On covering a ball by congruent subsets in normed spaces
OOn covering a ball by congruent subsets in normed spaces
Sergij V. Goncharov ∗ August 2017
Abstract
We consider the covering of a ball in certain normed spaces by its congruent subsets andshow that if the finite number of sets is not greater than the dimensionality of the space,then the centre of the ball either belongs to the interior of each set, or doesn’t belong to theinterior of any set. We also provide some examples when it belongs to the interior of exactlyone set. These are the specific cases of the modified problem originally posed for dissection.
MSC2010:
Pri 46B20, Sec 46C05, 52C17, 52C22
Keywords: ball, normed space, Hilbert space, cover, subset, congruence, motion, centre, interior, LSB
Introduction 11 Preliminaries 22 Main 7References 10
Introduction
In [7, C6, p. 87] the problem, attributed to S.K. Stein, is posed: “Whether it is possible topartition the unit circle into congruent pieces so that the center is in the interior of one of thepieces?”. At present, for arbitrary number of pieces it is considered to be unsolved ([26]). It can begeneralized and varied in many ways, as stated in same place ([7, p. 88]), not only dimensionality.Some related questions were studied and answered more or less fully, — [10], [11], [13], [16],[22], [25] to name a few. A problem of this kind may depend greatly on the meaning of involvedterms like “piece”, “partition”, “congruence”: do we allow the pieces to intersect at boundaries?does congruence include reflection? should the piece be connected? measurable? For example, itis shown in [25] that the ball in R m cannot be “strictly” dissected into n ∈ [2; m ] topologicallycongruent pieces, to say nothing of the centre; see also [24], [12, 25.A.6, p. 599], [11].Hereinafter, we distinguish between 3 types of “decomposition” of the set B (in particular,the ball) into the congruent (sub)sets { A i } i ∈ I , so that B = (cid:83) i ∈ I A i (cf. [7, p. 79], [14, p. 49]): • partition : { A i } are pairwise disjoint; • dissection : interiors of { A i } are pairwise disjoint; • covering (or intra covering to emphasize A i ⊆ B ): no additional constraints are required .These terms aren’t “standardized”, and may have quite different meaning in other works.Any partition is a dissection, and any dissection is a covering. Therefore, the impossibility ofcovering satisfying certain additional conditions (e.g. relating to the centre) implies that dissectionand partition satisfying the same conditions are not possible as well. However, when such coveringexists, the corresponding dissection or partition may not exist. ∗ Faculty of Mechanics and Mathematics, Oles Honchar Dnipro National University, 72 Gagarin Avenue, 49010Dnipro, Ukraine.
E-mail: [email protected] a r X i v : . [ m a t h . F A ] A ug ere we consider the “decomposition” of (intra)covering type, in certain specific cases, whilethe original problem almost surely belongs to dissection type; and the majority of referencedpapers, temporally ordered from [24] to [16], deals with partition. The routinism of inference suggests few/some/most/all of the presented “results” to be wellknown, even if not claimed explicitly or publicly, and the aim is rather to remove the delusion thatthere’s no such well-knowness...
