Gromov-Hausdorff Distance Between Segment and Circle
aa r X i v : . [ m a t h . M G ] J a n Gromov–Hausdorff Distance Between Segment and Circle
Yibo Ji, Alexey A. Tuzhilin
Abstract
We calculate the Gromov–Hausdorff distance between a line segment and a circle in theEuclidean plane. To do that, we introduced a few new notions like round spaces and nonlinearitydegree of a metric space.
The Gromov–Hausdorff distance measures the difference between any two metric spaces. There area few possibilities to do that. One of them, to embed isometrically the both spaces into all possibleother metric spaces, then the least possible Hausdorff distance between the images will be just thatcharacteristic. Another way is to establish correspondences between these spaces and to measurethe least possible distortion of the spaces metrics produced by these correspondences.The Gromov–Hausdorff distance has many applications, for example, it helps to investigate thegrowth of groups, or it can be used in image recognition. However, to get concrete values of thedistance, even in the cases of “simple spaces” like line segment and the standard circle, is a verynon-trivial task. In this paper we discuss our original technique enabled to obtain the exact valuesin the latter case.
In what follows, we work with various non-empty metric spaces, and the distance between points p and q of a metric space we denote by | pq | , independently on the choice of the space. For a metricspace X , x ∈ X , non-empty A, B ⊂ X , r >
0, and s ≥
0, we use the following notions and notations: • U r ( x ) = (cid:8) y ∈ X : | xy | < r (cid:9) ( open ball with center x and radius r ); • B s ( x ) = (cid:8) y ∈ X : | xy | ≤ s (cid:9) ( closed ball with center x and radius s ); • | xA | = inf a ∈ A | xa | ( distance between x and A ); • U r ( A ) = (cid:8) y ∈ X : | yA | < r (cid:9) ( open r -neighborhood of A ); • B s ( A ) = (cid:8) y ∈ X : | yA | ≤ s (cid:9) ( closed s -neighborhood of A ); • d H ( A, B ) = inf (cid:8) t : A ⊂ U t ( B ) and U t ( A ) ⊃ B (cid:9) ( Hausdorff distance between A and B ).For any non-empty set X , we denote by P ( X ) the set of all non-empty subsets of X . If X isa metric space, then the Hausdorff distance on P ( X ) satisfies the triangle inequality, but can beequal ∞ , and can vanish for different subsets. However, if we restrict d H to the set H ( X ) of allnon-empty closed bounded subsets of X , then d H becomes a metric, see [1].1 . Preliminaries Theorem 2.1 ([1]) . Given a metric space X , the following properties are simultaneously presentedor not in the both X and H ( X ): completeness, total boundness, compactness. Given r >
0, a subset of a metric space is called r -separated if the distance between any itsdifferent points is at least r .Given two sets X and Y , a correspondence between X and Y is each subset R ⊂ X × Y suchthat for any x ∈ X there exists y ∈ Y with ( x, y ) ∈ R and, vise versa, for any y ∈ Y there exists x ∈ X with ( x, y ) ∈ R . Let R ( X, Y ) denote the set of all correspondences between X and Y . If X and Y are metric spaces, and R ∈ R ( X, Y ), then we define the distortion dis R of R as follows:dis R = sup n(cid:12)(cid:12) | x x | − | y y | (cid:12)(cid:12) : ( x , y ) , ( x , y ) ∈ R o . Further, the Gromov–Hausdorff distance d GH ( X, Y ) between metric spaces X and Y , or for short GH-distance , is the value d GH ( X, Y ) = 12 inf (cid:8) dis R : R ∈ R ( X, Y ) (cid:9) . An equivalent definition: it is the infimum of Hausdorff distances d H ( X ′ , Y ′ ) between all possiblesubsets X ′ and Y ′ of the metric spaces Z , provided X ′ is isometric to X , and Y ′ is isometric to Y .It is well-known [1] that d GH is a metric on the set of isometry classes of compact metric spaces, inparticular, two compact metric spaces are isometric if and only if the GH-distance between themvanishes. In general situation, d GH satisfies the triangle inequality, is bounded for bounded spaces(for non-bounded spaces it can be infinite), and can vanish for non-isometric spaces. Also, forany A, B ∈ P ( X ), it holds d GH ( A, B ) ≤ d H ( A, B ). In particular, if ¯ A is the closure of A , then d GH ( A, ¯ A ) = d H ( A, ¯ A ) = 0.If X , X , . . . and X are some metric spaces such that d GH ( X i , X ) →
