aa r X i v : . [ m a t h . M G ] M a r NOTES ON POINTED GROMOV-HAUSDORFFCONVERGENCE
DOROTHEA JANSEN
Abstract.
The present article addresses to everyone who starts work-ing with (pointed) Gromov-Hausdorff convergence. In the major part,both Gromov-Hausdorff convergence of compact and of pointed met-ric spaces are introduced and investigated. Moreover, the relation ofsublimits occurring with pointed Gromov-Hausdorff convergence andultralimits is discussed.
Contents
1. The compact case 22. The non-compact case 162.1. Comparison with the compact case 182.2. Properties as in the compact case 212.3. Convergence of points 282.4. Convergence of maps 323. Ultralimits 39References 47Gromov-Hausdorff distance is an often used tool for measuring how fartwo compact metric spaces are from being isometric. This distance, whichwas introduced by Gromov in [Gro81], leads to the notion of Gromov-Haus-dorff convergence which can be extended to non-compact metric spaces andallows to draw conclusions about the properties of the spaces ‘near’ to thelimit space, if the limit space is well understood. Many textbooks suchas [BBI01, sections 7.3-7.5], [Pet06, section 10.1] and [BH99, p. 70ff.] give a(more or less) detailed introduction to the distance of compact metric spaces.Some even more detailed proofs can be found in [Ron10]. Since the literatureon convergence of non-compact metric spaces usually is less comprehensive,this article treats the latter in detail. For the sake of completeness, it also
Date : March 29, 2017.2010
Mathematics Subject Classification.
Key words and phrases.
Gromov-Hausdorff convergence, ultralimits.These notes are based on the appendix of the author’s PhD thesis [Jan16].This work was supported by the Gottfried Wilhelm Leibniz-Preis of Prof. Dr. BurkhardWilking and the SFB 878: Groups, Geometry & Actions, at the University of Münster. contains a detailed introduction to the compact case, which is built on theliterature cited above.The first section deals with Gromov-Hausdorff distance of compact metricspaces. In addition, so called Gromov-Hausdorff approximations are intro-duced and the relation between those two terms is described. For both terms,a pointed and a non-pointed version is introduced, and it will be proven thatthese terms result in the same notion of convergence.The second section deals with convergence of non-compact metric spaces,and consists of three parts: First, for compact length spaces it will be proventhat this notion of convergence coincides with the one for compact spaces.Secondly, several properties of pointed Gromov-Hausdorff convergence willbe verified. After that, a convergence notion for points will be introducedand studied. Finally, convergence of (Lipschitz) maps will be investigated.The third and final section deals with ultralimits, a more general tool tocreate ‘limit spaces’, and states some properties of those. In particular, astrong correspondence between ultralimits and subsequences converging inthe pointed Gromov-Hausdorff sense will be established.1.
The compact case
Given a metric space, an interesting question is whether it is possible toassign each two subsets a distance such that this distance in turn defines ametric. In [Hau65, Chapter VIII §6], Hausdorff answered this question bydescribing what nowadays is called the Hausdorff distance: For two subsets ofa metric space, this is the minimal radius such that each subset is containedin the ball (with this radius) of the other subset. This was extended byGromov in [Gro81, section 6] to describe how far two compact metric spacesare from being isometric by mapping two such spaces isometrically into athird one and measuring the Hausdorff distance of the images. (In fact, onecan restrict to embedding the two spaces isometrically into their disjointunion.) This is the so called Gromov-Hausdorff distance.
Definition 1.1.
For bounded subsets A and B of a metric space ( X, d ) , the Hausdorff distance of A and B is defined as d d H ( A, B ) := inf { ε > | A ⊆ B Xε ( B ) and B ⊆ B Xε ( A ) } where B Xε ( B ) := { x ∈ X | ∃ b ∈ B : d ( x, b ) < ε } . For two compact metricspaces ( X, d X ) and ( Y, d Y ) , the Gromov-Hausdorff distance of X and Y isdefined as d GH ( X, Y ) := inf { d d H ( X, Y ) | d admissible metric on X ∐ Y } , where a metric d on the disjoint union X ∐ Y is called admissible if it satisfies d | X × X = d X and d | Y × Y = d Y .On the space of (non-empty) compact subspaces of X , this d H defines ametric, while d GH defines a metric on the set of isometry classes of (non-empty) compact metric spaces. This will be proven below. From now on, all OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 3 metric spaces are assumed to be non-empty. In order to compare two metricspaces with respect to some fixed base points, the pointed Gromov-Hausdorffdistance is used.
Definition 1.2.
Let ( X, d ) be a metric space, A, B ⊆ X bounded subsetsand a ∈ A , b ∈ B base points. The pointed Hausdorff distance of ( A, a ) and ( B, b ) is given by d d H (( A, a ) , ( B, b )) := d d H ( A, B ) + d ( a, b ) and the pointed Gromov-Hausdorff distance between two pointed compactmetric spaces ( X, x ) and ( Y, y ) is defined as d GH (( X, x ) , ( Y, y )) := inf { d d H (( X, x ) , ( Y, y )) | d adm. on X ∐ Y } . As in the non-pointed case, the pointed Gromov-Hausdorff distance definesa metric on the set of isometry classes of (non-empty) pointed compact metricspaces. In order to prove this, a notion strongly related to the one of Gro-mov-Hausdorff distance is used.
Definition 1.3.
Let ( X, d X ) and ( Y, d Y ) be metric spaces and ε > . Apair of (not necessarily continuous) maps f : X → Y and g : Y → X iscalled ( ε -)Gromov-Hausdorff approximations or ε -approximations if for all x, x , x ∈ X and y, y , y ∈ Y , | d X ( x , x ) − d Y ( f ( x ) , f ( x )) | < ε, d X ( g ◦ f ( x ) , x ) < ε, | d Y ( y , y ) − d X ( g ( y ) , g ( y )) | < ε, d Y ( f ◦ g ( y ) , y ) < ε . The set of all such pairs is denoted by
Appr ε ( X, Y ) . In the pointed case, onerestricts to pointed maps: For p ∈ X and q ∈ Y , Appr ε (( X, p ) , ( Y, q )) := { ( f, g ) ∈ Appr ε ( X, Y ) | f ( p ) = q and g ( q ) = p } . Remark.
In the literature, Gromov-Hausdorff approximations often are notdefined as pairs of maps but as one map f : X → Y where f has distortionless than ε , i.e. for all x , x ∈ X , f satisfies | d Y ( f ( x ) , f ( x )) − d X ( x , x ) | < ε, and B ε ( f ( X )) = Y . Observe that ( f, g ) ∈ Appr ε ( X, Y ) already implies that f has these properties (for the same ε ).In the following it will be seen that Gromov-Hausdorff distance less than ε corresponds to ε -approximations (up to a factor). The next propositionshows that (up to another factor) the definition of Gromov-Hausdorff ap-proximations used here can be replaced by the one described above. Proposition 1.4.
Let f : ( X, d X ) → ( Y, d Y ) be a map between metric spaceswith distortion smaller than ε > . Then there exists a map g : f ( X ) → X satisfying ( f, g ) ∈ Appr ε ( X, f ( X )) . Moreover, if Y = B ε ( f ( X )) , then thereexists a map h : Y → X such that ( f, h ) ∈ Appr ε ( X, Y ) . D. JANSEN
Proof.
For each y ∈ f ( X ) choose some g ( y ) ∈ f − ( y ) . In particular, thesuch defined map g satisfies f ◦ g = id | f ( X ) . For y , y ∈ f ( X ) , | d X ( g ( y ) , g ( y )) − d Y ( y , y ) | = | d X ( g ( y ) , g ( y )) − d Y ( f ( g ( y )) , f ( g ( y ))) | < ε, and for x ∈ X , d ( x, g ◦ f ( x )) = | d ( x, g ◦ f ( x ))) − d ( f ( x ) , f ( g ◦ f ( x ))) | < ε . Thus, ( f, g ) ∈ Appr ε ( X, f ( X )) .Now assume Y = B ε ( f ( X )) . For y ∈ f ( X ) , define h ( y ) := g ( y ) , otherwise,choose y ′ ∈ f ( X ) with d Y ( y, y ′ ) < ε and define h ( y ) := y ′ . By construction, h ◦ f = g ◦ f , i.e. for all x ∈ X , d X ( h ◦ f ( x ) , x ) < ε . For arbitrary y ∈ Y , using f ◦ g = id | f ( X ) , f ◦ h ( y ) = f ◦ g ( y ′ ) = y ′ for y ′ ∈ f ( X ) ∩ B ε ( y ) as in the definition of h . Hence, d Y ( f ◦ h ( y ) , y ) = d Y ( y ′ , y ) < ε . Finally, for arbitrary y , y ∈ Y , | d X ( h ( y ) , h ( y )) − d Y ( y , y ) |≤ | d X ( h ( y ) , h ( y )) − d Y ( f ( h ( y )) , f ( h ( y ))) | + | d Y ( f ( h ( y )) , f ( h ( y ))) − d Y ( y , y ) | < ε + d Y ( f ◦ h ( y ) , y ) + d Y ( f ◦ h ( y ) , y ) < ε . (cid:3) Next, a strong connection between existence of Gromov-Hausdorff approx-imations and the Gromov-Hausdorff distance will be proven.
Proposition 1.5.
Let X and Y be compact metric spaces with base points p ∈ X and q ∈ Y , respectively, and ε > . a) If d GH ( X, Y ) < ε , then Appr ε ( X, Y ) = ∅ . b) If Appr ε ( X, Y ) = ∅ , then d GH ( X, Y ) ≤ ε . c) If d GH (( X, p ) , ( Y, q )) < ε , then Appr ε (( X, p ) , ( Y, q )) = ∅ . d) If Appr ε (( X, p ) , ( Y, q )) = ∅ , then d GH (( X, p ) , ( Y, q )) ≤ ε .Proof. As the proofs of a) and b), respectively, are very similar to, butslightly easier than those of c) and d), respectively, only the latter two areproven here.c) Let < δ < ε − d GH (( X, p ) , ( Y, q )) and choose an admissible metric d with d d H (( X, p ) , ( Y, q )) < d GH (( X, p ) , ( Y, q )) + δ < ε .
Then d ( p, q ) < ε on the one hand and d d H ( X, Y ) < ε on the other, i.e. forall x ∈ X there exists y x ∈ Y that satisfies d ( x, y x ) < ε . Analogously, for OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 5 each y ∈ Y there is x y ∈ X satisfying d ( y, x y ) < ε . Define f : X → Y and g : Y → X by f ( x ) := ( q if x = p,y x otherwise, g ( y ) := ( p if y = q,x y otherwise.As seen above, d ( f ( x ) , x ) < ε for all x ∈ X . Thus, for all x, x ′ ∈ X , | d Y ( f ( x ) , f ( x ′ )) − d X ( x, x ′ ) | ≤ d ( f ( x ) , x )) + d ( f ( x ′ ) , x ′ ) < ε . Analogously, | d X ( g ( y ) , g ( y ′ )) − d Y ( y, y ′ ) | < ε for all y, y ′ ∈ Y . Similarly,for x ∈ X , d X ( g ◦ f ( x ) , x ) = d ( g ◦ f ( x ) , x ) ≤ d ( g ( f ( x )) , f ( x )) + d ( f ( x ) , x ) < ε, as well as d Y ( f ◦ g ( y ) , y ) < ε for all y ∈ Y . Thus, ( f, g ) ∈ Appr ε (( X, p ) , ( Y, q )) . This proves c).d) Fix an arbitrary pair ( f, g ) ∈ Appr ε (( X, p ) , ( Y, q )) . The definition ofan admissible metric d : ( X ∐ Y ) × ( X ∐ Y ) → R requires d | X × X := d X , d | Y × Y := d Y and d ( y, x ) := d ( x, y ) for x ∈ X and y ∈ Y . Hence, it suffices todefine d ( x, y ) for x ∈ X and y ∈ Y . Then d is positive definite and symmetricby definition. Thus, in order to prove that d is a metric, it remains to checkthe triangle inequality. If done so, then d is in fact an admissible metric.Define d : ( X ∐ Y ) × ( X ∐ Y ) → R via d ( x, y ) := ε { d X ( x, x ′ ) + d Y ( f ( x ′ ) , y ) | x ′ ∈ X } for x ∈ X and y ∈ Y . It remains to check the triangle inequality. For x , x ∈ X and y ∈ Y , d ( x , x ) + d ( x , y )= d X ( x , x ) + ε { d X ( x , x ′ ) + d Y ( f ( x ′ ) , y ) | x ′ ∈ X } = ε { d X ( x , x ) + d X ( x , x ′ ) + d Y ( f ( x ′ ) , y ) | x ′ ∈ X }≥ ε { d X ( x , x ′ ) + d Y ( f ( x ′ ) , y ) | x ′ ∈ X } = d ( x , y ) and d ( x , y ) + d ( y, x )= ε + inf { d X ( x , x ′ ) + d Y ( f ( x ′ ) , y )= ε + inf { + d X ( x , x ′′ ) + d Y ( f ( x ′′ ) , y ) | x ′ , x ′′ ∈ X }≥ ε + inf { d X ( x , x ′ ) + d Y ( f ( x ′ ) , f ( x ′′ )) + d X ( x , x ′′ ) | x ′ , x ′′ ∈ X } D. JANSEN ≥ ε + inf { d X ( x , x ′ ) + ( d X ( x ′ , x ′′ ) − ε ) + d X ( x , x ′′ ) | x ′ , x ′′ ∈ X }≥ inf { d X ( x , x ) | x ′ , x ′′ ∈ X } = d ( x , x ) . The two remaining triangle inequalities d ( x, y ) + d ( y , y ) ≥ d ( x, y ) and d ( y , x ) + d ( x, y ) ≥ d ( y , y ) , where x ∈ X and y , y ∈ Y , can be provenanalogously.Using this metric d , d ( p, q ) = ε { d X ( p, x ′ ) + d Y ( f ( x ′ ) , q ) | x ′ ∈ X } = ε since ≤ inf { d X ( p, x ′ ) + d Y ( f ( x ′ ) , q ) | x ′ ∈ X } ≤ d X ( p, p ) + d Y ( f ( p ) , q ) = 0 .Furthermore, for x ∈ X , d ( x, f ( x )) = ε { d X ( x, x ′ ) + d Y ( f ( x ′ ) , f ( x )) | x ′ ∈ X } = ε using x ′ = x . For y ∈ Y , this implies d ( y, g ( y )) ≤ d ( y, f ◦ g ( y )) + d ( f ◦ g ( y ) , g ( y )) < ε + ε ε . Thus, X ⊆ B dε/ ( f ( X )) ⊆ B d ε/ ( Y ) and Y ⊆ B d ε/ ( X ) , i.e. d d H ( X, Y ) ≤ ε and d GH (( X, p ) , ( Y, q )) ≤ d d H (( X, p ) , ( Y, q )) = d d H ( X, Y ) + d ( p, q ) ≤ ε . This proves d). (cid:3)
Using these approximations, one can prove that the pointed Gromov-Hausdorff distance defines a metric. Two pointed metric spaces ( X, p ) and ( Y, q ) are called isometric if there exists an isometry f : X → Y with f ( p ) = q . Proposition 1.6.
On the space of isometry classes of (pointed) compactmetric spaces, d GH defines a metric.Proof. In order to prove that the Gromov-Hausdorff distance indeed definesa metric, one needs that the Hausdorff distance defines a metric. Therefore,this proof splits into several steps: First, the Hausdorff distance will beinvestigated. Then it will be proven that the Gromov-Hausdorff distancedefines a pseudo-metric on the class of (pointed) compact metric spaces,i.e. it is not definite, but satisfies all the other properties of a metric. Finally,it will be proven that this already defines a metric up to isometry.