There are works concerning the original centre-in-interior dissec-tion problem, under “natural” (or “physical”) assumptions (such as the space being Euclidean,boundaries of parts being rectifiable, parts being connected): cf. [13], [15], [3].In our opinion, the most similar negative result relating to covering is obtained for pre-Hilbertspaces in [10, Th. 1.1]: in spite of “indivisibility” and “partitioning” terms, it is actually a coveringconsidered there, under the assumption that exactly one set contains the centre. See Rem. 1 below.We try to attain the generality by considering spaces and coverings with as few additionalproperties and constraints as possible. Thereby few different interpretations of the problem (theball is closed/open etc.) are aggregated.Hereinafter, we consider a normed linear space X over the field of reals R , θ is the zero of X .Where we need the specific space such as R m , we will note it.Completeness of X is not assumed. (cid:107) x (cid:107) is the norm of x ∈ X , inducing the metric ρ ( x, y ) = (cid:107) x − y (cid:107) .The balls: open B ( x, r ) = { y ∈ X : (cid:107) y − x (cid:107) < r } , closed B ( x, r ) = { y ∈ X : (cid:107) y − x (cid:107) (cid:54) r } ; the(closed) sphere S ( x, r ) = { y ∈ X : (cid:107) y − x (cid:107) = r } . r > A ⊆ X and B ⊆ X congruent , A ∼ = B , iff there is an isometric surjectivemapping ( motion ) f : X ↔ X : ∀ x, y ∈ X : (cid:107) f ( x ) − f ( y ) (cid:107) = (cid:107) x − y (cid:107) (surjectivity implies that f − : X ↔ X is a motion too), and f ( A ) = B . The identity map I : I ( x ) = x is a motion.Int A = { x ∈ A | ∃ ε > B ( x, ε ) ⊆ A } and A = { x ∈ X | ∀ ε > B ( x, ε ) ∩ A (cid:54) = ∅ } are theinterior and the closure of A , respectively.We assume that X has these additional properties: • dim X > ∃ a, b ∈ X , which are linearly independent. • NCS: (cid:107) · (cid:107) is strictly convex, that is, ∀ x, y ∈ S ( θ, x (cid:54) = y : λ ∈ (0; 1) ⇒ (cid:107) λx + (1 − λ ) y (cid:107) < R m and R m ∞ when m (cid:62)
2, or L and L ∞ . Watery Warning: some of the following lemmas seem “folkloric”, with proofs included for thesake of integrity and probably present elsewhere.
Lemma 1. If f : X ↔ X is a motion, then ∀ S ( x, r ) : f (cid:0) S ( x, r ) (cid:1) = S ( f ( x ) , r ) .Proof. a) ∀ y ∈ S ( x, r ): (cid:107) f ( y ) − f ( x ) (cid:107) = (cid:107) y − x (cid:107) = r ⇒ f (cid:0) S ( x, r ) (cid:1) ⊆ S ( f ( x ) , r ). b) ∀ z ∈ S ( f ( x ) , r ): z = f ( y ), (cid:107) y − x (cid:107) = (cid:107) f ( y ) − f ( x ) (cid:107) = (cid:107) z − f ( x ) (cid:107) = r ⇒ y ∈ S ( x, r ) ⇒ S ( f ( x ) , r ) ⊆ f (cid:0) S ( x, r ) (cid:1) . Lemma 2. If f : X ↔ X is a motion, then ∀ B ( x, r ) : f (cid:0) B ( x, r ) (cid:1) = B ( f ( x ) , r ) .Proof. f (cid:0) B ( x, r ) (cid:1) = f (cid:0) { x } ∪ (cid:83) u ∈ (0; r ) S ( x, u ) (cid:1) = { f ( x ) } ∪ (cid:83) u ∈ (0; r ) f (cid:0) S ( x, u ) (cid:1) Lemma 1 == { f ( x ) } ∪ (cid:83) u ∈ (0; r ) S (cid:0) f ( x ) , u (cid:1) = B (cid:0) f ( x ) , r (cid:1) Lemma 3.
Let f : X ↔ X be a motion. Then f = h ◦ g (that is, f ( x ) = h (cid:0) g ( x ) (cid:1) ), where g : X ↔ X and h : X ↔ X are uniquely determined motions such that1) ∀ x ∈ X : (cid:107) g ( x ) (cid:107) = (cid:107) x (cid:107) ( ⇔ g ( θ ) = θ );2) ∃ a ∈ X : ∀ x ∈ X : h ( x ) = x + a . roof. Consider g ( x ) = f ( x ) − f ( θ ) and h ( x ) = x + f ( θ ). Obviously, h ◦ g = f . (cid:107) g ( x ) (cid:107) = (cid:107) f ( x ) − f ( θ ) (cid:107) = (cid:107) x − θ (cid:107) = (cid:107) x (cid:107) . (Implied by g ( θ )= θ : (cid:107) g ( x ) (cid:107) = (cid:107) g ( x ) − g ( θ ) (cid:107) = (cid:107) x − θ (cid:107) .) g and h are motions: (cid:107) g ( x ) − g ( y ) (cid:107) = (cid:107) f ( x ) − f ( y ) (cid:107) = (cid:107) x − y (cid:107) and (cid:107) h ( x ) − h ( y ) (cid:107) = (cid:107) x − y (cid:107) (isometry), inverse maps g − ( x ) = f − ( x + f ( θ )) and h − ( x ) = x − f ( θ ) imply surjectivity.Uniqueness: f ( θ ) = h ( g ( θ )) = h ( θ ) = a , g ( x ) = h − ( f ( x )) = f ( x ) − a = f ( x ) − f ( θ ).Here, we call h shift and g non-shift components of the motion f . If h = I or g = I , therespective component is called trivial . It is easy to see that if f has trivial shift or non-shiftcomponent, then the respective component of f − = g − ◦ h − is trivial as well. Theorem 1. (Mazur-Ulam, [21]; [18, 5.3, Th. 12]). The motion that maps θ to θ is linear. Remark.