0, then we say that thesequence X i is Gromov–Hausdorff convergent to X and write X i GH −−→ X .It is easy to see that for any correspondence R ∈ R ( X, Y ), its closure in X × Y has the samedistortion, thus, to achieve d GH ( X, Y ), it suffices to consider only closed correspondences. In otherwords, if R c ( X, Y ) is the set of all closed correspondences between X and Y , then we have d GH ( X, Y ) = 12 inf (cid:8) dis R : R ∈ R c ( X, Y ) (cid:9) . For any metric space X and a real number λ >
0, we denote by λX the metric space whichdiffers from X by multiplication of all its distances by λ . For λ = 0, we define λX as a single pointspace. Theorem 2.2 ([1]) . Let X and Y be metric spaces. Then (1) if X is a single-point metric space, then d GH ( X, Y ) = diam Y ;(2) if diam X < ∞ , then d GH ( X, Y ) ≥ | diam X − diam Y | ;(3) d GH ( X, Y ) ≤ max { diam X, diam Y } , in particular, d GH ( X, Y ) < ∞ for bounded X and Y ;(4) for any metric spaces X , Y and any λ > , we have d GH ( λX, λY ) = λd GH ( X, Y ) . . Homogeneous and round metric spaces Fix a real 0 < b ≤ diam X and an integer n ≥
2. A metric space X is called ( b, n ) -homogeneous iffor any point x ∈ X there exists a b -separated n -point subset S ⊂ X such that x ∈ S . Theorem 3.1.
Let X be a ( b, n ) -homogeneous metric space. Suppose that Y is not ( a, n ) -homogeneousfor some < a < b . Then d GH ( X, Y ) ≥ b − a .Proof. Put c = b − a and suppose to the contrary that 2 d GH ( X, Y ) < c , then there exists R ∈R ( X, Y ) such that dis
R < c , therefore, for any ( x , y ) , ( x , y ) ∈ R with | x x | ≥ b , it holds | y y | > | x x | − c ≥ b − c = a . Since Y is not a -homogeneous, there exists y ∈ Y such thatthere is no ( a, n )-separated n -point subset T ⊂ Y with y ∈ T . Since R is a correspondence,there exists x ∈ X such that ( x, y ) ∈ R . Since X is ( b, n )-homogeneous, there exists b -separated S = { x , . . . , x n } ⊂ X such that x = x . Let T = { y , . . . , y n } with y = y and ( x i , y i ) ∈ R for any i . Then T is an a -separated n -point subset Y containing y , a contradiction.A metric space X is called round if it is ( b, < b < diam X . Example 3.2.
Let X ⊂ R n be a sphere of radius r > X is round, butnot ( a, a > r . Corollary 3.3.
Let X be a round metric space, < a < diam X , and suppose that a metric space Y is not ( a, -homogeneous. Then d GH ( X, Y ) ≥ diam X − a .Proof. Theorem 3.1 implies that 2 d GH ( X, Y ) ≥ b − a for each b , 0 < a < b < diam X , and theresult follows from the arbitrariness of b . Example 3.4.
Let X ⊂ R m and Y ⊂ R n be spheres of radii b and a w.r.t. some norms, then2 d GH ( X, Y ) ≥ b − a. In this section, we introduce the notion of nonlinearity degree and investigate several its properties.To start with, we denote by Lip a ( X ) the set of all real-valued a -Lipschitz functions defined on ametric space X . Definition 4.1.
The nonlinearity degree of a metric space X is defined as follows: c ( X ) := inf f ∈ Lip ( X ) sup (cid:8) | xy | − | f ( x ) − f ( y ) | : x, y ∈ X (cid:9) . Remark 4.2.
Clearly, c ( X ) ≥
0, and for any X ⊂ R we have c ( X ) = 0 (indeed, to achieve the value0, we can take the inclusion mapping as f ). Also, since c ( X ) ≤ diam X , then for each boundedspace X , the value c ( X ) is finite.If f : X → R is a -Lipschitz, then for any b ∈ R the function f + b is a -Lipschitz as well, and forany x, y ∈ X the value | xy | − | f ( x ) − f ( y ) | remains the same. In addition, for a bounded metricspace, each a -Lipschitz function is bounded, thus, to calculate the value c ( X ) in this case, we canrestrict ourselves by those f ∈ Lip ( X ) that satisfy inf f ( X ) = 0. The set of all such functions wedenote by Lip ( X ). Hence, we got the following . Nonlinearity degree of a metric space Theorem 4.3.