Step 1: d H defines a metric in the non-pointed case. Let ( X, d ) be a metricspace and A, B, C ⊆ X be compact. First, prove that d H is a metric in thenon-pointed case:By definition, d d H ( B, A ) = d d H ( A, B ) , d d H ( A, B ) ≥ and d d H ( A, A ) = 0 .In order to prove the triangle inequality, define r := d d H ( A, B ) ≥ and r := d d H ( B, C ) ≥ and let ε > be arbitrary. For a ∈ A there exists b ∈ B OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 7 with d ( a, b ) < r + ε . Furthermore, there is c ∈ C with d ( b, c ) < r + ε . Hence, d ( a, c ) < r + r + 2 ε and this proves A ⊆ B r + r +2 ε ( C ) . An analogousargumentation proves C ⊆ B r + r +2 ε ( A ) . Therefore, d d H ( A, C ) ≤ r + r +2 ε .Since ε > was arbitrary, d d H ( A, C ) ≤ r + r = d d H ( A, B ) + d d H ( B, C ) . Assume that A = B and d d H ( A, B ) = 0 . Without loss of generality, assumethere exists a ∈ A with a / ∈ B . In particular, d ( a, b ) > for all b ∈ B .Because B is compact, < inf { d ( a, b ) | b ∈ B } ≤ d d H ( A, B ) , and this is acontradiction. Step 2: d H defines a metric in the pointed case. Now fix a ∈ A , b ∈ B and c ∈ C . Since d H is a metric in the non-pointed case, d d H (( A, a ) , ( B, b )) = d d H ( A, B ) + d ( a, b ) ≥ and equality holds if and only if A = B and a = b . Obviously, d H issymmetric and d d H (( A, a ) , ( B, b )) + d d H (( B, b ) , ( C, c ))= d d H ( A, B ) + d d H ( B, C ) + d ( a, b ) + d ( b, c ) ≥ d d H ( A, C ) + d ( a, c )= d d H (( A, a ) , ( C, c )) . Thus, d H defines a metric. Step 3: d GH defines a pseudo-metric. From now on, the proof restricts tothe case of pointed metric spaces since the other one can be done completelyanalogously. Obviously, d GH is non-negative and symmetric. It remains toprove the triangle inequality. Let ( X, x ) , ( Y, y ) and ( Z, z ) be pointedcompact metric spaces. For arbitrary ε > , choose admissible metrics d XY on X ∐ Y and d Y Z on Y ∐ Z such that d d XY H (( X, x ) , ( Y, y )) < d GH (( X, x ) , ( Y, y )) + ε and d d Y Z H (( Y, y ) , ( Z, z )) < d GH (( Y, y ) , ( Z, z )) + ε . Define an admissible metric d XZ on X ∐ Z by d XZ ( x, z ) = inf { d XY ( x, y ) + d Y Z ( y, z ) | y ∈ Y } . This actually defines a metric: Since everything else is obvious, only thetriangle inequality needs to be checked. If all regarded points are containedin X or all in Z , there is nothing to prove. For x , x ∈ X and z ∈ Z , d XZ ( x , x ) + d XZ ( x , z )= d X ( x , x ) + inf { d XY ( x , y ′ ) + d Y Z ( y ′ , z ) | y ′ ∈ Y } = inf { d XY ( x , x ) + d XY ( x , y ′ ) + d Y Z ( y ′ , z ) | y ′ ∈ Y } D. JANSEN ≥ inf { d XY ( x , y ′ ) + d Y Z ( y ′ , z ) | y ′ ∈ Y } = d XZ ( x , z ) and d XZ ( x , z ) + d XZ ( z, x )= inf { d XY ( x , y ′ ) + d Y Z ( y ′ , z ) + d Y Z ( z, y ′′ ) + d XY ( y ′′ , x ) | y ′ , y ′′ ∈ Y }≥ inf { d XY ( x , y ′ ) + d Y ( y ′ , y ′′ ) + d XY ( y ′′ , x ) | y ′ , y ′′ ∈ Y }≥ inf { d XY ( x , y ′ ) + d XY ( y ′ , x ) | y ′ ∈ Y }≥ d X ( x , x )= d XZ ( x , x ) . The remaining triangle inequalities d XZ ( z , z ) + d XZ ( z , x ) ≥ d XZ ( z , x ) and d ( z , x ) + d XZ ( x, z ) ≥ d XZ ( z , z ) , where x ∈ X and z , z ∈ Z , can beproven analogously.With similar arguments, one can prove that d XY Z defines an admissiblemetric on X ∐ Y ∐ Z where d XY Z ( x, y ) := d XY ( x, y ) if x, y ∈ X ∐ Y,d XZ ( x, y ) if x, y ∈ X ∐ Z,d
Y Z ( x, y ) if x, y ∈ Y ∐ Z. With those admissible metrics, d GH (( X, x ) , ( Z, z )) ≤ d d XY Z H ( X, Z ) + d XY Z ( x , z ) ≤ d d XY Z H ( X, Y ) + d d XY Z H ( Y, Z ) + d XY Z ( x , y ) + d XY Z ( y , z ) ≤ d d XY H ( X, Y ) + d d Y Z H ( Y, Z ) + d XY ( x , y ) + d Y Z ( y , z ) < d GH (( X, x ) , ( Y, y )) + d GH (( Y, y ) , ( Z, z )) + 2 ε, where in the second last inequality the fact is used that for every r > theinclusion X ⊆ B d XY r ( Y ) implies the inclusion X ⊆ B d XY Z r ( Y ) . Now letting ε → proves the triangle inequality for d GH . Step 4: d GH defines a metric up to isometry. It is easy to see that thedistance of isometric pointed compact spaces vanishes: Let ( X, p ) and ( Y, q ) be isometric via isometries f and g . Then ( f, g ) ∈ Appr ε / (( X, p ) , ( Y, q )) for arbitrary ε > . By Proposition 1.5, d GH (( X, p ) , ( Y, q )) ≤ ε . Hence, d GH (( X, p ) , ( Y, q )) = 0 .Conversely, let ( X, p ) and ( Y, q ) be two pointed compact metric spacessatisfying d GH (( X, p ) , ( Y, q )) = 0 . By definition, for each n ≥ there isan admissible metric d n on X ∐ Y with d d n H ( X, Y ) + d n ( p, q ) < n . Since X is compact and thus separable, there exists a countable dense subset X ′ = { x i | i ∈ N } ⊆ X with x = p . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 9
Define y n := q . The constant sequence ( y n ) n ∈ N converges to q , and foreach n , d n ( x , y n ) = d n ( p, q ) < n .Because of d d n H ( X, Y ) < n , there exists some y n ∈ Y with d n ( x , y n ) < n .Since Y is compact, ( y n ) n has a convergent subsequence ( y n i ) i ∈ N with somelimit y ∈ Y . Then d n i ( x , y ) ≤ d n i ( x , y n i ) + d n i ( y n i , y ) → as i → ∞ . The same argument for x gives a subsequence d n ij of d n i and a point y ∈ Y with d n ij ( x , y ) → as j → ∞ . By a diagonal argument, there is asubsequence d l of d n and a sequence ( y i ) i ∈ N with y = q with d l ( x i , y i ) → as l → ∞ for all i .Define f : X ′ → Y by f ( x i ) := y i . Since the d l are admissible metrics, foreach l , d Y ( f ( x i ) , f ( x j )) = d l ( f ( x i ) , f ( x j )) = d l ( y i , y j ) and d X ( x i , x j ) = d l ( x i , x j ) . Therefore, | d Y ( f ( x i ) , f ( x j )) − d X ( x i , x j ) | = | d l ( y i , y j ) − d l ( x i , x j ) |≤ d l ( y i , x i ) + d l ( x j , y j )) → as l → ∞ . Hence, f is an isometry. Since X ′ is dense, f can be extended uniquely to anisometric embedding f : X → Y with f ( p ) = q . With a similar constructionand using a subsequence of d l , there is an isometric embedding g : Y → X with g ( q ) = p . After passing to this subsequence, for each x , d l ( g ◦ f ( x ) , x ) ≤ d l ( g ( f ( x )) , f ( x )) + d l ( f ( x ) , x ) → as l → ∞ . Thus, f is an isometry with f ( p ) = q , i.e. ( X, p ) and ( Y, q ) are isometric. (cid:3) The definitions of pointed and non-pointed Gromov-Hausdorff distanceessentially give the same notion of convergence. This will be proven next.
Proposition 1.7.
Let X and Y be compact metric spaces. a) For each x ∈ X and y ∈ Y , d GH ( X, Y ) ≤ d GH (( X, x ) , ( Y, y )) . b) For any x ∈ X there exists y ∈ Y such that d GH (( X, x ) , ( Y, y )) ≤ · d GH ( X, Y ) . Proof.
Both statements follow easily from the definitions:a) First, let x ∈ X and y ∈ Y be arbitrary. Then d GH ( X, Y ) = inf { d d H ( X, Y ) | d admissible metric on X ∐ Y }≤ inf { d d H ( X, Y ) + d ( x, y ) | d admissible metric on X ∐ Y } = inf { d d H (( X, x ) , ( Y, y )) | d admissible metric on X ∐ Y } = d GH (( X, x ) , ( Y, y )) . b) Now let r := d GH ( X, Y ) ≥ . For arbitrary n ∈ N , let d n be an admissiblemetric on X ∐ Y satisfying d d n H ( X, Y ) < d GH ( X, Y ) + 1 n = r + 1 n . Thus, X ⊆ ¯ B d n r +1 /n ( Y ) , i.e. for given x ∈ X there exists y n ∈ Y such that d n ( x, y n ) ≤ r + n . Since Y is compact, there exists a convergent subsequence ( y n m ) m ∈ N of ( y n ) n ∈ N with limit y ∈ Y . Then d d nm H (( X, x ) , ( Y, y ))= d d nm H ( X, Y ) + d n m ( x, y ) ≤ r + 1 n m + d n m ( x, y n m ) + d n m ( y n m , y ) ≤ r + 2 n m + d Y ( y n m , y ) and d GH (( X, x ) , ( Y, y ))= inf { d d H (( X, x ) , ( Y, y )) | d admissible metric on X ∐ Y }≤ inf { d d nm H (( X, x ) , ( Y, y )) | m ∈ N }≤ inf { r + 2 n m + d Y ( y n m , y ) | m ∈ N } = 2 r. (cid:3) It is not hard to give examples where the inequality in Proposition 1.7 a)is strict or where equality in b) holds for either all or none of the points. Inorder to improve readability of the example, the following two statementsare proven first.
Lemma 1.8.
Let ( X, d ) be a metric space, a ∈ A ⊆ X and b ∈ X . Then d H (( A, a ) , ( { b } , b )) ≥ d H (( A, a ) , ( { a } , a )) = sup { d ( a, a ′ ) | a ′ ∈ A } .Proof. First, recall d H ( A, { b } ) = inf { r > | A ⊆ B r ( b ) , b ∈ B r ( A ) } = max { inf { r > | A ⊆ B r ( b ) } , inf { r > | b ∈ B r ( A ) }} = max { sup { d ( a ′ , b ) | a ′ ∈ A } , d ( A, b ) } . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 11
In particular, d H (( A, a ) , ( { a } , a )) = d H ( A, { a } ) = sup { d ( a ′ , a ) | a ′ ∈ A } . Moreover, d H (( A, a ) , ( { b } , b )) = d H ( A, { b } ) + d ( a, b )= max { sup { d ( a ′ , b ) | a ′ ∈ A } , d ( A, b ) } + d ( a, b ) ≥ sup { d ( a ′ , b ) + d ( a, b ) | a ′ ∈ A }≥ sup { d ( a ′ , a ) | a ′ ∈ A } = d H (( A, a ) , ( { a } , a )) . (cid:3) Proposition 1.9.
Let ( X, d X ) be a compact metric space and x ∈ X . Then d GH (( X, x ) , ( { pt } , pt)) = sup { d X ( x, p ) | p ∈ X } .Proof. By Lemma 1.8, d GH (( X, x ) , ( { pt } , pt)) = inf { d d H (( X, x ) , ( { pt } , pt)) | d adm. on X ∐ { pt }}≥ inf { sup { d ( x, p ) | p ∈ X } | d adm. on X ∐ { pt }} = sup { d X ( x, p ) | p ∈ X } = d d X H (( X, x ) , ( { x } , x )) . On the other hand, d GH (( X, x ) , ( { pt } , pt)) ≤ d d X H (( X, x ) , ( { x } , x )) and this proves the claim. (cid:3) The following examples proves that d GH (( X, x ) , ( Y, y )) may attain anyvalue between d GH ( X, Y ) and · d GH ( X, Y ) . Example 1.10.
Let D = { x ∈ R | k x k ≤ } denote the disk of radius in R . By Proposition 1.9, for arbitrary x ∈ D , d GH (( D , x ) , ( { pt } , pt)) = sup { d D ( x, p ) | p ∈ D } = k x k + 1 . Hence, for any λ ∈ [1 , , every point x with k x k = λ − satisfies d GH (( D , x ) , ( { pt } , pt)) = λ · d GH ( D , { pt } ) . In particular, two extreme cases occur in the situation of Proposition 1.7:For X = D , Y = { pt } and x = (0 , ∈ R , there is no y ∈ Y with d GH (( X, x ) , ( Y, y )) = 2 · d GH ( X, Y ) . On the contrary, in this case, d GH (( X, x ) , ( Y, y )) = d GH ( X, Y ) for all y ∈ Y .On the other hand, if x = (1 , ∈ R , then d GH (( X, x ) , ( Y, y )) = 2 · d GH ( X, Y ) for all y ∈ Y . Definition 1.11.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed compactmetric spaces. a) If d GH ( X i , X ) → as i → ∞ , then X i converges to X .b) If d GH (( X i , p i ) , ( X, p )) → as i → ∞ , then ( X i , p i ) converges to ( X, p ) .If X i converges to X , this is denoted by X i → X . If ( X i , p i ) converges to ( X, p ) , this is denoted by ( X i , p i ) → ( X, p ) . Corollary 1.12.
Let ( X, d X ) and ( X i , d X i ) , i ∈ N , be compact metric spaces. a) If ( X i , x i ) → ( X, x ) for some x i ∈ X i and x ∈ X , then X i → X aswell. b) If X i → X and x ∈ X , then there exist points x i ∈ X i such that ( X i , x i ) → ( X, x ) .Proof. This is a direct consequence of Proposition 1.9. (cid:3)
Recall that a metric space ( X, d X ) is called length space if d ( x, y ) = inf { L ( c ) | c continuous curve from x to y } for any x, y ∈ X , where L ( c ) denotes the length of c . Proposition 1.13.
A complete compact Gromov-Hausdorff limit of compactlength spaces is a length space.
In the proof, the following statement is used.
Lemma 1.14 (cf. [BBI01, Theorem 2.4.16]) . Let ( X, d ) be a complete metricspace. Then ( X, d ) is a length space if and only if for all x, y ∈ X and ε > there exists an ε -midpoint, i.e. a point z ∈ X with | d ( x, z ) − d ( x, y ) | ≤ ε and | d ( y, z ) − d ( x, y ) | ≤ ε ,Proof. First, let ( X, d ) be a length space and x, y ∈ X and ε > be arbitrary.Since X is a length space, there exists a curve c : [0 , L ] → X with c (0) = x , c ( L ) = y and length L ( c ) ≤ d ( x, y ) + ε . Without loss of generality, assume c to be parametrised by arc length. In particular, L = L ( c ) ≤ d ( x, y ) + ε .Define z := c ( L ) . Clearly, d ( x, z ) ≤ · L ( c | [0 , L ] ) = L ≤ d ( x, y ) + ε, and analogously, d ( y, z ) ≤ d ( x, y ) + ε . Now assume d ( y, z ) − d ( x, z ) > ε .Then d ( x, y ) ≤ d ( x, z ) + d ( z, y ) < d ( y, z ) − ε ≤ d ( x, y ) , and this is a contradiction. Hence, d ( x, y ) − d ( x, z ) ≤ d ( y, z ) − d ( x, z ) ≤ ε . Analogously, | d ( y, z ) − d ( x, y ) | ≤ ε .Now let X be a metric space such that for all pairs of points and ε > there exists an ε -midpoint, and let x, y ∈ X be arbitrary. If for every ε > there is a curve γ connecting x and y of length L ( γ ) ≤ d ( x, y ) + ε , then inf { L ( γ ) | γ connects x and y } = d ( x, y ) OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 13 and this proves that ( X, d ) is a length space.So, let L := d ( x, y ) , ε > be arbitrary and define γ inductively as follows:First, let γ (0) = x and γ (1) = y . Now, assume γ ( k m ) to be defined for some m ∈ N and all k ∈ N with ≤ k ≤ m . For odd ≤ k ≤ m +1 − , let γ ( k m +1 ) be an ε m +1 -midpoint of γ ( k − m +1 ) and γ ( k +12 m +1 ) .Inductively, d ( γ ( k m ) , γ ( k +12 m )) ≤ L m + ε m · P mi =1 12 i : For m = 0 , by defini-tion, d ( γ (0) , γ (1)) = L . Let the statement be true for some m ∈ N , and let ≤ k ≤ m +1 − . First assume k = 2 l + 1 to be odd. Then · d (cid:0) γ (cid:16) k m +1 (cid:17) , γ (cid:16) k + 12 m +1 (cid:17)(cid:1) ≤ ε m +1 + d (cid:0) γ (cid:16) l m (cid:17) , γ (cid:16) l + 12 m (cid:17)(cid:1) ≤ ε m +1 + L m + ε m · m X i =1 i = L m + ε m · m +1 X i =1 i . The proof for even k can be done analogously. Observe d (cid:0) γ (cid:16) k m (cid:17) , γ (cid:16) k + 12 m (cid:17)(cid:1) ≤ L m + ε m · m X i =1 i ≤ L + ε m . Hence, for all m ∈ N and ≤ k < l ≤ m , d (cid:0) γ (cid:16) k m (cid:17) , γ (cid:16) l m (cid:17)(cid:1) ≤ l − X j = k d (cid:0) γ (cid:16) j m (cid:17) , γ (cid:16) j + 12 m (cid:17)(cid:1) ≤ l − X j = k L + ε m = ( L + ε ) · (cid:16) l m − k m (cid:17) . In particular, defined as a function on the dyadic numbers in [0 , L ] , γ isLipschitz. Thus, it can be extended to a Lipschitz, hence continuous, curve γ : [0 , L ] → X where γ ( t ) is defined as the limit of γ ( t n ) for dyadic numbers t n → t . For such ≤ s < t ≤ L and dyadic numbers s n → s and t n → t , d ( γ ( s ) , γ ( t )) = lim n →∞ d ( γ ( s n ) , γ ( t n )) ≤ lim n →∞ ( L + ε ) · | t n − s n | = ( L + ε ) · ( t − s ) . Therefore, L ( γ ) = sup n N − X i =0 d ( γ ( t i ) , γ ( t i +1 )) | N ∈ N , t < t < . . . < t N = 1 o ≤ sup n N − X i =0 ( L + ε ) · ( t i +1 − t i ) | N ∈ N , t < t < . . . < t N = 1 o = L + ε . (cid:3) Proof of Proposition 1.13.
