imposing the comparison distance inequality for pairs (w, z) placed as above, while still being able to obtain the above non-branching result.

Uniformly totally bounded families
As in the case of metric spaces we will give criteria for when a set of bounded Lorentzian-metric spaces is precompact with respect to the Gromov-Hausdorff semi-distance.
Remark 7.1. In the next proofs we shall use the semi-distance d GH between bounded spaces (X, d) whose distance d satisfies the reverse triangle inequality as in Definition 1.1.
For a bounded Lorentzian-metric space we define diam X := max X×X d. We already observed that it positive and bounded.
Definition 7.2. Let D > 0, α := {α k } k∈N ⊂ (0, ∞) be a decreasing sequence with α k → 0 and β := {β k } k∈N ⊂ (0, ∞) be an increasing sequence with β k → ∞. The class X(D, α, β) of bounded Lorentzian-metric spaces consists of all those X such that (1) diam X ≤ D for all X ∈ X, (2) every X ∈ X contains for each k an α k -net consisting of no more than β k points, We say that X is uniformly totally bounded with respect to (D, α, β).
Observe that if α i = α ki , β i = β ki , are subsequences, then X(D, α, β) ⊂ X(D, α , β ). As a consequence, if X satisfies (1) and (2) then (3) can be satisfies by enlarging X, that is by passing to suitable subsequences of α k and β k . Remark 7.3. If X is uniformly totally bounded with respect to (D, α, β) the family also satisfies the following conditions which are closer to the notion in metric geometry: (1) There is a constant D such that diam X ≤ D for all X ∈ X.
(2) For every > 0 there exists a natural number N = N ( ) such that every X ∈ X contains an -net consisting of no more than N points.
Our objective is to prove that (X, d GH ) is a compact metric space. For a metric space this property is equivalent to separability and sequential compactness. Property (3) will only enter the proof of separability, that is, the proof of sequential compactness will only use (1), (2) and (4). Definition 7.4. Causets and -nets are said to be rational if the distance function takes only rational values.
Proposition 7.5. The family of rational causets is denumerable. Moreover, it is Gromov-Hausdorff dense in the family of causets.
Remark 7.6. Together with Corollary 4.28 the proposition implies that rational causets are dense w.r.t. the Gromov-Hausdorff topology in the family of bounded Lorentzian metric spaces.
Proof. Let us prove the latter statement. Let n be the (finite) cardinality of the causet S = {x 1 , · · · , x n }, then the n(n − 1)/2 numbers d(x i , x j ), for i, j = 1, · · · , n, completely describe the causet. They are subject to the constraints (i) and (iii) of Def. 1.1. Each causet of cardinality n can then be represented by a point in R n(n−1)/2 . Let k ≤ n(n − 1)/2 be the number of distances different from zero. We are going to consider the space R k as we are not going to perturb distances that are zero. Definition 1.1(iii) is an open condition. Indeed, the distinction metric γ takes a minimal positive value α among pairs of distinct points. If the distances d(x i , x j ) are perturbed by less than half this value, (iii) still holds. Let > 0. We shall perturb each positive d(x i , x j ) by less than min(α/2, ). Observe that the inequalities in Definition 1.1(i) involve only those distances which are positive.
We first perturb it so as to satisfy all inequalities (i) in a strict sense. Each distance d(x i , x j ) > 0 can be thought as a link connecting x i to x j . We can move from x i to x j by a chain with a maximal number of links (observe that each chain has a finite number of elements due to chronology, i.e. boundedness of d, and that by the same reason no chain passes twice from the same point). If t ij is the maximal length of the chain we replace d(x i , x j ) with d(x i , x j ) + δt 2 ij , where δ ≤ min(α/2, )/[n(n − 1)/2] 2 . In this way all inequalities (i) are satisfies in a strict sense. Those inequality are now an open condition on R k , thus taking into account that condition (iii) is also open we can indeed find a point in Q k arbitrary close to our values. In conclusion, the positive distances can be perturbed to become rational while preserving properties (i) and (iii). Moreover, the perturbation of each distance can be chosen to be less that . As a consequence the newly obtained rational causetS satisfies d GH (S,S) ≤ .
For the former statement, observe that Q n(n−1)/2 has the cardinality of N and the cardinality of N copies of N is that of N, which proves the claim. Proposition 7.7. For a uniformly totally bounded family X the metric space (X, d GH ) is separable.
Proof. We want to show that the closure of the rational causets contained in X is the whole space. Let X ∈ X be given. Then for each k there is an α k -net S k ⊂ X. Consider the setS k = ∪ i≤k S i , then by Remark 4.27, point 2, for i ≤ k, S i is an α i -net inS k (notice that S i counts no more than β i points), while for j > k the setS k itself is a α j -net inS k which, by (3), counts no more than β j points.
By Remark 4.27, point 4, d GH (S k , X) ≤ 2α k asS k is a α k -net of X as it contains the α k -net S k of X.
By Remark 4.27, point 6 and Proposition 4.12, we can find a subsetŠ k ofS k which is a causet, hence a bounded Lorentzian metric space, with d GH (Š k ,S k ) = 0. It is still true thatŠ k admits for every j an α j -net inŠ k which counts no more than β j points (they are obtained from Proposition 4.12 applied to the analogous sets forS k ). This means thatŠ k belongs to X(D, α, β). Now we can decrease all positive distances so as to preserve the α j -nets, via a redefinition of the following form d(x i , x j )−δ √ t ij , δ > 0, with t ij interpreted as in Proposition 7.5 and δ sufficiently small. This operation brings the inequality of the type of Definition 1.1(i) to a strict form, and all distances can be replaced by a rational value by preserving conditions Definition 1.1(i),(iii) and the α j -nets.
In conclusion, as α k → 0, we find a rational causet in X(D, α, β) at arbitrarily small Gromov-Hausdorff distance from X, which proves the desired result as the rational causets are countable.

Compactness of uniformly totally bounded families
Theorem 7.8. For any uniformly totally bounded family X the metric space (X, d GH ) is compact.
Remark 7.9. Due to the equivalence between the Lorentzian diameter and the distinction-metric diameter cf. Prop. 4.23, a uniformly totally bounded family in our sense is uniformly totally bounded in the classical sense [5]. Thus it is possible to use the classical precompactness theorem [5,Thm. 7.4.15] to infer that a subsequence converges to a metric space. If we were able to show that that limit metric space is also a bounded Lorentzian metric space one could use the equivalence of GH-convergences, Thm. 4.30, to obtain a Lorentzian precompactness theorem. We could not follow this strategy, so we prove directly that the limit is a bounded Lorentzian metric space by reworking the precompactness theorem.
The remaining of this section devoted to the proof of the sequential compactness of X.
In this proof environment we shall also obtain some useful lemmas/theorems/corollaries. Let X be uniformly totally bounded with respect to (D, α, β), see Definition 7.2. Let {X m } ⊆ X be a sequence. We know that (X m , T m ) is compact. For each positive integer k, X m admits an α k -net S (k) m of at most β k points. By adding points arbitrarily to each net we can assume without loss of generality that each α k -net consists of exactly β k points. Set N k := k s=1 β s . Let us order the (distinct) elements of S (k) m from N k−1 + 1 to N k in some (arbitrary) way, so that we can denote S