For any bounded metric space X , we have c ( X ) := inf f ∈ Lip ( X ) sup (cid:8) | xy | − | f ( x ) − f ( y ) | : x, y ∈ X (cid:9) . The nonlinearity degree can be used to estimate the minimal Gromov–Hausdorff distance be-tween metric space X and non-empty subsets of R . Theorem 4.4.
For any metric space X , we have (cid:12)(cid:12) X P ( R ) (cid:12)(cid:12) = inf Z ∈P ( R ) d GH ( X, Z ) ≤ c ( X ) / . If X is bounded, then there exists Z ∈ H ( R ) such that d GH ( X, Z ) ≤ c ( X ) / .Proof. If c ( X ) = ∞ then the result holds.Now, suppose that c ( X ) < ∞ . Take f k ∈ Lip ( X ) such thatsup n | xy | − (cid:12)(cid:12) f k ( x ) − f k ( y ) (cid:12)(cid:12) : x, y ∈ X o ≤ c ( X ) + 1 k , and put X k = f k ( X ). Let R k ∈ R ( X, X k ) be the graph of f k , namely, R k = n(cid:0) x, f k ( x ) (cid:1) : x ∈ X o . Since f k is 1-Lipschitz, then for any (cid:0) x, f k ( x ) (cid:1) , (cid:0) y, f k ( y ) (cid:1) ∈ R k we have | f k ( x ) − f k ( y ) | ≤ | xy | ,therefore, dis R k = sup x,y ∈ X n | xy | − (cid:12)(cid:12) f k ( x ) − f k ( y ) (cid:12)(cid:12)o ≤ c ( X ) + 1 k . Due to the arbitrariness of k ∈ N , we getinf Z ∈P ( R ) d GH ( X, Z ) ≤ inf k d GH ( X, X k ) ≤ c ( X ) / . Now we prove the existence of Z for bounded X . By Theorem 4.3, we can take f k from Lip ( X )instead of from Lip ( X ). Let X k be the closure of X k = f k ( X ). Since d GH ( X k , X k ) = 0, thenwe have inf k d GH ( X, X k ) ≤ c ( X ) / X implies d = diam X < ∞ , and by Theorem 2.2, we havediam X k = diam X k ≤ diam X + 2 d GH ( X, X k ) ≤ d + c ( X ) < ∞ . Since X k ≥ ∈ X k for each k ∈ N , then X k ⊂ (cid:2) , d + c ( X ) (cid:3) =: I for all k ∈ N , and, bycompactness of H ( I ) due to Theorem 2.1, there exists a convergent subsequence X k i ∈ H ( I ). Put Z = lim i →∞ X k i . Since d GH ( X k i , Z ) ≤ d H ( X k i , Z ), we obtain X k i GH −−→ Z , therefore d GH ( X, Z ) ≤ c ( X ) / Example 4.5.
Now we show that the inequality in Theorem 4.4 cannot be changed to equality,even in the case of compact X . Let X be the standard unit circle in the Euclidean plane, endowedwith the intrinsic metric. We shall find a compact Z ⊂ R such that d GH ( X, Z ) < c ( X ) /
2. Let ustake as Z the segment [0 , π ]. Below, in Lemma 6.1, we shall prove that d GH ( X, Z ) = π . On . Nonlinearity degree of a metric space f ∈ Lip ( S , R ), the function g ( x ) = f ( x ) − f ( − x ) has at least one zeropoint x . Thensup n | xy | − (cid:12)(cid:12) f ( x ) − f ( y ) (cid:12)(cid:12) : x, y ∈ X o ≥ | x ( − x ) | − (cid:12)(cid:12) f ( x ) − f ( − x ) (cid:12)(cid:12) = π. Since the choice of f is arbitrary, we know that c ( X ) ≥ π > d GH ( X, Z ).The next result characterizes all compact metric spaces with c ( X ) = 0. Corollary 4.6.
Each compact metric space X with c ( X ) = 0 is isometric to a compact subset of R .Proof. By Theorem 4.4, there exists a compact subset Z ⊂ R with d GH ( X, Z ) = 0, thus X and Z are isometric. Proposition 4.7.
For a compact metric space X , there exists a function f ∈ Lip ( X ) such that sup (cid:8) | xy | − | f ( x ) − f ( y ) | : x, y ∈ X (cid:9) = c ( X ) . Proof.
By Theorem 4.3, we can choose a sequence f n ∈ Lip ( X ) such that for each n ∈ N it holdssup (cid:8) | xy | − | f n ( x ) − f n ( y ) | : x, y ∈ X (cid:9) < c ( X ) + 1 n . The next lemma is a direct consequence of Arzel`a-Ascoli theorem [1].
Lemma 4.8.