Let x, y ∈ X and ε > be arbitrary. ApplyingLemma 1.14, it is enough to find an ε -midpoint z of x and y .Choose i ∈ N such that d GH ( X i , X ) < ε . By Proposition 1.5, there exist ( f, g ) ∈ Appr ε / ( X i , X ) . Let z ′ be an ε -midpoint of g ( x ) and g ( y ) , anddefine z := f ( z ′ ) . Then | d X ( x, z ) − d X ( x, y ) | ≤ | d X ( x, z ) − d X i ( g ( x ) , g ( z )) | + | d X i ( g ( x ) , g ( z )) − d X i ( g ( x ) , z ′ ) | + | d X i ( g ( x ) , z ′ ) − d X i ( g ( x ) , g ( y )) | + | d X i ( g ( x ) , g ( y )) − d X ( x, y ) | < · ε · d X i ( g ◦ f ( z ′ ) , z ′ ) | + ε ε < ε . Analogously, | d X ( y, z ) − d X ( x, y ) | < ε . (cid:3) In general, the Gromov-Hausdorff distance of two subsets of the samemetric space, equipped with the induced metric, can be estimated by theirHausdorff distance. If this metric space is a length space and the subsets areballs, this estimate can be expressed by using the radii and the distance ofthe base points. This uses the property of length spaces that r -ball arounda ball of radius s coincides with the r + s ball (around the same base point). Lemma 1.15.
Let ( X, d ) be a length space, p ∈ X and r, s > . Then B r ( B s ( p )) = B r + s ( p ) . Proof.
Let q ∈ B r ( B s ( p )) , i.e. there exists x ∈ B s ( p ) with d ( x, q ) < r . Then d ( q, p ) ≤ d ( q, x ) + d ( x, p ) < r + s proves B r ( B s ( p )) ⊆ B r + s ( p ) . In fact, this inclusion holds in every metricspace.Conversely, let q ∈ B r + s ( p ) . Since B s ( p ) ⊆ B r ( B s ( p )) , without loss ofgenerality, assume q ∈ B r + s ( p ) \ B s ( p ) . Let l := d ( p, q ) denote the distanceof p and q . In particular, s ≤ l < r + s . Fix a shortest geodesic γ : [0 , l ] → X with γ (0) = p and γ ( l ) = q . Define ε := · min { s, r + s − l } > and t := s − ε ∈ (0 , s ) ⊆ [0 , l ] . Then d ( γ ( t ) , p ) = t < s and d ( γ ( t ) , q ) = l − t = l − s + ε < l − s + r + s − l = r. Hence, γ ( t ) ∈ B s ( p ) and q ∈ B r ( γ ( t )) . Thus, B r + s ( p ) ⊆ B r ( B s ( p )) . (cid:3) Lemma 1.16.
Let ( X, d ) be a length space, p, q ∈ X , r, s > . Then d d H ( ¯ B r ( p ) , ¯ B s ( q )) ≤ d ( p, q ) + | r − s | . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 15
Proof.
Let ε := d ( p, q ) + | r − s | . If ε = 0 , the claim holds due to p = q and r = s . Hence, assume ε > . Then, applying Lemma 1.15, B r ( p ) ⊆ B d ( p,q )+ r ( q ) ⊆ B d ( p,q )+ | r − s | + s ( q ) = B ε + s ( q ) = B ε ( B s ( q )) . Analogously, B s ( q ) ⊆ B ε ( B r ( p )) . Therefore, d d H ( ¯ B r ( p ) , ¯ B s ( q )) = d d H ( B r ( p ) , B s ( q )) ≤ ε . (cid:3) Corollary 1.17.
Let ( X, d ) be a length space, p, q ∈ X , r, s > . Then a) d GH (( ¯ B Xr ( p ) , p ) , ( ¯ B Xs ( p ) , p )) ≤ | r − s | , b) d GH (( ¯ B Xr ( p ) , p ) , ( ¯ B Xr ( q ) , q )) ≤ d ( p, q ) . The diameters of metric spaces with small Gromov-Hausdorff distance arealmost the same. In particular, for a convergent sequence of metric spaces,their diameters converge to the diameter of the limit space.
Proposition 1.18.
For compact metric spaces ( X, d X ) and ( Y, d Y ) , | diam( X ) − diam( Y ) | ≤ d GH ( X, Y ) . In particular, if X i → X for compact metric spaces ( X i , d X i ) , i ∈ N , then diam( X i ) → diam( X ) . Proof.
Let ε := d GH ( X, Y ) , δ > and d be an admissible metric on X ∐ Y such that d d H ( X, Y ) < d GH ( X, Y ) + δ = ε + δ. This implies Y ⊆ B dε + δ ( X ) . Thus, for any y , y ∈ Y there are x , x ∈ X with d ( x i , y i ) < ε + δ for ≤ i ≤ . Hence, d Y ( y , y ) ≤ d ( y , x ) + d X ( x , x ) + d ( x , y ) < ε +2 δ + diam( X ) . Therefore, diam( Y ) = sup { d Y ( y , y ) | y , y ∈ Y } ≤ diam( X ) + 2 ε +2 δ. Since δ > was arbitrary, diam( Y ) ≤ diam( X ) + 2 ε . The other inequalitycan be proven analogously. (cid:3) Corollary 1.19. If ( X, d ) is a compact metric space and { pt } the spaceconsisting of only one point, then d GH ( X, { pt } ) = · diam( X ) .Proof. By Proposition 1.18, diam( X ) ≤ · d GH ( X, { pt } ) . Thus, only theother inequality has to be proven.Let δ := · diam( X ) , and define an admissible metric d on the disjointunion X ∐ { pt } by d ( x, pt) := δ . As usually, only the triangle inequalityneeds to be checked. For arbitrary x , x ∈ X , d ( x , x ) + d ( x , pt) = d ( x , x ) + δ ≥ δ = d ( x , pt) and d ( x , pt) + d (pt , x ) = 2 δ = diam( X ) ≥ d ( x , x ) . Using this metric, d GH ( X, { pt } ) ≤ d d H ( X, { pt } ) = δ. (cid:3) For a metric space ( X, d X ) , let λX denote the rescaled metric space ( λX, d λX ) := ( X, λd X ) . Rescaling of compact metric spaces behaves nicelyunder Gromov-Hausdorff distance. For any p ∈ X and r > , observe B Xr ( p ) = { q ∈ X | d X ( q, p ) < r } = { q ∈ X | λd X ( q, p ) < λr } = B λXλr ( p ) . Lemma 1.20.
Let ( X, d X ) and ( Y, d Y ) be compact metric spaces.For the Hausdorff distance, d λX H = λ · d X H (both in the standard and in thepointed case).For the Gromov-Hausdorff distance, both d GH ( λX, λY ) = λ · d GH ( X, Y ) and, for all x ∈ X and y ∈ Y , d GH (( λX, x ) , ( λY, y )) = λ · d GH (( X, x ) , ( Y, y )) .Proof. First, let
A, B ⊆ X . Then d λX H ( A, B ) = inf { ε > | A ⊆ B λXε ( B ) and B ⊆ B λXε ( A ) } = inf { λ ˜ ε > | A ⊆ B X ˜ ε ( B ) and B ⊆ B X ˜ ε ( A ) } = λ · inf { ˜ ε > | A ⊆ B X ˜ ε ( B ) and B ⊆ B X ˜ ε ( A ) } = λ · d X H ( A, B ) . Furthermore, for a ∈ A and b ∈ B , d λX H (( A, a ) , ( B, b )) = d λX H ( A, B ) + d λX ( a, b )= λ · d X H ( A, B ) + λ · d X ( a, b )= λ · d X H (( A, a ) , ( B, b )) . By definition, an admissible metric ˜ d on λX ∐ λY is a metric on X ∐ Y satisfying ˜ d | X × X = d λX = λ · d X and ˜ d | Y × Y = d λY = λ · d Y . Furthermore, d := λ · ˜ d is a metric if and only if ˜ d is a metric. In addition, this metric d satisfies d | X × X = λ · ˜ d | X × X = d X and d | Y × Y = d Y . Thus, d is an admissiblemetric on X ∐ Y . On the other hand, using similar arguments, if d is anadmissible metric on X ∐ Y , then ˜ d := λ · d is an admissible metric on λX ∐ λY . Hence, d GH ( λX, λY ) = inf { d ˜ d H ( λX, λY ) | ˜ d admissible metric on λX ∐ λY } = inf { d λd H ( λX, λY ) | λ · d admissible metric on λX ∐ λY } = inf { λ · d d H ( λX, λY ) | d admissible metric on X ∐ Y } = λ · d GH ( X, Y ) . Analogously, d GH (( λX, x ) , ( λY, y )) = λ · d GH (( X, x ) , ( Y, y )) . (cid:3) The non-compact case
For non-compact metric spaces, the above way of defining a metric (upto isometry) does not work: Using the Hausdorff distance as before on un-bounded sets may give distance infinity. Thus, instead of defining a notionof distance for non-compact metric spaces, convergence is defined by using
OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 17 compact subspaces of these spaces only. On these, the previous definitionscan be applied.A metric space is called proper if all closed balls are compact. Throughoutthe remaining section, all metric spaces will assumed to be proper. Noticethat proper metric spaces are complete.For a metric space ( X, d X ) , p ∈ X and r > , let ¯ B r ( p ) := { q ∈ X | d X ( p, q ) ≤ r } denote the closed ball of radius r around p . Definition 2.1.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed propermetric spaces. If d GH (( ¯ B X i r ( p i ) , p i ) , ( ¯ B Xr ( p ) , p )) → as i → ∞ for all r > , where the balls are equipped with the restricted metric, then ( X i , p i ) converges to ( X, p ) (in the pointed Gromov-Hausdorff sense) . If ( X i , p i ) converges to ( X, p ) , this is denoted by ( X i , p i ) → ( X, p ) and ( X, p ) is called the (pointed Gromov-Hausdorff) limit of ( X i , p i ) .Frequently, a sequence ( X i , p i ) does not converge itself but has a con-verging subsequence. The limit of such a subsequence is called sublimit of ( X i , p i ) , and ( X i , p i ) is said to subconverge to this limit.Naturally, the question arises under which conditions a given sequence ofmetric spaces converges in the pointed Gromov-Hausdorff sense. For mani-folds, the following theorem by Gromov states that in some cases at leasta (Gromov-Hausdorff) sublimit exists. In section 3, another, more generalconcept of creating and guaranteeing ‘limits’ will be introduced. It will turnout that these limits in fact are Gromov-Hausdorff sublimits as well. Theorem 2.2 (Gromov’s Pre-compactness Theorem, [Pet06, Cor. 1.11]) . For n ≥ , κ ∈ R and D > , the following classes are pre-compact, i.e. everysequence in the class has a convergent subsequence whose limit lies in theclosure of this class: a) The collection of n -dimensional closed Riemannian manifolds with Ric ≥ ( n − · κ and diam ≤ D . b) The collection of n -dimensional pointed complete Riemannian mani-folds with Ric ≥ ( n − · κ . The section is structured as follows: In subsection 2.1, the compabilityof the definition of pointed Gromov-Hausdorff convergence in Definition 2.1with the notion of convergence induced by the Gromov-Hausdorff distanceof compact metric (length) spaces (Definition 1.1 and Definition 1.2) is ver-ified. Subsequently, subsection 2.2 deals with stating and verifying severalproperties of pointed Gromov-Hausdorff convergence. In this context, con-vergence of points and convergence of maps, respectively, are introduced insubsection 2.3 and subsection 2.4, respectively.
Comparison with the compact case.
Applied to compact lengthspaces, the convergence in the pointed Gromov-Hausdorff sense coincideswith the convergence of compact metric spaces in the pointed sense defined inthe previous section. Conversely, given (non-pointed) convergence as definedfor compact metric spaces and a fixed base point in the limit space, thereexist base points such that the spaces converge in the pointed Gromov-Haus-dorff sense.In order to prove this, one uses the fact that approximations can be re-stricted to smaller balls. This is shown in the following lemma. Anotherstatement of the lemma is that base points can be changed in a certain way.This will be useful later on as well.
Lemma 2.3.
Let ( X, d X ) and ( Y, d Y ) be length spaces, a) Let p, p ′ ∈ X , q, q ′ ∈ Y and R ≥ r > satisfy ¯ B Xr ( p ′ ) ⊆ ¯ B XR ( p ) and ¯ B Yr ( q ′ ) ⊆ ¯ B YR ( q ) . Moreover, let ε > , ( f, g ) ∈ Appr ε (( ¯ B XR ( p ) , p ) , ( ¯ B YR ( q ) , q )) and δ := max { d ( f ( p ′ ) , q ′ ) , d ( p ′ , g ( q ′ )) } ≥ . ThenAppr ε + δ (( ¯ B Xr ( p ′ ) , p ′ ) , ( ¯ B Yr ( q ′ ) , q ′ )) = ∅ and d GH (( ¯ B Xr ( p ′ ) , p ′ ) , ( ¯ B Yr ( q ′ ) , q ′ )) ≤ ε +2 δ. b) For p ∈ X , q ∈ Y and R ≥ r > , d GH (( ¯ B Xr ( p ) , p ) , ( ¯ B Yr ( q ) , q )) ≤ · d GH (( ¯ B XR ( p ) , p ) , ( ¯ B YR ( q ) , q )) . Proof. a) For simplicity, let δ f := d ( f ( p ′ ) , q ′ ) and δ g := d ( p ′ , g ( q ′ )) . Inparticular, δ = max { δ f , δ g } . Let ˜ ε := 4 ε + δ . As ¯ B Xr ( p ′ ) ⊆ ¯ B XR ( p ) , one canrestrict f to ¯ B Xr ( p ′ ) . For x ∈ ¯ B Xr ( p ′ ) , d Y ( f ( x ) , q ′ ) ≤ d Y ( f ( x ) , f ( p ′ )) + d Y ( f ( p ′ ) , q ′ ) ≤ ( d X ( x, p ′ ) + ε ) + δ f < r + ε + δ f . Hence, f ( ¯ B Xr ( p ′ )) ⊆ ¯ B Yr + ε + δ f ( q ′ ) . Analogously, one can prove the inclusion g ( ¯ B Yr ( q ′ )) ⊆ ¯ B Xr + ε + δ g ( p ′ ) . Now modify f and g in order to obtain maps ˜ f and ˜ g , respectively, whose images are contained in ¯ B Yr ( q ′ ) and ¯ B Xr ( p ′ ) ,respectively, such that ( ˜ f , ˜ g ) are ˜ ε -approximations:For y ∈ ¯ B Yr + ε + δ f ( q ′ ) \ ¯ B Yr ( q ′ ) choose a shortest geodesic c : [0 , l ] → Y with c (0) = q ′ and c (1) = y where r < l := d Y ( y, q ′ ) ≤ r + ε + δ f . Then d Y ( c ( r ) , q ′ ) = r , in particular, c ( r ) ∈ ¯ B Yr ( q ′ ) , and for ˆ y := c ( r ) , d ( y, ˆ y ) = d Y ( y, q ′ ) − d Y (ˆ y, q ′ ) < ( r + ε + δ f ) − r = ε + δ f . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 19
Using this, define ˜ f : ¯ B Xr ( p ′ ) → ¯ B Yr ( q ′ ) by ˜ f ( x ) := q ′ if x = p ′ ,f ( x ) if x = p ′ and f ( x ) ∈ ¯ B Yr ( q ′ ) , d f ( x ) if x = p ′ and f ( x ) / ∈ ¯ B Yr ( q ′ ) . Since d Y ( ˜ f ( p ′ ) , f ( p ′ )) = d Y ( q ′ , f ( p ′ )) = δ f < ε + δ f and by construction, d Y ( ˜ f ( x ) , f ( x )) < ε + δ f for all x ∈ ¯ B Xr ( p ′ ) . Similarly, define ˜ g : ¯ B Yr ( q ′ ) → ¯ B Xr ( p ′ ) . Using analogousarguments proves d X (˜ g ( y ) , g ( y )) < ε + δ g for all y ∈ ¯ B Yr ( q ′ ) .By definition, ˜ f ( p ′ ) = q ′ and ˜ g ( q ′ ) = p ′ , so it remains to prove that ( ˜ f , ˜ g ) are ˜ ε -approximations. By construction, | d X ( x , x ) − d Y ( ˜ f ( x ) , ˜ f ( x )) |≤ | d X ( x , x ) − d Y ( f ( x ) , f ( x )) | + | d Y ( f ( x ) , f ( x )) − d Y ( ˜ f ( x ) , ˜ f ( x )) | < ε +( d Y ( f ( x ) , ˜ f ( x )) + d Y ( f ( x ) , ˜ f ( x ))) < ε +2( ε + δ f ) < ˜ ε, where x , x ∈ ¯ B Xr ( p ′ ) . Analogously, | d Y ( y , y ) − d X (˜ g ( y ) , ˜ g ( y )) | < ˜ ε forarbitrary y , y ∈ ¯ B Yr ( q ′ ) . Furthermore, for x ∈ ¯ B Xr ( p ′ ) , d X ( x, ˜ g ◦ ˜ f ( x )) ≤ d X ( x, g ◦ f ( x )) + d X ( g ◦ f ( x ) , g ◦ ˜ f ( x )) + d X ( g ◦ ˜ f ( x ) , ˜ g ◦ ˜ f ( x )) < ε +( ε + d Y ( f ( x ) , ˜ f ( x ))) + ( ε + δ g ) < ε + δ f + δ g = ˜ ε. Analogously, d Y ( y, ˜ f ◦ ˜ g ( y )) < ˜ ε for all y ∈ ¯ B Yr ( q ′ ) . Hence, ( ˜ f , ˜ g ) ∈ Appr ˜ ε (( ¯ B Xr ( p ′ ) , p ′ ) , ( ¯ B Yr ( q ′ ) , q ′ )) , and by Proposition 1.5, d GH (( ¯ B Xr ( p ′ ) , p ′ ) , ( ¯ B Yr ( q ′ ) , q ′ )) ≤ ε. b) Let δ > be arbitrary and ε := d GH (( ¯ B XR ( p ) , p ) , ( ¯ B YR ( q ) , q )) + δ > . ByProposition 1.5, Appr ε (( ¯ B XR ( p ) , p ) , ( ¯ B YR ( q ) , q )) = ∅ , and by a), d GH (( ¯ B Xr ( p ) , p ) , ( ¯ B Yr ( q ) , q )) ≤ ε = 16 · d GH (( ¯ B XR ( p ) , p ) , ( ¯ B YR ( q ) , q )) + 16 δ. Since δ > was arbitrary, this implies the claim. (cid:3) In order to avoid confusion, for the next two statements, let X i GH → X and ( X i , p i ) GH → ( X, p ) , respectively, denote the convergence of compact metricspaces in the sense of Definition 1.1 and Definition 1.2, respectively. Further,denote by ( X i , p i ) pGH → ( X, p ) the convergence in the pointed Gromov-Haus-dorff sense of Definition 2.1. Proposition 2.4.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed com-pact length spaces with ( X i , p i ) pGH → ( X, p ) . Then X i GH → X , in particular, diam( X i ) → diam( X ) .Proof. Assume (diam( X i )) i ∈ N is not bounded. Let r > diam( X ) . Withoutloss of generality, assume diam( X i ) > r for all i ∈ N .Let < ε < r − diam( X ) and choose points x i , y i ∈ B X i r ( p i ) satisfying d X i ( x i , y i ) ≥ r − ε . Let ε i := 2 · d GH (( X i , p i ) , ( X, p )) and fix approximations ( f i , g i ) ∈ Appr ε i (( X i , p i ) , ( X, p )) . Then diam( X ) ≥ d X ( f i ( x i ) , f i ( y i )) ≥ r − ε − ε i . Since this holds for all i ∈ N , diam( X ) ≥ r − ε > diam( X ) + ε . This is a contradiction. Thus, there is an
R > diam( X ) with diam( X i ) < R for all i ∈ N . Then d GH ( X i , X ) = d GH ( ¯ B X i R ( p i ) , ¯ B XR ( p )) ≤ d GH (( ¯ B X i R ( p i ) , p i ) , ( ¯ B XR ( p ) , p )) → as i → ∞ . Hence, X i → X . Proposition 1.18 implies the second part of the claim. (cid:3) Corollary 2.5.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed compactlength spaces. Then ( X i , p i ) GH → ( X, p ) if and only if ( X i , p i ) pGH → ( X, p ) .Proof. The proof is done by proving both implications separately. First,assume ( X i , p i ) GH → ( X, p ) and let r > be arbitrary.By Proposition 1.18, diam( X i ) → diam( X ) , i.e. without loss of generality,assume a strict diameter bound D on all spaces X i and X . In particular, forall r ≥ D , ( ¯ B X i r ( p i ) , p i ) = ( X i , p i ) converges to ( X, p ) = ( ¯ B Xr ( p ) , p ) . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 21
For < r < D , d GH (( ¯ B X i r ( p i ) , p i ) , ( ¯ B Xr ( p ) , p )) ≤ · d GH (( ¯ B X i D ( p i ) , p i ) , ( ¯ B XD ( p ) , p ))= 16 · d GH (( X i , p i ) , ( X, p )) → by Lemma 2.3. Hence, ( X i , p i ) pGH → ( X, p ) .Now let ( X i , p i ) pGH → ( X, p ) . By Proposition 2.4, diam( X i ) → diam( X ) .Without loss of generality, assume diam( X i ) ≤ X ) =: r . Thus, d GH (( X i , p i ) , ( X, p )) = d GH (( ¯ B X i r ( p i ) , p i ) , ( ¯ B Xr ( p ) , p )) → . (cid:3) In particular, if X i , X are compact and p ∈ X , then, by Corollary 1.12,there exist p i ∈ X i such that ( X i , p i ) GH → ( X, p ) . Hence, ( X i , p i ) pGH → ( X, p ) .From now on, let ( X i , p i ) → ( X, p ) denote convergence in the pointedGromov-Hausdorff sense.2.2. Properties as in the compact case.
This subsection deals with sev-eral properties which are familiar from the compact case. First of all, theGromov-Hausdorff distance defines a metric on the set of the isometry classesof compact metric spaces. In the non-compact case, the limit of pointed Gro-mov-Hausdorff convergence still is unique up to isometry.
Proposition 2.6.
Let ( X, d X , p ) , ( Y, d Y , q ) and ( X i , d X i , p i ) , i ∈ N , bepointed length spaces. Assume ( X i , p i ) → ( X, p ) and ( X i , p i ) → ( Y, q ) . Then ( X, p ) and ( Y, q ) are isometric.Proof. For every r > , both ¯ B Xr ( p ) and ¯ B Yr ( q ) are limits of ¯ B X i r ( p i ) , andthus, there exists a (bijective) isometry f r : ¯ B Xr ( p ) → ¯ B Yr ( q ) with f r ( p ) = q .Choose a countable dense subset X ′ := { x , x , x , . . . } of X with x = p , fix i ∈ N and let N i be the minimal natural number with d ( x i , q ) < N i . Define y ni := q if n < N i and y ni := f n ( x i ) otherwise. For n ≥ N i , d Y ( y ni , q ) = d Y ( f n ( x i ) , f n ( p )) = d X ( x i , p ) , i.e. ( y ni ) n ∈ N is a sequence in the compact subset ¯ B Yd X ( x i ,p ) ( q ) . By a diagonalargument, there exists a subsequence ( n m ) m ∈ N of the natural numbers suchthat for every i ∈ N the sequence ( y n m i ) m ∈ N has a limit y i ∈ ¯ B Yd X ( x i ,p ) ( q ) . Inparticular, y n = f n ( p ) = q for all n ∈ N implies y = q . For i, j ∈ N , byconstruction, d Y ( y i , y j ) = lim m →∞ d Y ( y n m i , y n m j )= lim m →∞ d Y ( f n m ( x i ) , f n m ( x j ))= d X ( x i , x j ) , i.e. the map ˜ f : X ′ → Y defined by ˜ f ( x i ) := y i is an isometry with ˜ f ( p ) = q . As Y is complete, there exists an extension of ˜ f to an isometry f : X → Y with f ( p ) = q : Let x ∈ X be arbitrary. Since X ′ was chosen to be dense,there exists a sequence ( x i j ) j ∈ N in X ′ converging to x . This is a Cauchysequence, hence, ( ˜ f ( x i j )) j ∈ N is a Cauchy sequence as well and has a limit y =: f ( x ) .This defines indeed an isometry f : X → Y : Let x, x ′ ∈ X be arbitraryand x i j and x i l , respectively, be sequences in X ′ converging to x and x ′ ,respectively. Then d Y ( f ( x ) , f ( x ′ )) = lim j,l →∞ d Y ( ˜ f ( x i j ) , ˜ f ( x i l ))= lim j,l →∞ d X ( x i j , x i l )= d X ( x, x ′ ) . Thus, f is an isometry. It remains to prove that f is bijective:Using a further subsequence n m a and the inverse maps f − n ma , an isometry g : Y → X can be constructed analogously. For arbitrary x ∈ X , let ( y k l ) l ∈ N be the sequence in the dense subset Y ′ ⊆ Y used in the construction of g converging to f ( x ) ∈ Y . Then d X ( g ◦ f ( x ) , x ) = lim a →∞ lim l,j →∞ d X ( f − n ma ( y k l ) , x i j )= lim a →∞ lim l,j →∞ d Y ( y k l , f n ma ( x i j ))= d Y ( f ( x ) , f ( x )) = 0 . Analogously, f ◦ g = id . Thus, f is bijective. (cid:3) As in the compact case, Gromov-Hausdorff convergence preserves being alength space.
Proposition 2.7.
Let ( X i , d X i , p i ) , i ∈ N , be pointed length spaces and ( X, d X , p ) be a pointed metric space. If ( X i , p i ) → ( X, p ) , then X is a lengthspace.Proof. Let x, y ∈ X and ε > be arbitrary. For r := max { d X ( x, p ) , d X ( y, p ) } choose n ∈ N with d GH (( ¯ B X n r ( p n ) , p n ) , ( ¯ B Xr ( p ) , p )) < ε . The rest of theproof can be done completely analogously to the one of Proposition 1.13. (cid:3) As in the compact case, in the non-compact case there is a correspondencebetween (pointed) Gromov-Hausdorff convergence and approximations. Inorder to prove this, the following lemma is needed.
Lemma 2.8.
For all r > , let ( ε rn ) n ∈ N be a monotonically decreasing nullsequence, and h : R > → R > a function with lim x → h ( x ) = 0 . Then thereexists a sequence ( r n ) n ∈ N with lim n →∞ r n = ∞ and ε r n n ≤ h (cid:0) r n (cid:1) for almostall n ∈ N .Proof. Let A := { n ∈ N | ∀ r > ε rn > h (cid:0) r (cid:1) } denote the set of all naturalnumbers n for which no such ‘ r n ’ can exist. This set is finite: Fix r > . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 23
Then ε rn > h (cid:0) r (cid:1) for all n ∈ A , but, since ( ε rn ) n ∈ N is a null sequence, thisinequality only holds for finitely many n . Hence, A is finite.Without loss of generality, assume that for each n there is at least one r > such that ε rn ≤ h (cid:0) r (cid:1) .Let R n := { r > | ε rn ≤ h (cid:0) r (cid:1) } 6 = ∅ denote the set of all radii whichare possible candidates for ‘ r n ’. Then ( R n ) n ∈ N is an increasing sequence:Fix r ∈ R n . Since ( ε rn ) n ∈ N is monotonically decreasing, ε rn +1 ≤ ε rn ≤ h (cid:0) r (cid:1) .Thus, r ∈ R n +1 .Suppose that these sets are uniformly bounded, i.e. there exists C > such that S n ∈ N R n ⊆ [0 , C ] . Then ε rn > h (cid:0) r (cid:1) for all n and all r > C .Consequently, for all r > C the sequence ( ε rn ) n ∈ N is bounded below by h (cid:0) r (cid:1) .This is a contradiction to ( ε rn ) n ∈ N being a null sequence.Therefore, S n ∈ N R n is unbounded, i.e. for all C > there exists some N ∈ N such that R j [0 , C ] for all j ≥ N . In particular, for all k ∈ N thereis a minimal N k ∈ N such that for all j ≥ N k there is some r kj ∈ R j with r kj > k . There are two cases:1. Let N k → ∞ . For every n ∈ N , n ≥ N , there is some k ∈ N with N k ≤ n < N k +1 . Fix this k and define r n := r kn for some r kn ∈ R n satisfying r kn > k . Then, for arbitrary k ∈ N and all n ≥ N k , r n > k .Thus, r n → ∞ . Furthermore, by choice, ε r n n ≤ h (cid:0) r n (cid:1) .2. Let k ∈ N such that N k = N k for all k ≥ k . For n < N k , define r n as in the first case. For n = N + m ≥ N k = N k + m , choose any r n := r k + mn ∈ R n ∩ ( k + m, ∞ ) . Then r n → ∞ and ε r n n ≤ h (cid:0) r n (cid:1) . (cid:3) Proposition 2.9.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be length spaces.Then the following statements are equivalent. a) ( X i , p i ) → ( X, p ) . b) For all functions g : R > → R > with lim x → g ( x ) = 0 there exists r i → ∞ with d GH (( ¯ B X i r i ( p i ) , p i ) , ( ¯ B Xr i ( p ) , p )) ≤ g (cid:16) r i (cid:17) . c) There exist r i → ∞ and ε i → with d GH (( ¯ B X i r i ( p i ) , p i ) , ( ¯ B Xr i ( p ) , p )) ≤ ε i . Proof.
The proof is done by verifying the implications a) ⇒ b), b) ⇒ c) andc) ⇒ a). First, let ( X i , p i ) → ( X, p ) and g : R > → R > with lim x → g ( x ) = 0 be arbitrary. For fixed r > , define ˜ ε ri := d GH (( ¯ B X i r ( p i ) , p i ) , ( ¯ B Xr ( p ) , p )) → as i → ∞ and ε ri := sup { ˜ ε rj | j ≥ i } → as i → ∞ . This sequence ( ε ri ) i ∈ N is monotonically decreasing and satisfies ε ri ≥ ˜ ε ri . ByLemma 2.8, there exists r i → ∞ such that ε r i i ≤ g (cid:0) r i (cid:1) for all i ∈ N . In particular, d GH (( ¯ B X i r i ( p i ) , p i ) , ( ¯ B Xr i ( p ) , p )) = ˜ ε r i i ≤ ε r i i ≤ g (cid:16) r i (cid:17) , and this proves b). Obviously, b) implies c) via choosing g := id and ε i := r i .Finally, let d GH (( ¯ B X i r i ( p i ) , p i ) , ( ¯ B Xr i ( p ) , p )) ≤ ε i for some r i → ∞ and ε i → . Fix r > . Let i ∈ N be large enough such that r < r i . Then d GH (( ¯ B X i r ( p i ) , p i ) , ( ¯ B Xr ( p ) , p )) ≤ ε i , by Lemma 2.3, and this implies the claim. (cid:3) Corollary 2.10.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces. Then the following statements are equivalent. a) ( X i , p i ) → ( X, p ) . b) There is ε i → such that Appr ε i (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) = ∅ forall i . c) There is ε i → such that d GH (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) ≤ ε i forall i .Proof. This is a direct implication of Proposition 1.5 and Proposition 2.9. (cid:3)
Similarly to the compact case, the Gromov-Hausdorff distance and con-vergence, respectively, is related to the diameters of the spaces: On the onehand, the distance of balls in X and X × Y are bounded from above by thediameter of Y . Recall that in the special case of X = { pt } , the (non-pointed)distance equals diam( Y ) . On the other hand, in the compact case it wasproven that convergence of spaces implies convergence of the diameters. Forlength spaces, an analogous statement will be established. Proposition 2.11.