Each uniformly bounded sequence of L -Lipschitz functions on a compact metric spacecontains a subsequence uniformly convergent to an L -Lipschitz function. Since any collection of functions from Lip ( X ) is uniformly bounded, we can apply Lemma 4.8,thus, w.l.o.g. we can suppose that the sequence f n itself uniformly converges to some f ∈ Lip ( X ).Take arbitrary ε > N ∈ N such that for any n ≥ N and any x ∈ X we have (cid:12)(cid:12) f ( x ) − f n ( x ) (cid:12)(cid:12) < ε/ /n < ε/ , hence for any x, y ∈ X and any n ≥ N it holdssup (cid:8) | xy | − | f ( x ) − f ( y ) | : x, y ∈ X (cid:9) ≤ sup (cid:8) | xy | − | f n ( x ) − f n ( y ) | : x, y ∈ X (cid:9) + 2 ε/ < c ( X ) + ε. Due to the arbitrariness of ε , we have sup (cid:8) | xy | − | f ( x ) − f ( y ) | : x, y ∈ X (cid:9) ≤ c ( X ), that concludesthe proof. Remark 4.9.
Proposition 4.7 gives another proof of Theorem 4 . X is compact. Theorem 4.10.
Suppose we have a connected compact metric space X equipped with a homeomor-phism α of order such that (cid:12)(cid:12) α ( x ) x (cid:12)(cid:12) = diam X for each x ∈ X , and a compact metric space Y such that c ( Y ) < diam X . Then we have d GH ( X, Y ) ≥ m , where m = (cid:0) diam X − c ( Y ) (cid:1) . . Nonlinearity degree of a metric space Proof.
Suppose to the contrary that 2 d GH ( X, Y ) < m , thus there is a closed correspondence R ∈R c ( X, Y ) such that dis
R < m .By Proposition 4 .
7, there exists v ∈ Lip ( Y ) such thatsup (cid:8) | y y | − | v ( y ) − v ( y ) | : y , y ∈ Y (cid:9) = c ( Y ) . In the following text, we will abbreviate c ( Y ) as c and α ( x ) as α x .Since X × Y is compact and R is its closed subset, then R is compact, together with R ( x ) = R ∩ (cid:0) { x } × Y (cid:1) and R ( α x ) = R ∩ (cid:0) { α x } × Y (cid:1) . Since the function v is continuous, then the restrictionsof v to R ( x ) and R ( α x ) are bounded and attain their maximal and minimal values, thus the followingfunction f : X → R is correctly defined: f ( x ) = max v (cid:0) R ( x ) (cid:1) − min v (cid:0) R ( α x ) (cid:1) . Lemma 4.11.
For any sequence x i ∈ X such that x i → ˜ x ∈ X it holds lim sup i → + ∞ f ( x i ) ≤ f (˜ x ) . Proof.
Let us put R v := (cid:8)(cid:0) x, v ( y ) (cid:1) : ( x, y ) ∈ R (cid:9) ⊂ X × R , then we can rewrite f as follows: f ( x ) = max R v ( x ) − min R v ( α x ) . Since R is compact, then R v is compact as the image of R under the continuous mapping id × v .Thus R v is closed and bounded.We put z i = max R v ( x i ), w i = min R v ( α x i ), then f ( x i ) = z i − w i , both the sequences z i , w i are bounded, and both the z := lim sup i → + ∞ z i , w := lim inf i → + ∞ w i are finite. Now we choosesubsequences z k i and w l i such that z = lim i → + ∞ z k i and w = lim i → + ∞ w l i . Since R v is closed, ( x k i , z k i ) , ( α x li , w l i ) ∈ R v , and z , w are finite, then(˜ x, z ) = lim i → + ∞ ( x k i , z k i ) ∈ R v and ( α ˜ x , w ) = lim i → + ∞ ( α x li , w l i ) ∈ R v , therefore, z ∈ R v (˜ x ) and w ∈ R v ( α ˜ x ). Thus, we getmax R v (˜ x ) ≥ z = lim sup i → + ∞ z i and min R v ( α ˜ x ) ≤ w = lim inf i → + ∞ w i . In account, f (˜ x ) ≥ z − w = lim sup i → + ∞ z i − lim inf i → + ∞ w i == lim sup i → + ∞ z i + lim sup i → + ∞ ( − w i ) ≥ lim sup i → + ∞ ( z i − w i ) = lim sup i → + ∞ f ( x i ) , and the proof is completed. . Nonlinearity degree of a metric space R < m and | x α x | = diam