Let ( X, d X , x ) and ( Y, d Y , y ) be pointed metric spaces.If Y is compact, then d GH (( ¯ B Xr ( x ) , x ) , ( ¯ B X × Yr (( x , y )) , ( x , y ))) ≤ diam( Y ) for all r > .Proof. It suffices to define an admissible metric and to estimate the Hausdorffdistance with respect to this metric.Let δ > be arbitrary. Define an admissible metric d on ( X × Y ) ∐ X by d (( x, y ) , x ′ ) := p d X ( x, x ′ ) + d Y ( y, y ) + δ . As usual, the only tricky part is to prove the triangle inequality: By theMinkowski inequality, for x , x ′ , x , x ′ ∈ X and y , y ∈ Y , d (( x , y ) , x ′ ) + d ( x ′ , x ′ )= q d X ( x , x ′ ) + d Y ( y , y ) + δ + d X ( x ′ , x ′ ) ≥ q d X ( x , x ′ ) + d X ( x ′ , x ′ ) + d Y ( y , y ) + δ OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 25 ≥ q d X ( x , x ′ ) + d Y ( y , y ) + δ = d (( x , y ) , x ′ ) . With completely analogous argumentation, one can prove the remaining in-equalities d ( x ′ , ( x , y )) + d (( x , y ) , x ′ ) ≥ d ( x ′ , x ′ ) ,d (( x , y ) , ( x , y )) + d (( x , y ) , x ′ ) ≥ d (( x , y ) , x ′ ) and d (( x , y ) , x ′ ) + d ( x ′ , ( x , y )) ≥ d (( x , y ) , ( x , y )) . Now fix r > and let ( x, y ) ∈ ¯ B X × Yr (( x , y )) be arbitrary. In particular, x ∈ ¯ B Xr ( x ) . Thus, d (( x, y ) , ¯ B Xr ( x )) ≤ d (( x, y ) , x )= p d Y ( y, y ) + δ ≤ p diam( Y ) + δ . Hence, ¯ B X × Yr (( x , y )) ⊆ ¯ B d √ diam( Y ) + δ ( ¯ B Xr ( x )) . For arbitrary x ∈ ¯ B Xr ( x ) , one has d (( x, y ) , ( x , y )) = d X ( x, x ) < r , andtherefore, ( x, y ) ∈ ¯ B X × Yr (( x , y )) . Thus, d ( x, ¯ B X × Yr ( x , y )) ≤ d ( x, ( x, y )) = δ and ¯ B Xr ( x ) ⊆ ¯ B dδ ( ¯ B X × Yr ( x , y )) . Hence, d GH (( ¯ B Xr ( x ) , x ) , ( ¯ B X × Yr (( x , y )) , ( x , y ))) ≤ d d H ( ¯ B Xr ( x ) , ¯ B X × Yr ( x , y )) ≤ max { p diam( Y ) + δ , δ } = p diam( Y ) + δ . Since δ was arbitrary, this proves the claim. (cid:3) In order to prove the convergence of diameters, one needs the followingproperty of length spaces of infinite diameter: Any ball of radius r hasdiameter at least r . Though it is easy to see this, for the sake of completeness,the proof is given first. Lemma 2.12.
Let ( X, d, p ) be a pointed length space and < r < diam( X )2 .Then diam( ¯ B Xr ( p )) ≥ r .Proof. Assume that d ( q, p ) ≤ r for all q ∈ X . Hence, ¯ B r ( p ) = X , and thisimplies diam( X ) ≤ r < diam( X ) , which is a contradiction. Hence, there is q r ∈ X such that l r := d ( q r , p ) > r . Fix a minimisinggeodesic γ : [0 , l r ] → X with γ (0) = p and γ ( l r ) = q r . Then d ( p, γ ( r )) = r ,hence, γ ( r ) ∈ ¯ B r ( p ) . In particular, diam( ¯ B r ( p )) ≥ d ( p, γ ( r )) = r . (cid:3) Proposition 2.13.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces. If ( X i , p i ) → ( X, p ) , then diam( X i ) → diam( X ) . (Here, both diam( X i ) tending to infinity as well as the notion ∞ → ∞ are allowed.)Proof. Let ε i → be as in Corollary 2.10 with d GH (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) ≤ ε i . By Proposition 1.18, | diam( ¯ B X i / ε i ( p i )) − diam( ¯ B X / ε i ( p )) | ≤ ε i → . Dis-tinguish the two cases of X being bounded and unbounded, respectively. Case 1: diam( X ) < ∞ . Without loss of generality, assume diam( X ) < ε i for all i ∈ N . Then X = B X / ε i ( p ) and | diam( ¯ B X i / ε i ( p i )) − diam( X ) | = | diam( ¯ B X i / ε i ( p i )) − diam( ¯ B X / ε i ( p )) | → , in particular, diam( ¯ B X i / ε i ( p i )) → diam( X ) as i → ∞ . Without loss of gen-erality, assume diam( ¯ B X i / ε i ( p i )) ≤ · diam( X ) for all i ∈ N .Let r i := min (cid:8) ε i , · diam( X i ) (cid:9) < · diam( X i ) . By Lemma 2.12, r i ≤ diam( B X i r i ( p i )) ≤ diam( B X i / ε i ( p i )) ≤ · diam( X ) < ε i . Hence, diam( X i ) = 3 r i ≤ · diam( X ) , the X i are compact and Proposition 1.18implies the claim. Case 2: diam( X ) = ∞ . Assume there is a subsequence ( i j ) j ∈ N and C > with diam( X i j ) < C for all j ∈ N . Pass to this subsequence. After passingto a further subsequence, C < ε i for all i ∈ N . Then X i = B X i / ε i ( p i ) andthis implies diam( ¯ B X i / ε i ( p i )) = diam( X i ) < C . Further, by Lemma 2.12, diam( ¯ B X / ε i ( p )) ≥ ε i and | diam( ¯ B X i / ε i ( p i )) − diam( ¯ B X / ε i ( p )) | ≥ ε i − C → ∞ . This is a contradiction. Hence, diam( X i ) → ∞ . (cid:3) Gromov-Hausdorff convergence is compatible with rescaling: Given a con-verging sequence of length spaces and a converging sequence of rescalingfactors, the rescaled sequence converges and the limit space is the originalone rescaled by the limit of the rescaling sequence. More generally, given aconverging sequence of metric spaces and some bounded sequence of rescal-ing factors, the sublimits of the rescaled sequence correspond exactly to thesublimits of the rescaling sequence.For a metric space ( X, d ) , recall that αX denotes the rescaled metric space ( X, α d ) . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 27
Proposition 2.14.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces and r i , r, α i , α > . a) If ( X i , p i ) → ( X, p ) and r i → r , then ( ¯ B X i r i ( p i ) , p i ) → ( ¯ B Xr ( p ) , p ) . b) If α i → α , then ( α i X, p ) → ( αX, p ) . c) If ( X i , p i ) → ( X, p ) and α i → α , then ( α i X i , p i ) → ( αX, p ) . d) If ( X i , p i ) → ( X, p ) and ( α i X i , p i ) → ( Y, q ) , then there is α such that α i → α and ( Y, q ) ∼ = ( αX, p ) .Proof. a) By Corollary 1.17, d GH (( ¯ B X i r i ( p i ) , p i ) , ( ¯ B X i r ( p i ) , p i )) ≤ | r − r i | → , and the triangle inequality implies d GH ( ¯ B X i r i ( p i ) , ¯ B Xr ( p )) ≤ d GH ( ¯ B X i r i ( p i ) , ¯ B X i r ( p i )) + d GH ( ¯ B X i r ( p i ) , ¯ B Xr ( p )) → . b) Without loss of generality, let α = 1 . First, let X be compact. Define f i : X → α i X and g i : α i X → X by f i ( x ) := x and g i ( x ) := x for all x ∈ X .Furthermore, let < ε i := 2 · | α i − | · diam( X ) → . For any x, x ′ ∈ X , | d α i X ( f i ( x ) , f i ( x ′ )) − d X ( x, x ′ ) | = | α i − | · d X ( x, x ′ ) < ε i . Analogously, | d α i X ( x, x ′ ) − d X ( g i ( x ) , g i ( x ′ )) | < ε i . Furthermore, d X ( x, g i ◦ f i ( x )) = 0 < ε i and d X ( f i ◦ g i ( x ) , x ) = 0 < ε i .Thus, ( f i , g i ) ∈ Appr ε i (( α i X, p ) , ( X, p )) and ( α i X, p ) → ( X, p ) . Now let X be non-compact and r > . Then, using a) and the compact case, d GH (( ¯ B α i Xr ( p ) , p ) , ( ¯ B Xr ( p ) , p )) ≤ d GH (( ¯ B α i Xr ( p ) , p ) , ( ¯ B α i Xα i r ( p ) , p )) + d GH (( ¯ B α i Xα i r ( p ) , p ) , ( ¯ B Xr ( p ) , p ))= α i · d GH (( ¯ B Xr/α i ( p ) , p ) , ( ¯ B Xr ( p ) , p )) + d GH (( α i ¯ B Xr ( p ) , p ) , ( ¯ B Xr ( p ) , p )) → . c) By the triangle inequality, for fixed r > , d GH (( ¯ B α i X i r ( p i ) , p i ) , ( ¯ B αXr ( p ) , p )) ≤ d GH (( ¯ B α i X i r ( p i ) , p i ) , ( ¯ B α i Xα i r/α ( p ) , p ))+ d GH (( ¯ B α i Xα i r/α ( p ) , p ) , ( ¯ B α i Xr ( p ) , p ))+ d GH (( ¯ B α i Xr ( p ) , p ) , ( ¯ B αXr ( p ) , p )) . By a), d GH (( ¯ B α i X i r ( p i ) , p i ) , ( ¯ B α i Xα i r/α ( p ) , p ))= α i · d GH (( ¯ B X i r/α i ( p i ) , p i ) , ( ¯ B Xr/α ( p ) , p )) → , by Corollary 1.17, d GH (( ¯ B α i Xα i r/α ( p ) , p ) , ( ¯ B α i Xr ( p ) , p )) ≤ | r − α i α · r | → , and by b), d GH (( ¯ B α i Xr ( p ) , p ) , ( ¯ B αXr ( p ) , p )) → . Hence, ( ¯ B α i X i r ( p i ) , p i ) → ( ¯ B αXr ( p ) , p ) for every r > .d) Let α be an arbitrary accumulation point of ( α i ) i ∈ N . Hence, for a subse-quence ( i j ) j ∈ N , both α i j → α , and by c), ( α i j X i j , p i j ) → ( αX, p ) as j → ∞ .On the other hand, ( α i j X i j , p i j ) → ( Y, q ) as j → ∞ . Thus, ( Y, q ) and ( αX, p ) are isometric (cf. Proposition 2.6). (cid:3) Corollary 2.15.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces and ( α i ) i ∈ N be a bounded sequence. If ( X i , p i ) → ( X, p ) , then thesublimits of ( α i X i , p i ) correspond to the ( αX, p ) for exactly the accumulationpoints α of ( α i ) i ∈ N .Proof. Let α be an accumulation point of ( α i ) i ∈ N and ( α i j ) j ∈ N be the subse-quence converging to α . Then ( X i j , p i j ) → ( X, p ) , and by Proposition 2.14, ( α i j X i j , p i j ) → ( αX, p ) . Now let ( Y, y ) be a sublimit of ( α i X i , p i ) , i.e. ( α i j X i j , p i j ) → ( Y, y ) forsome subsequence ( i j ) j ∈ N . Since ( α i j ) j ∈ N is a bounded sequence, there ex-ists a convergent subsequence ( α i jl ) l ∈ N with limit α . For this subsequence, ( α i jl X i jl , p i jl ) → ( Y, y ) , and ( α i jl X i jl , p i jl ) → ( αX, p ) by the first part. Thus, ( Y, y ) is isometric to ( αX, p ) for an accumulation point α of ( α i ) i ∈ N . (cid:3) Convergence of points.
In the previous section, convergent sequencesof pointed metric (length) spaces were studied. Given such a sequence andusing the corresponding approximations, a notion for convergence of pointscan be introduced.
Definition 2.16.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces. Assume ( X i , p i ) → ( X, p ) and let ε i → and ( f i , g i ) ∈ Appr ε i (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) as in Corollary 2.10. Let q i ∈ ¯ B X i / ε i ( p i ) and q ∈ X . Then q i converges to q ,denoted by q i → q , if f i ( q i ) converges to q (in X ).For ( X i , p i ) → ( X, p ) as in the definition, p i → p due to f i ( p i ) = p .Moreover, for each x ∈ X there exists such a sequence x i satisfying x i → x ,e.g. x i := g i ( x ) .Convergence q i → q depends on the choice of the underlying Gromov-Hausdorff approximations: Convergence with respect to one pair of approxi-mations does not necessarily imply convergence for another, as the followingexample shows. OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 29
Example 2.17.
For i ∈ N , let X i = X = S be the -dimensional sphere, p i = p = N the north pole and q i = q some fixed point on the equator. Let ϕ denote the rotation of S by π fixing p and define f i = g i = f ′ i = g ′ i = id S , f ′ i +1 = ϕ and g ′ i +1 = ϕ − .Then both ( f i , g i ) and ( f ′ i , g ′ i ) are pointed isometries between ( X i , p i ) and ( X, p ) satisfying f i ( q i ) = q , but f ′ i ( q i ) = q = ϕ ( q ) = f ′ i +1 ( q i +1 ) . Hence, f ′ i ( q i ) is not convergent at all, but subconvergent with limits q and ϕ ( q ) .In this example, after replacing the approximations, two sublimits occur:One sublimit is the limit corresponding to the original approximations, theother one is its image under an isometry of the limit space. Since Gromov-Hausdorff convergence distinguishes spaces only up to isometry, concretely ( X, p ) ∼ = ( h ( X ) , h ( p )) = ( X, h ( p )) for any isometry h , this can be interpretedas follows: If q is a sublimit of q i with respect to one Gromov-Hausdorff ap-proximation, then it is a sublimit for all Gromov-Hausdorff approximations.This is a general fact as the subsequent lemma shows. In order to provethis, the separability of a connected proper metric space is needed. Thoughit is easy to see that such a space is separable, for completeness, the proofis given first. Lemma 2.18.
A connected proper metric space is separable.Proof.
Let ( X, p ) be a connected proper metric space and let p ∈ X bearbitrary. Then X = [ q ∈ Q ∩ (0 , ∞ ) ¯ B q ( p ) . As a compact set, every ¯ B q ( p ) is separable where q ∈ Q is positive. Therefore,there exists a countable dense subset A q ⊆ ¯ B q ( p ) . Let A := S q ∈ Q ∩ (0 , ∞ ) A q .This A is countable, and for arbitrary x ∈ X there is a positive q ∈ Q suchthat x ∈ ¯ B q ( p ) , i.e. there exists a sequence x n ∈ A q ⊆ A converging to x .Thus, A is dense in X , hence, X is separable. (cid:3) Lemma 2.19.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces. Assume ( X i , p i ) → ( X, p ) and let ε i , ε ′ i → , r i , r ′ i → ∞ and ( f i , g i ) ∈ Appr ε i (( ¯ B X i r i ( p i ) , p i ) , ( ¯ B Xr i ( p ) , p )) , ( f ′ i , g ′ i ) ∈ Appr ε ′ i (( ¯ B X i r ′ i ( p i ) , p i ) , ( ¯ B Xr ′ i ( p ) , p )) . Let q i ∈ ¯ B X i min { r i ,r ′ i } ( p i ) and q ∈ X . If f i ( q i ) → q and q ′ is an accumulationpoint of f ′ i ( q i ) , then there exists an isometry h : X → X such that h ( q ) = q ′ .Proof. Without loss of generality, let r i = r ′ i : Otherwise, let R i := min { r i , r ′ i } and, by Lemma 2.3 and the construction in its proof, there are ( ˜ f i , ˜ g i ) ∈ Appr ε i (( ¯ B X i R i ( p i ) , p i ) , ( ¯ B XR i ( p ) , p ))( ˜ f ′ i , ˜ g ′ i ) ∈ Appr ε ′ i (( ¯ B X i R i ( p i ) , p i ) , ( ¯ B XR i ( p ) , p )) with ˜ f i ( q i ) → q if and only if f i ( q i ) → q, ˜ f ′ i ( q i ) → q if and only if f ′ i ( q i ) → q. Define h i , ¯ h i : ¯ B Xr i ( p ) → ¯ B Xr i ( p ) by h i := f ′ i ◦ g i and ¯ h i := f i ◦ g ′ i . In particular, h i ( p ) = ¯ h i ( p ) = p . For any x, x ′ ∈ ¯ B Xr i ( p ) , | d X ( h i ( x ) , h i ( x ′ )) − d X ( x, x ′ ) |≤ | d X ( f ′ i ( g i ( x )) , f ′ i ( g i ( x ′ ))) − d X i ( g i ( x ) , g i ( x ′ )) | + | d X i ( g i ( x ) , g i ( x ′ )) − d X ( x, x ′ ) |≤ ε ′ i + ε i → . Analogously, | d X (¯ h i ( x ) , ¯ h i ( x ′ )) − d X ( x, x ′ ) | → . Moreover, d X (¯ h i ◦ h i ( x ) , x )= d X ( f i ◦ g ′ i ◦ f ′ i ◦ g i ( x ) , x ) ≤ d X i ( g i ◦ f i ◦ g ′ i ◦ f ′ i ◦ g i ( x ) , g i ( x )) + ε i ≤ d X i ( g ′ i ◦ f ′ i ◦ g i ( x ) , g i ( x )) + 2 ε i ≤ d X i ( g i ( x ) , g i ( x )) + 2 ε i + ε ′ i → , and analogously, d X ( h i ◦ ¯ h i ( x ) , x ) → . Hence, if the h i and ¯ h i (sub)converge(in some sense), their corresponding (sub)limits are isometries fixing p with ¯ h = h − .The idea for proving subconvergence is to choose a countable dense subset A ⊆ X , to define the sublimit of all h i ( a ) where a ∈ A and to extend thislimit to a continuous map on X . Doing the same simultaneously for ¯ h i givesanother sublimit that turns out to be the inverse of the first. In the end,identifying X with itself using this isometry proves the claim.Choose a countable dense subset A = { a n | n ∈ N } ⊆ X (cf. Lemma 2.18)and, for i large enough such that d X ( a n , p ) ≤ r i , define z in := h i ( a n ) and ¯ z in := ¯ h i ( a n ) . Since d X ( z in , p ) = d X ( h i ( a n ) , h i ( p )) → d X ( a n , p ) , the sequence ( d ( z in , p )) i ∈ N is bounded from above by some R > . Hence, z in is contained in ¯ B XR ( p ) , and therefore, has a convergent subsequence. Ananalogous argument proves subconvergence for (¯ z in ) i ∈ N . Thus, using a diag-onal argument, there is a subsequence ( i j ) j ∈ N such that for any n ∈ N thesequences ( z i j n ) j ∈ N and (¯ z i j n ) j ∈ N , respectively, converge to some z n ∈ X and ¯ z n ∈ X , respectively.Define h ( a n ) := z n and ¯ h ( a n ) := ¯ z n . In particular, d X ( h ( a n ) , h ( a m )) = d X ( a n , a m ) = d X (¯ h ( a n ) , ¯ h ( a m )) OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 31 for all n, m ∈ N . For arbitrary x ∈ X , choose a Cauchy sequence ( a n k ) k ∈ N in A converging to x and let h ( x ) := lim k →∞ h ( a n k ) and ¯ h ( x ) := lim k →∞ ¯ h ( a n k ) . In fact, for any k ∈ N , d X ( h i j ( x ) , h ( x )) ≤ d X ( h i j ( x ) , h i j ( a n k )) + d X ( h i j ( a n k ) , h ( a n k )) + d X ( h ( a n k ) , h ( x )) ≤ d X ( x, a n k ) + ε i j + ε ′ i j + d X ( h i j ( a n k ) , h ( a n k )) + d X ( h ( a n k ) , h ( x )) → d X ( x, a n k ) + d X ( h ( a n k ) , h ( x )) as j → ∞ . Since this holds for every k ∈ N and d X ( x, a n k ) + d X ( h ( a n k ) , h ( x )) → as k → ∞ , h i j ( x ) → h ( x ) as j → ∞ . Analogously, ¯ h i j ( x ) → ¯ h ( x ) as j → ∞ . In particular, ¯ h i j ◦ h i j → ¯ h ◦ h andvice versa. Thus, h is an isometry on X with inverse ¯ h .Now let f i ( q i ) → q . Then d X ( f ′ i j ( q i j ) , h ( q )) ≤ d X ij ( g ′ i j ◦ f ′ i j ( q i j ) , g ′ i j ◦ h ( q )) + ε ′ i j ≤ d X ij ( q i j , g ′ i j ◦ h ( q )) + 2 ε ′ i j ≤ d X ( f i j ( q i j ) , f i j ◦ g ′ i j ◦ h ( q )) + 2 ε ′ i j + ε i j ≤ d X ( f i j ( q i j ) , q ) + d X ( q, ¯ h ′ i j ◦ h ( q )) + 2 ε ′ i j + ε i j → as j → ∞ . This proves f ′ i j ( q i j ) → h ( q ) as j → ∞ . (cid:3) The following statements allow to change the base points of a given con-vergent sequence.
Proposition 2.20.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces, and let q i ∈ X i and q ∈ X . If ( X i , p i ) → ( X, p ) and q i → q , then ( X i , q i ) → ( X, q ) .Proof. The proof is an immediate consequence of Lemma 2.3 and Proposition 2.9:Choose ε i → and ( f i , g i ) ∈ Appr ε i (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) as in Definition 2.16with f i ( q i ) → q . In particular, d X i ( q i , g i ( q )) ≤ ε i + d X ( f i ( q i ) , f i ( g i ( q ))) ≤ ε i + d X ( f i ( q i ) , q ) → . Hence, δ i := max { d X ( f i ( q i ) , q ) , d X i ( q i , g i ( q )) } → .Since f i ( p i ) = p , d X i ( p i , q i ) ≤ ε i + d X ( p, q ) + d X ( q, f i ( q i )) → d X ( p, q ) . Let r > be arbitrary. Fix i large enough such that r + d X ( p, q )) ≤ ε i and such that d X i ( p i , q i ) ≤ d X ( p, q ) or d X i ( p i , q i ) ≤ r , respectively, if p = q or p = q , respectively. In particular, ¯ B X i r ( q i ) ⊆ ¯ B X i r + d Xi ( p i ,q i ) ( p i ) ⊆ ¯ B X i / ε i ( p i ) , ¯ B Xr ( q ) ⊆ ¯ B Xr + d X ( p,q ) ( p ) ⊆ ¯ B X / ε i ( p ) and Appr ε i + δ i (( ¯ B X i r ( q i ) , q i ) , ( ¯ B Xr ( q ) , q )) = ∅ by Lemma 2.3. By Proposition 1.5, d GH (( ¯ B X i r ( q i ) , q i ) , ( ¯ B Xr ( q ) , q )) ≤ ε i +2 δ i → , and Proposition 2.9 implies the claim. (cid:3) Corollary 2.21.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces. Let q i ∈ X i with d X i ( p i , q i ) → . Assume ( X i , p i ) → ( X, p ) . Then ( X i , q i ) → ( X, p ) .Proof. Choose ε i → and ( f i , g i ) ∈ Appr ε i (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) asin Corollary 2.10. Then d X ( f i ( q i ) , p ) = d X ( f i ( q i ) , f i ( p )) ≤ d X i ( q i , p i ) + ε i → . Hence, q i → p , and Proposition 2.20 implies the claim. (cid:3) Corollary 2.22.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces. Let q i ∈ X i with d X i ( p i , q i ) ≤ C for some C > . If ( X i , p i ) → ( X, p ) ,then there exists q ∈ X such that ( X i , q i ) subconverges to ( X, q ) .Proof. Let ε i → and ( f i , g i ) ∈ Appr ε i (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) be asin Corollary 2.10. For R > C there is i > such that C + ε i ≤ R for all i ≥ i . Therefore, f i ( q i ) ∈ ¯ B R ( p ) for all i ≥ i . Since this ball is compact,there exists a convergent subsequence with limit q ∈ ¯ B R ( p ) . After passingto this subsequence, q i → q , and Proposition 2.20 implies the claim. (cid:3) Convergence of maps.
So far, statements about the convergence ofmetric spaces and points were made. But even statements about maps be-tween those convergent space are possible: In fact, Lipschitz maps (sub)con-verge (in some sense) to Lipschitz maps. The proof of this seems to be rathertechnical, but in fact essentially only uses the same methods one can use toprove convergence of compact subsets (without bothering Gromov’s Pre-compactness Theorem). Therefore, a proof of the latter is given in advanceafter establishing the following (technical) lemma.
Lemma 2.23.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be pointed lengthspaces. Assume ( X i , p i ) → ( X, p ) and let ε i → and ( f i , g i ) ∈ Appr ε i (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) be as in Corollary 2.10. Moreover, let A i ⊆ B X i / ε i ( p i ) and A ⊆ X be compactand f ′ i : A i → A , g ′ i : A → A i and δ i → satisfy d X ( f ′ i ( x i ) , f i ( x i )) ≤ δ i and d X i ( g ′ i ( x ) , g i ( x )) ≤ δ i for all x i ∈ A i and x ∈ A . Then A i → A . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 33
Proof.
Prove ( f ′ i , g ′ i ) ∈ Appr ε i + δ i ) ( A i , A ) : For x i , x i ∈ A i , | d X ( f ′ i ( x i ) , f ′ i ( x i )) − d X i ( x i , x i ) |≤ | d X ( f ′ i ( x i ) , f ′ i ( x i )) − d X ( f i ( x i ) , f i ( x i )) | + | d X ( f i ( x i ) , f i ( x i )) − d X i ( x i , x i ) | < d X ( f ′ i ( x i ) , f i ( x i )) + d X ( f ′ i ( x i ) , f i ( x i )) + ε i ≤ ε i +2 δ i . Analogously, | d X i ( g ′ i ( x ) , g ′ i ( x )) − d X ( x , x ) | < ε i +2 δ i for all x , x ∈ A .Moreover, for x i ∈ A i , d X i ( g ′ i ◦ f ′ i ( x i ) , x i )) ≤ d X i ( g ′ i ◦ f ′ i ( x i ) , g i ◦ f ′ i ( x i ))+ d X i ( g i ◦ f ′ i ( x i ) , g i ◦ f i ( x i )) + d X i ( g i ◦ f i ( x i ) , x i ) < δ i + ( d X i ( f ′ i ( x i ) , f i ( x i )) + ε i ) + ε i ≤ ε i + δ i ) , and analogously, d X ( f ′ i ◦ g ′ i ( x ) , x )) < ε i + δ i ) for all x ∈ A . (cid:3) Proposition 2.24.
Let ( X, d X , p ) and ( X i , d X i , p i ) , i ∈ N , be length spacessuch that ( X i , p i ) → ( X, p ) and let ε i → and ( f i , g i ) ∈ Appr ε i (( ¯ B X i / ε i ( p i ) , p i ) , ( ¯ B X / ε i ( p ) , p )) be as in Corollary 2.10. Let K i ∈ X i be compact with K i ⊆ ¯ B X i R ( p i ) for some R > . After passing to a subsequence, there exists K ⊆ ¯ B r ( p ) such that K i subconverges to K .Proof. Without loss of generality, assume R ≤ ε i and ε i ≤ for all i ∈ N .Let x i ∈ K i ⊆ ¯ B X i R ( p i ) be arbitrary. Then f i ( x i ) ∈ B XR + ε i ( p ) ⊆ ¯ B XR +1 ( p ) .Hence, the sequence ( f i ( x i )) i ∈ N is contained in a compact set, and there-fore has a convergent subsequent. Unfortunately, for different choices of x i different subsequences might converge. Therefore, a diagonal argument oncountable dense subsets of the K i will be used.Let A i = { a ni | n ∈ N } ⊆ K i be a countable dense subset. As seenabove, the sequence ( f i ( a ni )) i ∈ N , where n ∈ N , has a convergent subsequencewith limit y n ∈ ¯ B XR +1 ( p ) . Moreover, this subsequence can be chosen suchthat, after passing to this subsequence, d X ( f i ( a ni ) , y n ) < ε i . By a diagonalargument, there exists a common subsequence such that for every n ∈ N there is y n ∈ ¯ B R +1 ( p ) with d X ( f i ( a ni ) , y n ) < ε i for all i ∈ N . Pass to thissubsequence.Define A := { y n | n ∈ N } as the set of all these limits and let K := ¯ A denote its closure. In particular, K is compact. Define maps f ′ i : K i → K and g ′ i : K → K i in the following way: For x i ∈ A i , i.e. x i = a ni for some n ∈ N , define f ′ i ( x i ) := y n ∈ A ⊆ K . If x i ∈ K i \ A i , choose a ni ∈ A i with d X i ( x i , a ni ) < ε i and define f ′ i ( x i ) := y n ∈ A ⊆ K . In particular, d X ( f ′ i ( x i ) , f i ( x i )) ≤ d X ( y n , f i ( a ni )) + d X ( f i ( a ni ) , f i ( x i )) < ε i ε i + d X i ( a ni , x i )) < ε i (cid:16) ε i + ε i (cid:17) = 32 ε i . For x ∈ A , i.e. x = y n for some n ∈ N , define g ′ i ( y n ) := a ni ∈ A i ⊆ K i . For x ∈ X \ A , choose y n ∈ A with d X ( x, y n ) < ε i and let g ′ i ( x ) := a ni ∈ A i ⊆ K i .Then d X i ( g ′ i ( x ) , g i ( x )) = d X i ( a ni , g i ( x )) < ε i + d X ( f i ( a ni ) , x ) ≤ ε i + d X ( f i ( a ni ) , y n ) + d X ( y n , x ) < ε i . Now Lemma 2.23 implies the claim. (cid:3)
Lemma 2.25.
Let ( X, d X ) , ( Y, d Y ) , ( X i , d X i ) and ( Y i , d Y i ) , i ∈ N , be com-pact length spaces such that X i → X and Y i → Y . Moreover, let α > , K i ⊆ X i be compact subsets and f i : K i → Y i be α -bi-Lipschitz. Afterpassing to a subsequence, the following holds: a) There exist compact subsets K ⊆ X and K ′ ⊆ Y which are Gromov-Hausdorff limits of K i and f i ( K i ) , respectively, and an α -bi-Lipschitzmap f : K → K ′ with f ( K ) = K ′ . b) For any compact subset L ⊆ K ⊆ X there are compact subsets L i ⊆ K i such that L i → L and f i ( L i ) → f ( L ) in the Gromov-Hausdorff sense.Proof. a) In order to prove the first part, pass to the subsequence of Proposition 2.24.Then there are compact sets K ⊆ X and K ′ ⊆ Y such that K i → K and f i ( K i ) → K ′ . For these, fix ε i → , ( f Xi , g Xi ) ∈ Appr ε i ( K i , K ) and ( f Yi , g Yi ) ∈ Appr ε i ( f i ( K i ) , K ′ ) , cf. Figure 1.The idea is to define f as a limit of h i := f Yi ◦ f i ◦ g Xi : K → K ′ : For x, x ′ ∈ K , d Y ( h i ( x ) , h i ( x ′ )) = d Y ( f Yi ◦ f i ◦ g Xi ( x ) , f Yi ◦ f i ◦ g Xi ( x ′ )) ≤ ε i + d Y i ( f i ◦ g Xi ( x ) , f i ◦ g Xi ( x ′ )) ≤ ε i +( α · d X i ( g Xi ( x ) , g Xi ( x ′ ))) ≤ ε i +( α · ( ε i + d X ( x, x ′ )))= α · d X ( x, x ′ ) + ( α + 1) · ε i . As in the proof of Proposition 2.24, the h i ( x ) do not have to converge.Therefore, a diagonal argument on a dense subset of K will be used toconstruct a limit map which can be extended using the completeness of thelimit space. OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 35 X i ⊆ K i f i ( K i ) ⊇ Y i X ⊆ KK ′ ⊇ Yf i f h i = f Yi ◦ f i ◦ g Xi f Xi g Xi f Yi g Yi Figure 1.