X , then for any y ∈ R ( x ) and y ∈ R ( α x ) we have | y y | > diam X − m , thus (cid:12)(cid:12) v ( y ) − v ( y ) (cid:12)(cid:12) ≥ | y y | − c > diam X − m − c =: d. So, either v ( y ) − v ( y ) > d or v ( y ) − v ( y ) > d . Since the values max v (cid:0) R ( x ) (cid:1) and min v (cid:0) R ( α x ) (cid:1) are attained, then for each x ∈ X it holds either f ( x ) > d or − f ( x ) > d , and the latter is equivalentto f ( x ) < − d .We put A := (cid:8) x ∈ X : f ( x ) > d (cid:9) and B := (cid:8) x ∈ X : f ( x ) < − d (cid:9) . Since c ≤ diam X and m = (diam X − c ), we get d = diam X − m − c = 13 (diam X − c ) ≥ ≥ −
13 (diam X − c ) = c + m − diam X = − d, hence A is the complement of B . Moreover, since f ( x ) = ± d , we have A = (cid:8) x ∈ X : f ( x ) ≥ d (cid:9) and B = (cid:8) x ∈ X : f ( x ) ≤ − d (cid:9) . Now we show that A is closed and, thus, B is open. Indeed, consider an arbitrary sequence x i ∈ A converging to some x ∈ X . Then, by Lemma 4.11, we have f ( x ) ≥ lim sup x i → x f ( x i ) ≥ d , thus x ∈ A and, therefore, A is closed. Lemma 4.12.
We have α ( B ) ⊂ A .Proof. Take arbitrary x ∈ B , then, by definition, max v (cid:0) R ( x ) (cid:1) − min v (cid:0) R ( α x ) (cid:1) ≤ − d , thus − f ( x ) = min v (cid:0) R ( x ) (cid:1) − max v (cid:0) R ( α x ) (cid:1) ≤ max v (cid:0) R ( x ) (cid:1) − min v (cid:0) R ( α x ) (cid:1) ≤ − d, so f ( x ) ≥ d and, hence, α x ∈ A . Lemma 4.13.
It holds that α ( A ) ⊂ B .Proof. If not, there exists x ∈ A such that α x ∈ A . Without loss of generality, suppose thatmax v (cid:0) R ( α x ) (cid:1) ≥ max v (cid:0) R ( x ) (cid:1) . Since x ∈ A , we have, by definition, max v (cid:0) R ( x ) (cid:1) ≥ d +min v (cid:0) R ( α x ) (cid:1) .Recall that for any y ∈ R ( x ) and y ′ ∈ R ( α x ) it holds (cid:12)(cid:12) v ( y ) − v ( y ′ ) (cid:12)(cid:12) > d , thus we getmax v (cid:0) R ( α x ) (cid:1) ≥ d + max v (cid:0) R ( x ) (cid:1) ≥ d + min v (cid:0) R ( α x ) (cid:1) . As we mentioned above, R ( α x ) is compact and v is continuous, hence there exist y , y ∈ R ( α x )such that v ( y ) = max v (cid:0) R ( α x ) (cid:1) and v ( y ) = min v (cid:0) R ( α x ) (cid:1) . Since v is 1-Lipschitz, we have | y y | ≥ v ( y ) − v ( y ), therefore, m > dis R ≥ diam R ( α x ) ≥ | y y | ≥ v ( y ) − v ( y ) == max v (cid:0) R ( α x ) (cid:1) − min v (cid:0) R ( α x ) (cid:1) ≥ d = 2 (cid:0) diam X − m − c (cid:1) . Thus m > (diam X − c ) which achieves contradiction and, hence, we have α ( A ) ⊂ B .Since X = A ⊔ B and α : X → X is a homeomorphism, Lemmas 4.12 and 4.13 imply that therestriction of α to A is a homeomorphism onto B , hence both A and B are proper closed-open,therefore, X is not connected, a contradiction which completes the proof of the theorem. . Geometric calculation of distortion Consider the special case we are interesting in, namely, let X = I λ ⊂ R be a segment of the length λ , and Y = S = (cid:8) ( x, y ) ∈ R : x + y = 1 (cid:9) be the standard circle endowed with the intrinsicmetric. Since we are interesting in the Gromov–Hausdorff distance between I λ and S , it does notmatter where we place I λ on the line R . We will use two possible placements: [0 , λ ] and [ − λ/ , λ/ R ∈ R ( I λ , S ). Our goal is to calculate geometrically the distortion of R . To do that, we parameterize the circle S by the standard angular coordinate − π ≤ ϕ ≤ π , thusthe interior distance between points ϕ , ϕ ∈ S equals min (cid:8) | ϕ − ϕ | , π − | ϕ − ϕ | (cid:9) . Let − λ/ ≤ t ≤ λ/ I λ . Now we represent the correspondence R asa subset of rectangle Q := [ − λ/ , λ/ × [ − π, π ] ⊂ R .To calculate the distortion, we need to maximize the function f (cid:0) ( t , ϕ ) , ( t, ϕ ) (cid:1) = ((cid:12)(cid:12) | t − t | − | ϕ − ϕ | (cid:12)(cid:12) if | ϕ − ϕ | ≤ π, (cid:12)(cid:12) | t − t | − π + | ϕ − ϕ | (cid:12)(cid:12) if | ϕ − ϕ | ≥ π over all ( t , ϕ ) , ( t, ϕ ) ∈ R . In other words, we need to find the least possible a such that f (cid:0) ( t , ϕ ) , ( t, ϕ ) (cid:1) ≤ a for all ( t , ϕ ) , ( t, ϕ ) ∈ R .Take arbitrary a >
0, and denote by D a ( t , ϕ ) the subset of the plane R consisting of all( t, ϕ ) ∈ R such that f (cid:0) ( t , ϕ ) , ( t, ϕ ) (cid:1) ≤ a . Clearly that D a ( t , ϕ ) can be obtained from D a := D a (0 ,
0) by shifting by ( t , ϕ ), i.e., D a ( t , ϕ ) = ( t , ϕ ) + D a . Evident observation shows that D a looks like the blue domain in Figure 1, left-hand side. aa-a -a π-π λ/2- /2λ tφ D a Q QD a D (t , ) a φ φ t (t , ) φ Figure 1: Geometric calculation of distortion. . Calculating GH-distance between circle and segment
Theorem 5.1.
Under the notations introduced above, dis R ≤ a if and only if for any ( t , ϕ ) ∈ R it holds R ⊂ D a ( t , ϕ ) . In particular, the distortion of R ⊂ Q equals the least possible a satisfyingthe previous condition. Remark 5.2.
If a graph P satisfies the above property with respect to a , then any subgraph of italso satisfies the above property.Let us note that the domain Q ∩ D a ( t , ϕ ) is depicted on the right-hand side of Figure 1 ingreen, as the intersection of blue and yellow domains. Now we apply the technique described above to calculate the Gromov–Hausdorff distance betweenthe standard circle S = (cid:8) ( x, y ) ∈ R : x + y = 1 (cid:9) with intrinsic metric and the segment I λ = [0 , λ ] ⊂ R . We start from the following Lemma 6.1.
For each ≤ λ < π we have d GH ( I λ , S ) ≥ π − λ . Proof.
Notice that the metric space S is round, diam S = π , and for any λ ′ > λ the segment I λ is not ( λ ′ / , π > λ ′ > λ we can use Corollary 3.3 and get d GH ( I λ , S ) ≥ π − λ ′ . The result follows from arbitrariness of λ ′ . Proposition 6.2.
For each ≤ λ ≤ π/ we have d GH ( I λ , S ) = π − λ . Proof.
By Lemma 6.1, we get the necessary lower bound for the d GH ( I λ , S ). It remains to get thesame upper bound. To do that, we will construct a special correspondence R λ ∈ R ( I λ , S ) withdis R λ = π − λ/
2, which completes the proof.For λ = 0 we put R λ = I λ × S . Since the space I λ consists of a single point, we getdis R λ = diam S = π = π − λ/ . Now, suppose that λ >
0. Identify R with C , and let R λ be the graph of the mapping γ λ : I λ → S defined as γ : t e πi t/λ . Notice that the image of the mapping t πt/λ , t ∈ I λ ,equals { e it : t ∈ [0 , π ] } , thus R λ is a correspondence, and for any t, t ′ ∈ I λ , we have (cid:12)(cid:12) γ ( t ) γ ( t ′ ) (cid:12)(cid:12) = min n πλ | t − t ′ | , π − πλ | t − t ′ | o . . Calculating GH-distance between circle and segment (cid:12)(cid:12) γ ( t ) γ ( t ′ ) (cid:12)(cid:12) depends only on s := | t − t ′ | , we put f ( s ) = (cid:12)(cid:12) γ ( t ) γ ( t ′ ) (cid:12)(cid:12) − | t − t ′ | , thendis R λ = max ≤ s ≤ λ (cid:12)(cid:12) f ( s ) (cid:12)(cid:12) . Since the function f ( s ) = ( πλ s − s for 0 ≤ πλ s ≤ π, π − πλ s − s for π ≤ πλ s ≤ π, is linear on [0 , λ/