Sets and maps used to construct f : K → K ′ .Let A = { x j | j ∈ N } be a countable dense subset of K . Then h i ( x j ) ∈ K ′ for all i, j ∈ N , and since K ′ is compact, by a diagonal argument, there isa subsequence ( i n ) n ∈ N such that ( h i n ( x j )) n ∈ N converges for every j ∈ N .Define f : A → K ′ by f ( x j ) = lim n →∞ h i n ( x j ) . This map is α -bi-Lipschitz:For arbitrary j, l ∈ N , with the above estimate, d Y ( f ( x j ) , f ( x l )) = lim n →∞ d Y ( h i n ( x j ) , h i n ( x l )) ≤ lim n →∞ ( α + 1) · ε i n + α · d X ( x j , x l )= α · d X ( x j , x l ) . Analogously, d Y ( f ( x j ) , f ( x l )) ≥ α · d X ( x j , x l ) . Since A is a countable dense subset of K , f can be extended to an α -bi-Lipschitz map f : K → K ′ (cf. Lemma 2.26) where f ( x ) = lim l →∞ f ( x j l ) for x ∈ K and x j l ∈ A with x j l → x . In particular, for n ∈ N and l ∈ N , d Y ( f ( x ) , h i n ( x )) ≤ d Y ( f ( x ) , f ( x j l )) + d Y ( f ( x j l ) , h i n ( x j l )) + d Y ( h i n ( x j l ) , h i n ( x )) ≤ d Y ( f ( x ) , f ( x j l )) + d Y ( f ( x j l ) , h i n ( x j l )) + α · d X ( x j l , x ) + ( α + 1) · ε i n → d Y ( f ( x ) , f ( x j l )) + α · d X ( x j l , x ) as n → ∞→ as l → ∞ . Hence, f ( x ) = lim n →∞ h i n ( x ) .Moreover, observe the following: Since f i is α -bi-Lipschitz, it is injective.Therefore, the inverse f − i of f i exists on f i ( K i ) ⊇ im( g Xi ) and is α -bi-Lipschitz as well. Hence, for x ∈ K and y ∈ K ′ , d Y ( h i ( x ) , y ) = d Y ( f Yi ◦ f i ◦ g Xi ( x ) , y ) ≤ ε i + d Y i ( f i ◦ g Xi ( x ) , g Yi ( y )) ≤ ε i + α · d X i ( g Xi ( x ) , f − i ◦ g Yi ( y )) ≤ ε i + α · (2 ε i + d X i ( x, f Xi ◦ f − i ◦ g Yi ( y )))= 2( α + 1) ε i + α · d X i ( x, h ′ i ( y )) where h ′ i := f Xi ◦ f − i ◦ g Yi . With analogous arguments and using a furthersubsequence ( i n m ) m ∈ N of ( i n ) n ∈ N , there is an α -bi-Lipschitz map g : K ′ → K with g ( y ) = lim m →∞ h ′ i nm ( y ) for all y ∈ K ′ . In particular, for all y ∈ K ′ , d Y ( f ◦ g ( y ) , y ) = lim m →∞ d Y ( h i nm ( g ( y )) , y ) ≤ lim m →∞ α + 1) ε i nm + α · d X ( g ( y ) , h ′ i nm ( y ))= 0 . Thus, f ◦ g = id K ′ . Hence, K ′ ⊆ im( f ) which proves K ′ = f ( K ) . In fact,with analogous argumentation, one can prove g ◦ f = id K , i.e. g is the inverseof f . This proves the first part.b) The proof of the second statement is based on the first part and is donewith very similar methods.Let ( f Xi , g Xi ) ∈ Appr ε i ( K i , K ) and ( f Yi , g Yi ) ∈ Appr ε i ( f i ( K i ) , K ′ ) be asbefore. Then L i := g Xi ( L ) ⊆ K i is a compact subset of K i . The proof ofthe subconvergences will be done in two steps: First, prove L i → L , then f i ( L i ) → f ( L ) . For the maps defined below, cf. Figure 2.First, define ( ˜ f Xi , ˜ g Xi ) ∈ Appr ε i ( L i , L ) as follows: For x i ∈ g Xi ( L ) , choosea point y ∈ L with x i = g Xi ( y ) ; for x i ∈ L i \ g Xi ( L ) , choose y ∈ L with d X i ( x i , g Xi ( y )) < ε i . Then define ˜ f Xi ( x i ) := y . Finally, set ˜ g Xi := g Xi . Bydefinition, d X i (˜ g Xi ◦ ˜ f Xi ( x i ) , x i ) = d X i ( g Xi ◦ ˜ f Xi ( x i ) , x i ) < ε i for all x i ∈ L i . Conversely, for x ∈ L and by applying this inequality, d X ( ˜ f Xi ◦ ˜ g Xi ( x ) , x ) = d X ( ˜ f Xi ◦ g Xi ( x ) , x ) ≤ d X i ( g Xi ◦ ˜ f Xi ( g Xi ( x ))) , g Xi ( x )) + ε i ≤ ε i . Now let x i , x ′ i ∈ L i be arbitrary. Then | d X ( ˜ f Xi ( x i ) , ˜ f Xi ( x ′ i )) − d X i ( x i , x ′ i ) |≤ | d X ( ˜ f Xi ( x i ) , ˜ f Xi ( x ′ i )) − d X i ( g Xi ( ˜ f Xi ( x i )) , g Xi ( ˜ f Xi ( x ′ i ))) | + | d X i ( g Xi ( ˜ f Xi ( x i )) , g Xi ( ˜ f Xi ( x ′ i ))) − d X i ( x i , x ′ i ) | < ε i + d X i ( g Xi ◦ ˜ f Xi ( x i ) , x i ) + d X i ( g Xi ◦ ˜ f Xi ( x ′ i ) , x ′ i ) < ε i . For x, x ′ ∈ L , by definition, | d X i (˜ g Xi ( x ) , ˜ g Xi ( x ′ )) − d X ( x, x ′ ) | < ε i < ε i , OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 37 X i ⊆ K i ⊆ L i = g Xi ( L ) f i ( L i ) ⊇ f i ( K i ) ⊇ Y i X ⊆ K ⊆ Lf ( L ) ⊇ f ( K ) = K ′ ⊇ Yf Xi g Xi ˜ f Xi ˜ g Xi f i | L i f L ˜ f Yi ˜ g Yi f Yi g Yi Figure 2.
Sets and maps used to construct L i → L .and this proves ( ˜ f Xi , ˜ g Xi ) ∈ Appr ε i ( L i , L ) .In order to prove the subconvergence of f i ( L i ) to f ( L ) , observe that thecompactness of L i and L , respectively, and the continuity of f i and f , re-spectively, prove the compactness of f i ( L i ) and f ( L ) , respectively.Let δ i ( x ) := d Y ( h i ( x ) , f ( x )) for x ∈ L and δ i := sup x ∈ L δ i ( x ) . For the subsequence ( i n ) n ∈ N of the first part, δ i n ( x ) converges to . Then δ i n converges to as well: Assume this is not the case, i.e. there is ǫ > such that for all l ∈ N there exists i n l ∈ N and x n l ∈ X with δ i nl ( x n l ) ≥ ε .After passing to a subsequence, there is x ∈ X such that x n l → x as l → ∞ .Then ε ≤ δ i nl ( x n l )= d Y ( h i nl ( x n l ) , f ( x n l )) ≤ d Y ( h i nl ( x n l ) , h i nl ( x )) + d Y ( h i nl ( x ) , f ( x )) + d Y ( f ( x ) , f ( x n l )) ≤ ( α · d X ( x n l , x ) + ( α + 1) · ε i nl ) + δ i nl ( x ) + α · d X ( x, x n l ) → as l → ∞ . This is a contradiction.
Construct ( ˜ f Yi , ˜ g Yi ) ∈ Appr ˜ ε i ( f i ( L i ) , f ( L )) for ˜ ε i := (4 α + 1) ε i +2 δ i asfollows: Define ˜ f Yi := f ◦ ˜ f Xi ◦ f − i and ˜ g Yi := f i ◦ g Xi ◦ f − (recall that f − i exists on f i ( L i ) ⊆ f i ( K i ) and that f : K → K ′ is bijective).First, let y i ∈ L i and y ∈ L be arbitrary. Then d Y i (˜ g Yi ◦ ˜ f Yi ( y i ) , y i ) = d Y i ( f i ◦ g Xi ◦ ˜ f Xi ◦ f − i ( y i ) , y i ) ≤ α · d X i ( g Xi ◦ ˜ f Xi ( f − i ( y i )) , f − i ( y i )) < α · ε i ≤ ˜ ε i , and completely analogously, d Y ( ˜ f Yi ◦ ˜ g Yi ( y ) , y ) = d Y ( f ◦ ˜ f Xi ◦ g Xi ◦ f − ( y ) , y ) < α ε i ≤ ˜ ε i . For y, y ′ ∈ L , | d Y i (˜ g Yi ( y ) , ˜ g Yi ( y ′ )) − d Y ( y, y ′ ) |≤ | d Y i (˜ g Yi ( y ) , ˜ g Yi ( y ′ )) − d Y ( f Yi ◦ ˜ g Yi ( y ) , f Yi ◦ ˜ g Yi ( y ′ )) | + | d Y ( f Yi ◦ f i ◦ g Xi ◦ f − ( y ) , f Yi ◦ f i ◦ g Xi ◦ f − ( y ′ )) − d Y ( y, y ′ ) | < ε i + d Y ( h i ◦ f − ( y ) , f ◦ f − ( y )) + d Y ( h i ◦ f − ( y ′ ) , f ◦ f − ( y ′ )) ≤ ε i +2 δ i ≤ ˜ ε i . Finally, let y i , y ′ i ∈ Y i . Using the above estimates, | d Y ( ˜ f Yi ( y i ) , ˜ f Yi ( y ′ i )) − d Y i ( y i , y ′ i ) |≤ | d Y ( ˜ f Yi ( y i ) , ˜ f Yi ( y ′ i )) − d Y i (˜ g Yi ( ˜ f Yi ( y i )) , ˜ g Yi ( ˜ f Yi ( y ′ i ))) | + | d Y i (˜ g Yi ( ˜ f Yi ( y i )) , ˜ g Yi ( ˜ f Yi ( y ′ i ))) − d Y i ( y i , y ′ i ) | < ε i +2 δ i + d Y i (˜ g Yi ( ˜ f Yi ( y i )) , y i ) + d Y i (˜ g Yi ( ˜ f Yi ( y ′ i )) , y ′ i ) ≤ ε i +2 δ i + 2 · α ε i = ˜ ε i . Thus, ( ˜ f Yi , ˜ g Yi ) ∈ Appr ˜ ε i ( f i ( L i ) , f ( L )) . Since ˜ ε i n → as n → ∞ , this proves f i n ( L i n ) → f ( L ) as n → ∞ . (cid:3) Lemma 2.26.
Let ( X, d X ) and ( Y, d Y ) be metric spaces where Y is complete,let A ⊆ X and f : A → Y be α -(bi)-Lipschitz for some α > . Then f canbe extended to an α -(bi)-Lipschitz map ˆ f : ¯ A → Y .Proof. Let a ∈ ¯ A \ A be arbitrary. Then there exists a (Cauchy) sequence ( a n ) n ∈ N in A converging to a . By Lipschitz continuity of f , ( f ( a n )) n ∈ N is aCauchy sequence, and thus has a limit ˆ a in the complete metric space Y . Forany sequence (˜ a n ) n ∈ N converging to a , d Y ( f ( a n ) , f (˜ a n )) ≤ α · d X ( a n , ˜ a n ) → ,i.e. the limit ˆ a is independent of the choice of ( a n ) n ∈ N . Now define ˆ f ( a ) := ˆ a for a ∈ ¯ A \ A and ˆ f ( a ) := f ( a ) for a ∈ A . For arbitrary a, b ∈ A and OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 39 sequences a n → a , b n → b in A , d Y ( ˆ f ( a ) , ˆ f ( b )) = lim n →∞ d Y ( f ( a n ) , f ( b n )) ≤ lim n →∞ α · d X ( a n , b n )= α · d ( a, b ) . Hence, ˆ f is α -Lipschitz. Analogously, if f is α -bi-Lipschitz, ˆ f is α -bi-Lipschitz. (cid:3) Ultralimits
Since sequences of proper spaces do not necessarily converge in the pointedGromov-Hausdorff sense, a tool to enforce convergence can be useful. Sucha tool are the so called ultralimits since they always exist and are sublimitsin the pointed Gromov-Hausdorff sense. A basic reference from which thefollowing definitions are taken is [BH99, section I.5]. Another, more settheoretical, reference is [Jec06, chapter 7]. In the following, ultralimits willbe introduced and some properties will be investigated.
Definition 3.1 ([BH99, Definition I.5.47]) . A non-principal ultrafilter on N is a finitely additive probability measure ω on N such that all subsets S ⊆ N are ω -measurable with ω ( S ) ∈ { , } and ω ( S ) = 0 if S is finite. Remark.
If two sets have ω -measure , their intersection has ω -measure as well: Let ω ( A ) = ω ( B ) = 1 . Then ω ( N \ ( A ∩ B )) = ω ( N \ A ∪ N \ B ) ≤ ω ( N \ A ) + ω ( N \ B ) = 0 , hence, ω ( A ∩ B ) = 1 .Using Zorn’s Lemma, the existence of such a non-principal ultrafilter canbe proven. But even more is true: Given any infinite set, there exists anon-principal ultrafilter such that the set has measure with respect to thisultrafilter. Lemma 3.2.
Let A ⊆ N be an infinite set. Then there exists a non-principalultrafilter ω on N such that ω ( A ) = 1 .Proof. Let G := { B ⊆ N | B ⊇ A or N \ B is finite } . For any B , B ∈ G , the intersection B ∩ B is non-empty: This is obviouslycorrect if both B j ⊇ A or both N \ B j are finite. Thus, let B ⊇ A and N \ B be finite: Then A \ B is finite as well, hence, B ∩ B ⊇ A ∩ B = A \ ( A \ B ) is infinite since A is infinite. In particular, the intersection is non-empty.Using that G contains all sets with finite complement, it follows from[Jec06, Lemma 7.2 (iii)], [Jec06, Theorem 7.5] and the subsequent remarktherein that there exists a non-principal ultrafilter ω such that ω ( X ) = 1 forall X ∈ G . In particular, ω ( A ) = 1 . (cid:3) Given a bounded sequence of real numbers, a non-principal ultrafilterprovides a kind of ‘limit’. In fact, these ‘limits’ are accumulation points andnon-principal ultrafilters pick out convergent subsequences.
Lemma 3.3 ([BH99, Lemma I.5.49]) . Let ω be a non-principal ultrafilter on N . For every bounded sequence of real numbers ( a i ) i ∈ N there exists a uniquereal number l ∈ R such that ω ( { i ∈ N | | a i − l | < ε } ) = 1 for every ε > . Denote this l by lim ω a i . Lemma 3.4. If ω is a non-principal ultrafilter on N and ( a i ) i ∈ N a boundedsequence of real numbers, then lim ω a i is an accumulation point of ( a i ) i ∈ N .Moreover, there exists a subsequence ( a i j ) j ∈ N converging to lim ω a i such that ω ( { i j | j ∈ N } ) = 1 .Conversely, if ( a i ) i ∈ N is a bounded sequence of real numbers and a ∈ R any accumulation point, then there exists a non-principal ultrafilter ω on N such that a = lim ω a i .Proof. Let ( a i ) i ∈ N be any bounded sequence of real numbers.First, fix a non-principal ultrafilter ω , let a := lim ω a i and A ε := { i ∈ N | | a i − a | < ε } for ε > . By definition, ω ( A ε ) = 1 ; in particular, A ε has infinitely manyelements. Thus, a is an accumulation point.Next, prove that there exists I ⊆ N with ω ( I ) = 1 such that the subse-quence ( a i ) i ∈ I converges to a . Assume this is not the case, i.e. every I ⊆ N satisfies ω ( I ) = 0 or ( a i ) i ∈ I does not converge to a . Since ω ( N ) = 1 , ( a i ) i ∈ N does not converge to a . Hence, there exists ε > such that A ε is finite. Inparticular, ω ( A ε ) = 0 and this is a contradiction.Now let J ⊆ N be a set of indices such that ω ( J ) = 1 and the subsequence ( a j ) j ∈ J converges to a . By Lemma 3.2, there exists a non-principal ultrafilter ω such that ω ( J ) = 1 . By the first part, there exists a subsequence of indices I ⊆ N with ω ( I ) = 1 and a j → lim ω a i as j → ∞ for j ∈ I . Now ω ( I ∩ J ) = 1 and both a j → a and a j → lim ω a i as j → ∞ for j ∈ I ∩ J . This proves a = lim ω a i . (cid:3) An immediate consequence of the above lemma is the following: Given twobounded sequences of real numbers, investigating sublimits coming from acommon subsequence and investigating the ‘limits’ with respect to the samenon-principal ultrafilter is the same.
Lemma 3.5.
Let ( a i ) i ∈ N and ( b i ) i ∈ N be bounded sequences of real numbers. a) If ω is a non-principal ultrafilter on N , then there exists a subsequence ( i j ) j ∈ N such that both a i j → lim ω a i and b i j → lim ω b i as j → ∞ . b) If there are a, b ∈ R and a subsequence ( i j ) j ∈ N such that both a i j → a and b i j → b as j → ∞ , then there exists a non-principal ultrafilter ω on N such that a = lim ω a i and b = lim ω b i . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 41
Proof. a) By Lemma 3.4, there are subsequences of indices
I, J ⊆ N withmeasures ω ( I ) = ω ( J ) = 1 , a j → lim ω a i as j → ∞ for j ∈ I and b j → lim ω b i as j → ∞ for j ∈ J. In particular, I ∩ J has ω -measure . Hence, it is infinite and provides acommon subsequence which satisfies the claim.b) This follows directly from the second part of Lemma 3.4 since the non-principal ultrafilter constructed there depends only on the indices of theconvergent subsequence. (cid:3) Corollary 3.6.