2] and on [ λ/ , λ ], we getmax ≤ s ≤ λ (cid:12)(cid:12) f ( s ) (cid:12)(cid:12) = max n(cid:12)(cid:12) f (0) (cid:12)(cid:12) , (cid:12)(cid:12) f ( λ/ (cid:12)(cid:12) , (cid:12)(cid:12) f ( λ ) (cid:12)(cid:12)o = max (cid:8) , | π − λ/ | , λ (cid:9) . Since 0 ≤ λ ≤ π/
3, then | π − λ/ | = π − λ/ ≥ λ >
0, therefore dis R λ = π − λ/ Proposition 6.3. If π ≤ λ ≤ π , then we have d GH ( I λ , S ) = π .Proof. Applying Lemma 6.5, it suffices to construct a correspondence R λ ∈ R ( I λ , S ) with dis R λ =2 π/
3, which completes the proof.Again, we identify R with C , and let R λ be the graph of the mapping γ λ : I λ → S defined as γ : t e πi t . Notice that the image of the mapping t πt , t ∈ I λ , contains { e it : t ∈ [0 , π ] } ,thus R λ is a correspondence, and for any t, t ′ ∈ I λ , we have (cid:12)(cid:12) γ ( t ) γ ( t ′ ) (cid:12)(cid:12) = min n | t − t ′ | , π − | t − t ′ | , | t − t ′ | − π, π − | t − t ′ | o . Since the value (cid:12)(cid:12) γ ( t ) γ ( t ′ ) (cid:12)(cid:12) depends only on s := | t − t ′ | , we put f ( s ) = (cid:12)(cid:12) γ ( t ) γ ( t ′ ) (cid:12)(cid:12) − | t − t ′ | , thendis R λ = max ≤ s ≤ λ (cid:12)(cid:12) f ( s ) (cid:12)(cid:12) . Since the function f ( s ) = s for 0 ≤ s ≤ π , π − s for π ≤ s ≤ π, s − π for π ≤ s ≤ π, π − s for π ≤ s ≤ π, is linear on [0 , π/ π/ , π/ π/ , π ], [ π, π/ ≤ s ≤ π/ (cid:12)(cid:12) f ( s ) (cid:12)(cid:12) = max n(cid:12)(cid:12) f (0) (cid:12)(cid:12) , (cid:12)(cid:12) f ( π/ (cid:12)(cid:12) , (cid:12)(cid:12) f (2 π/ (cid:12)(cid:12) , (cid:12)(cid:12) f ( π ) (cid:12)(cid:12) , (cid:12)(cid:12) f (7 π/ (cid:12)(cid:12)o = max { , π/ } = 2 π/ , therefore, dis R λ = 2 π/ Proposition 6.4. If π/ ≤ λ ≤ π , then we have d GH ( I λ , S ) = λ − π .Proof. By Theorem 2.2, we have d GH ( I λ , S ) ≥ diam I λ − diam S λ − π . Thus, it suffices to construct a correspondence R ∈ R ( I λ , S ) with dis R ≤ λ − π =: a . We will usenotations from Section 5 and apply Theorem 5.1. . Calculating GH-distance between circle and segment λ 2/-λ/2 λ/2 /2- ππ /2 /2- λπ /2- π /2 π = - π π /2- / π AB DA ´ D ´ B ´ CC ´ O Figure 2: A correspondence R to prove Proposition 6.4.Consider the correspondence depicted in Figure 2.We have to show that for any ( t , ϕ ) ∈ R ⊂ [ − λ/ , λ/ × [ − π, π ] the domain D a ( t , ϕ ) covers R . In Figure 3 we show the corresponding domain D a endowed with its special sizes; also we writedown the corresponding sizes of segments forming R .Due to symmetry reasons, it suffices to verify the cases of ( t , ϕ ) lying on blue, green or cyansegments of R . We show that in these cases R is covered by orange cross and rose rectangle fromthe left-hand side of Figure 3.To simplify verification of the formulas below, let us explicitly write down the coordinates ofthe points A , B , C , and D from Figure 2 and 3 (points with the same names supplied by primesare symmetric w.r.t. the origin): A = (cid:16) λ − π , π (cid:17) , B = (cid:16) π , π (cid:17) , C = (cid:16) λ π , λ π (cid:17) , D = (cid:16) λ , π (cid:17) . It is easy to calculate that | AC | = 3 π − λ √ ≤ λ − π √ , | CD | = λ − π √ < λ − π √ , | B ′ C | = | BC ′ | = λ + 3 π √ < λ + π √ . Since the length and width of each of the two rectangles forming the orange cross are √ λ + π ) . Calculating GH-distance between circle and segment tφ D (t , ) a φ ( + π λ ) / √ ( - ) / √ λ π ( - ) / √ λ π ( π + ) / √ λ ( - ( ) π λ ) / √ ( - ( ) π λ ) / √ ( - ) / √ λ π ( ) ( - ) / ( √ ) λ π ( - ) / ( √ ) λ π ( - ) / √ λ π ( ) A C DBD´ B´A´C´ O