Let ( a i ) i ∈ N and ( b i ) i ∈ N be bounded sequences of real numbers. a) If a i ≤ b i for all i ∈ N , then lim ω a i ≤ lim ω b i . b) lim ω ( a i + b i ) = lim ω a i + lim ω b i .Proof. Observe that Lemma 3.5 holds not only for two but finitely manysequences for real numbers. Applying this and the corresponding statementsfor limits of sequences of real numbers implies the claim. (cid:3)
An ultralimit is a ‘limit space’ assigned to a (pointed) sequence of metricspaces by using a non-principal ultrafilter. The construction of this ultra-limit is related to Gromov-Hausdorff convergence in the sense that sucha limit space is a sublimit in the pointed Gromov-Hausdorff sense. On theother hand, given any sublimit in the pointed Gromov-Hausdorff sense, thereexists a non-principal ultrafilter such that the corresponding ultralimit is ex-actly this sublimit. This fact can be extended to a similar statement aboutfinitely many different sequences and corresponding sublimits coming froma common subsequence.
Definition 3.7 ([BH99, Definition I.5.50]) . Let ω be a non-principal ultra-filter on N , ( X i , d i , p i ) , i ∈ N , be pointed metric spaces and X ω := { [( x i ) i ∈ N ] | x i ∈ X i and sup i ∈ N d i ( x i , p i ) < ∞} where ( x i ) i ∈ N ∼ ( y i ) i ∈ N if and only if lim ω d i ( x i , y i ) = 0 . Furthermore, let d ω ([( x i ) i ∈ N ] , [( y i ) i ∈ N ]) := lim ω d i ( x i , y i ) . Then ( X ω , d ω ) is ametric space, called ultralimit of ( X i , d i , p i ) and denoted by lim ω ( X i , d i , p i ) . Remark.
Let ω be a non-principal ultrafilter on N , ( X i , d i , p i ) , i ∈ N , bepointed metric spaces and Y i ⊆ X i . The limit ( Y ω , d Y ω ) := lim ω ( Y i , d i , p i ) iscanonically a subset of ( X ω , d X ω ) := lim ω ( X i , d i , p i ) : Obviously, { ( y i ) i ∈ N | y i ∈ Y i and sup i d i ( y i , p i ) < ∞}⊆ { ( x i ) i ∈ N | x i ∈ X i and sup i d i ( x i , p i ) < ∞} . Since the metric is the same on both X i and Y i and since the equivalenceclasses are only defined by using the ultrafilter and the metric, Y ω ⊆ X ω . With the same argumentation, the metric coincides: For y i , y ′ i ∈ Y i , d Y ω ([( y i ) i ∈ N ] Y ω , [( y ′ i ) i ∈ N ] Y ω )= lim ω d i ( y i , y ′ i )= d X ω ([( y i ) i ∈ N ] X ω , [( y ′ i ) i ∈ N ] X ω ) . Lemma 3.8 ([BH99, Lemma I.5.53]) . The ultralimit of a sequence of metricspaces is complete.
In order to prove the correspondence of sublimits and ultralimits, first,compact metric spaces are investigated.
Proposition 3.9.
Let ω be a non-principal ultrafilter on N and ( X i , d i , p i ) , i ∈ N , be pointed compact metric spaces with compact ultralimit ( X ω , d ω ) anddefine p ω := [( p i ) i ∈ N ] ∈ X ω . Then lim ω d GH (( X i , p i ) , ( X ω , p ω )) = 0 .Proof. The statement will be proven by using ε -nets: First, finite ε -nets in X i will be fixed and it will be proven that their ultralimit is a finite ε -net in X ω . Then the Gromov-Hausdorff distance of these nets will be estimated.Finally, using the triangle inequality and ε → prove the claim.Fix ε > . For every i ∈ N , fix a finite ε -net A εi = { a i , . . . , a n i i } inthe compact space X i with a i = p i , i.e. d ( a ki , a li ) ≥ ε for all k = l and X i = S n i j =1 B ε ( a ji ) . Let A εω be the ultralimit of these A εi , i.e. A εω = { [( a i ) i ∈ N ] | ∀ i ∈ N ∃ ≤ j i ≤ n i : a i = a j i i } ⊆ X ω , and let p ω := [( p i ) i ∈ N ] ∈ A εω . Then A εω is again a finite ε -net in X ω :Let [( a k i i ) i ∈ N ] , [( a l i i ) i ∈ N ] ∈ A εω . By definition, [( a k i i ) i ∈ N ] = [( a l i i ) i ∈ N ] if and only if lim ω d i ( a k i i , a l i i ) = 0 . Since d i ( a k i i , a l i i ) = 0 exactly for those i with k i = l i and d i ( a k i i , a l i i ) ≥ ε otherwise, this implies [( a k i i ) i ∈ N ] = [( a l i i ) i ∈ N ] if and only if ω ( { i ∈ N | k i = l i } ) = 1 . In particular, for [( a k i i ) i ∈ N ] = [( a l i i ) i ∈ N ] , d X ω ([( a k i i ) i ∈ N ] , [( a l i i ) i ∈ N ]) = lim ω d i ( a k i i , a l i i ) ≥ ε . Furthermore, for arbitrary [( x i ) i ∈ N ] there are a j i i such that x i ∈ B ε ( a j i i ) .Thus, d ω ([( x i ) i ∈ N ] , [( a j i i ) i ∈ N ]) = lim ω d i ( x i , a j i i ) < ε . This proves that A εω is an ε -net in X ω . It remains to prove that A εω isfinite: Assume it is not. Then S p ∈ A εω B ε ( p ) is an open cover of X ω , andthus, has a finite subcover X ω = S kj =1 B ε ( q j ) with q j ∈ A εω . Hence, forany q ∈ A εω \ { q , . . . , q k } there exists q j such that q ∈ B ε ( q j ) . This is acontradiction to d ω ( q, q j ) ≥ ε .Let n ω < ∞ denote the cardinality of A εω and I := { i ∈ N | n i = n ω } bethose indices such that A εi and A εω have the same cardinality. Then ω ( I ) = 1 : OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 43
Let A εω = { z , . . . , z n ω } and z k = [( a j ki i ) i ∈ N ] where ≤ j ki ≤ n i for each ≤ k ≤ n ω . For k = l , one has ω ( { i ∈ N | j ki = j li } ) . Thus, ω (cid:0) \ ≤ k
Let ω be a non-principal ultrafilter on N . If the ultralimitof compact metric spaces is compact, it is a sublimit in the pointed Gro-mov-Hausdorff sense which comes from a subsequence with index set whose ω -measure is .Proof. Let ( X i , d i , p i ) , i ∈ N , be pointed compact metric spaces, ( X ω , d ω ) their compact ultralimit and p ω = [( p i ) i ∈ N ] . By the previous proposition, lim ω d GH (( X i , p i ) , ( X ω , p ω )) = 0 , and by Lemma 3.4, there exists a subsequence ( i j ) j ∈ N of natural numberssatisfying ω ( { i j | j ∈ N } ) = 1 such that d GH (( X i j , p i j ) , ( X ω , p ω )) → as j → ∞ . (cid:3) This result now gives a corresponding result for non-compact spaces.
Proposition 3.11.
Let ω be a non-principal ultrafilter on N . The ultralimitof a sequence of pointed proper length spaces is a sublimit in the pointedGromov-Hausdorff sense (which comes from a subsequence with index set of ω -measure ).Conversely, the sublimit of a sequence of pointed proper length spaces inthe pointed Gromov-Hausdorff sense is the ultralimit with respect to a non-principal ultrafilter.Proof. Let ( X i , d i , p i ) , i ∈ N , be pointed proper length spaces, ( X ω , d ω ) thecorresponding ultralimit and p ω := [( p i ) i ∈ N ] ∈ X ω . First it will be shownthat an r -ball in the ultralimit is the ultralimit of r -balls. Then applying thecorresponding statement for compact sets proves the claim.For r > , let X rω ⊆ X ω denote the ultralimit of ( ¯ B X i r ( p i ) , d i , p i ) . This isa closed subset of X ω : First, observe X rω = { [( q i ) i ∈ N ] | q i ∈ X i and d i ( q i , p i ) ≤ r } . OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 45
Let ( z n ) n ∈ N be a sequence in X rω which converges to a limit z ∈ X ω . Denote z n = [( q ni ) i ∈ N ] and z = [( q i ) i ∈ N ] where q ni , q i ∈ X i with d i ( q ni , p i ) ≤ r for all i, n ∈ N and sup i ∈ N d i ( q i , p i ) < ∞ . Moreover, d ω ( z n , z ) = lim ω d i ( q ni , q i ) → as n → ∞ . For all n ∈ N , d ω ( z n , p ω ) = lim ω d i ( q ni , p i ) ≤ r . Hence, d ω ( z, p ω ) ≤ lim n →∞ d ω ( z, z n ) + d ω ( z n , p ω ) ≤ r and z ∈ X rω . This proves that X rω is closed.In fact, X rω = ¯ B X ω r ( p ω ) : First, let [( q i ) i ∈ N ] ∈ X rω ⊆ X ω be arbitrary. Since d ω ([( q i ) i ∈ N ] , [( p i ) i ∈ N ]) = lim ω d i ( p i , q i ) ≤ r, [( q i ) i ∈ N ] ∈ ¯ B X ω r ( p ω ) .Now let [( q i ) i ∈ N ] ∈ B X ω r ( p ω ) and I := { i ∈ N | d i ( p i , q i ) < r } . Define ˜ q i := ( q i if i ∈ I,p i if i / ∈ I. By definition, [(˜ q i ) i ∈ N ] ∈ X rω . Furthermore, [( q i ) i ∈ N ] = [(˜ q i ) i ∈ N ] ∈ X rω : Since [( q i ) i ∈ N ] ∈ B X ω r ( p ω ) , ≤ l := lim ω d i ( q i , p i ) < r . For δ := r − l > , ω ( { i ∈ N | | d i ( q i , p i ) − l | < δ } ) ≤ ω ( { i ∈ N | d i ( q i , p i ) < l + δ = r } )= ω ( I ) . Thus, for arbitrary ε > , ω ( { i ∈ N | d i ( q i , ˜ q i ) < ε } ) ≥ ω ( { i ∈ N | q i = ˜ q i } )= ω ( I ) = 1 . Therefore, lim ω d i ( q i , ˜ q i ) = 0 and [( q i ) i ∈ N ] = [(˜ q i ) i ∈ N ] ∈ X rω . Consequently, B X ω r ( p ω ) ⊆ X rω . Since X rω is closed, this proves ¯ B X ω r ( p ω ) ⊆ X rω , and hence,equality, i.e. ¯ B X ω r ( p ω ) = lim ω ( ¯ B X i r ( p i ) , d i , p i ) .For any r > and ε ri := d GH (( ¯ B X i r ( p i ) , p i ) , ( ¯ B X ω r ( p ω ) , p ω )) , lim ω ε ri = 0 byProposition 3.9. By Lemma 3.12, there exists r i > with lim ω r i = 0 and ω (cid:16)n i ∈ N | ε r i i ≤ r i o(cid:17) = 1 . By Lemma 3.4, there is J = { i < i < . . . } ⊆ N such that ω ( J ) = 1 and r i j → ∞ . Let I := J ∩ n i ∈ N | ε r i i ≤ r i o . Then ω ( I ) = 1 and I = { i j < i j < . . . } ⊆ J . Thus, r i jl → ∞ and d GH (( ¯ B X ijl r ijl ( p i jl ) , p i jl ) , ( ¯ B X ω r ijl ( p ω ) , p ω )) = ε r ijl i jl ≤ r i jl → as l → ∞ . Now Corollary 2.10 proves ( X i jl , p i jl ) → ( X ω , p ω ) in the pointedGromov-Hausdorff sense where ω ( { i j l | l ∈ N } ) = 1 and this finishes theproof of the first part. The proof of the second statement can be done completely analogously tothe one of Lemma 3.4. (cid:3)
Lemma 3.12.
Let ω be an ultrafilter on N and for every r > let ( ε ri ) i ∈ N be a sequence such that lim ω ε ri = 0 . Then there exists a sequence ( r i ) i ∈ N ofpositive real numbers such that lim ω r i = 0 and ω ( { i ∈ N | ε r i i ≤ r i } ) = 1 .Proof. For i ∈ N , let R i := { r > | ε ri ≤ r } . The idea of this proof,similar to the one of Lemma 2.8, is to find a sequence r i ∈ R i with r i > i for a set of indices of ω -measure . Since the R i need to be non-empty, let I := { i ∈ N | R i = ∅} . Due to lim ω ε i = 0 , ω ( I ) = ω (cid:0)(cid:8) i ∈ N | ∃ r > ε ri ≤ r (cid:9)(cid:1) ≥ ω ( { i ∈ N | ε i ≤ } ) = 1 , i.e. ω ( I ) = 1 . Let J := { i ∈ N | ¬∃ C > R i ⊆ [0 , C ] } be the indicesof the unbounded sets. In particular, J ⊆ I . In the following, the cases of ω ( J ) = 0 and ω ( J ) = 1 will be distinguished.In advance, observe that for sets of indices of ω -measure the correspond-ing R i cannot have a uniform upper bound: Let A ⊆ N be any subset suchthat there exists C > with S i ∈ A R i ⊆ [0 , C ] and let r > C . Then i ∈ A implies r / ∈ R i , i.e. ε ri > r . Thus, ω ( A ) ≤ ω ( { i ∈ N | ε ri > r } ) = 0 .First, let ω ( J ) = 1 . For i ∈ J , choose r i ∈ R i ∩ ( i, ∞ ) . For i ∈ N \ J , let r i := 1 . Then ω (cid:16)n i ∈ N | ε r i i ≤ r i o(cid:17) ≥ ω ( { i ∈ N | r i ∈ R i } ) ≥ ω ( J ) = 1 . For arbitrary ε > , choose N ∈ N with N ≤ ε . For i ∈ J with i ≥ N , r i < i ≤ N ≤ ε and ω (cid:16)n i ∈ N | r i ≤ ε o(cid:17) ≥ ω ( J ∩ [ N, ∞ )) = 1 . Thus, lim ω r i = 0 and r i has the desired properties.Now let ω ( J ) = 0 . For i ∈ I ∩ J c , let s i := sup R i denote the least upperbound of R i and choose r i ∈ [ s i , s i ] ∩ R i . For i ∈ I c ∪ J , let s i := r i := 1 .Then ω (cid:16)n i ∈ N | ε r i i ≤ r i o(cid:17) ≥ ω ( { i ∈ N | r i ∈ R i } ) ≥ ω ( I ∩ J c ) = 1 . Let ε > and K ε := { i ∈ I ∩ J c | s i > ε } . Then [ i ∈ K ε R i ⊆ [ i ∈ K ε [0 , s i ] ⊆ h , ε i , OTES ON POINTED GROMOV-HAUSDORFF CONVERGENCE 47 and thus, by the above argumentation, ω ( K ε ) = 0 . Then, using ω ( I ∩ J c ) = 1 , ω (cid:0)(cid:8) i ∈ N | s i ≤ ε (cid:9)(cid:1) = 1 − ω (cid:0)(cid:8) i ∈ N | s i > ε (cid:9)(cid:1) = 1 − ω (cid:0)(cid:8) i ∈ I ∩ J c | s i > ε (cid:9)(cid:1) = 1 − ω ( K ε ) = 1 . Hence, lim ω s i = 0 and r i ≤ s i proves the claim. (cid:3) As for bounded sequences of real numbers, investigating sublimits comingfrom the same subsequence is the same as investigating ultralimits.
Lemma 3.13.
Let ( X i , d X i , p i ) and ( Y i , d Y i , q i ) , i ∈ N , be pointed properlength spaces. a) Let ω be a non-principal ultrafilter on N . Then there exists a subse-quence ( i j ) j ∈ N such that both ( X i j , p i j ) → lim ω ( X i , d X i , p i ) and ( Y i j , q i j ) → lim ω ( Y i , d Y i , q i ) in the pointed Gromov-Hausdorff sense as j → ∞ . b) Let ( X, d X , p ) and ( Y, d Y , q ) be pointed length spaces and ( i j ) j ∈ N be asubsequence such that both ( X i j , p i j ) → ( X, p ) and ( Y i j , q i j ) → ( Y, q ) in the pointed Gromov-Hausdorff sense as j → ∞ . Then there existsa non-principal ultrafilter ω on N such that there are isometries lim ω ( X i , d X i , p i ) ∼ = ( X, p ) and lim ω ( Y i , d Y i , q i ) ∼ = ( Y, q ) . Proof.
Using Proposition 3.11, the proof can be done completely analogouslyto the one of Lemma 3.5. (cid:3)
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