Figure 3: A correspondence R to prove Proposition 6.4.and √ λ − π ), respectively, then, once the point ( t , φ ) lies on the blue segment of the relation R , the orange cross covers R .Now, consider the case when ( t , φ ) lies on the green segment. Still, the whole correspondence R is covered by the orange cross. To see that, it suffices to notice the following evident formulas: | CC ′ | = λ + π √ , | CD | + | C ′ D ′ | = λ − π √ , | A ′ C ′ | − | CD | = 2 π − λ √ < λ − π √ , | BC | = λ − π √ < λ − π √ , | AD | = π √ < λ + π √ . At last, let ( t , φ ) lie on the cyan segment. Then, similarly, the whole correspondence R exceptthe magenta segment is covered by the orange cross. Let us show that the magenta segment iscovered by the both orange cross and the rose rectangle. This can be extracted from the followinginequalities: | AC | + | A ′ C ′ | = 3 π − λ √ < λ + π √ , | BB ′ | = √ π < √ π − λ − π √ , | CC ′ | = λ + π √ . The proof is completed.
Lemma 6.5.
For each π ≤ λ ≤ π we have d GH ( I λ , S ) ≥ π .Proof. Since I λ is a subset of R , we have c ( I λ ) = 0 by Remark 4.2.For x ∈ S , denote the antipodal point of x by α ( x ), then α : S → S is a homeomorphismof order 2 and (cid:12)(cid:12) xα ( x ) (cid:12)(cid:12) = diam S . Since S is connected and compact, and I λ is compact, and . Calculating GH-distance between circle and segment c ( I λ ) = 0 < π = diam S , we can apply Theorem 4.10 to the pair ( S , I λ ) and, thus, we get2 d GH ( I λ , S ) ≥ π . Proposition 6.6. If π ≤ λ ≤ π , then we have d GH ( I λ , S ) = π .Proof. Lemma 6.5 implies that it suffices to prove the inequality d GH ( I λ , S ) ≤ π . To do that, weconstruct a similar correspondence R ′ λ ∈ R ( I λ , S ) as in the proof of Proposition 6.4 as follows.Compared with the previous correspondence, the endpoints of the cyan (respectively, magenta)segment are ( π , π ) and ( π , π ) (respectively, ( − π , − π ) and ( − π , − π )).Let us put P = R ′ π . Note that P is exactly the same construction for λ = π as in Proposi-tion 6.4. Thus, P satisfies the condition of Theorem 5.1 with respect to a = π .Note that for λ ∈ [ π, π ], the correspondence R ′ λ is a subgraph of P since R ′ λ only has shortergreen and red segments compared with P . Since P satisfies Theorem 5.1 for a = π , we concludethat R ′ λ also satisfies Theorem 5.1 for a = π . The proof is completed. Proposition 6.7. If λ ≥ π , then we have d GH ( I λ , S ) = λ − π .Proof. Again, by Theorem 2.2, we have d GH ( I λ , S ) ≥ diam I λ − diam S λ − π . Let Y = { e it : 0 < t < } ⊂ C be the standard unit circle, and X = h − − λ − π , − i ∪ Y ∪ h , λ − π i ⊂ C is equipped with intrinsic metric. Then S is isometric to Y , and I λ is isometric to Z = h − − λ − π , − i ∪ { e it : π ≤ t ≤ π } ∪ h , λ − π i ⊂ X. Clearly that d H ( Y, Z ) = λ − π , thus d GH ( I λ , S ) ≤ λ − π . Gathering together Propositions 6.2–6.7, we get
Theorem 6.8.
Let S = (cid:8) ( x, y ) ∈ R : x + y = 1 (cid:9) be the standard circle with intrinsic metric,and I λ = [0 , λ ] ⊂ R the segment of the length λ . Then d GH ( I λ , S ) = π − λ f or ≤ λ ≤ π,π f or π ≤ λ ≤ π,λ − π f or λ ≥ π, see Figure . . Calculating GH-distance between circle and segment References [1] D.Burago, Yu.Burago, S.Ivanov,