We consider the isometries that map X onto itself, while the theorem holds true forany bijective isometry between two normed spaces X (with θ X ) and Y (with θ Y ). Corollary.
Non-shift component g of the motion f is linear: g ( λx + µy ) = λg ( x ) + µg ( y ). Lemma 4.
If the motion f : X ↔ X is such that ∃ x ∈ X : (cid:107) f ( x ) (cid:107) (cid:54) (cid:107) x (cid:107) and (cid:107) f ( − x ) (cid:107) (cid:54) (cid:107) x (cid:107) ,then the shift component of f is trivial.Proof. Using the notation of Lemma 3, let f = h ◦ g and y = g ( x ). For x = θ : y = θ , so f ( x ) = a ,and (cid:107) a (cid:107) (cid:54) ⇔ a = θ . Suppose x (cid:54) = θ .By Th. 1, − y = g ( − x ), so f ( x ) = y + a and (cid:107) y + a (cid:107) (cid:54) (cid:107) x (cid:107) = (cid:107) y (cid:107) , f ( − x ) = − y + a and (cid:107) − y + a (cid:107) (cid:54) (cid:107) y (cid:107) ⇔ (cid:107) y − a (cid:107) (cid:54) (cid:107) y (cid:107) . If (cid:107) y + a (cid:107) < (cid:107) y (cid:107) or (cid:107) y − a (cid:107) < (cid:107) y (cid:107) , then by triangle inequality2 (cid:107) y (cid:107) = (cid:107) y − ( − y ) (cid:107) (cid:54) (cid:107) y − a (cid:107) + (cid:107) a − ( − y ) (cid:107) < (cid:107) y (cid:107) , — a contradiction; thus (cid:107) y − a (cid:107) = (cid:107) y + a (cid:107) = (cid:107) y (cid:107) . y = ( y − a ) + ( y + a ). Assume a (cid:54) = θ ⇔ y − a (cid:54) = y + a . For s = ( y − a ) / (cid:107) y (cid:107) , t = y/ (cid:107) y (cid:107) , u = ( y + a ) / (cid:107) y (cid:107) : s, t, u ∈ S ( θ, s (cid:54) = u , (cid:107) s + u (cid:107) = (cid:107) t (cid:107) = 1, contradicting NCS. So a = θ .Let a , ..., a m be linear independent (LI) elements of X (thus dim X (cid:62) m ). We denote by M ( a , ..., a m ) = (cid:8) m (cid:80) i =1 x i a i | x i ∈ R (cid:9) the m -dimensional linear manifold generated by them. Itfollows from LI that ∀ x ∈ M ( a , ..., a m ) the coordinates { x i } are determined uniquely. Suppose x ( k ) , y ∈ M ( a , ..., a m ). Since (cid:107) x ( k ) − y (cid:107) (cid:54) m (cid:80) i =1 | x ( k ) i − y i |·(cid:107) a i (cid:107) by triangle inequality, we immediatelysee that x ( k ) i −−−−→ k →∞ y i for i = 1 , m implies x ( k ) −−−−→ k →∞ y , that is, (cid:107) x ( k ) − y (cid:107) −−−−→ k →∞ M ( a , ..., a m ) (making it a subspace of X ),though known well enough (see [6, 1.2.3], [18, 5.2, Ex. 4]), are obtained in the next lemma by“elementary” reasonings, without resort to norm equivalence or functionals. Lemma 5. If x ( k ) ∈ M ( a , ..., a m ) and x ( k ) −−−−→ k →∞ x ∈ X , then x ∈ M ( a , ..., a m ) , which is closedtherefore, and x ( k ) i −−−−→ k →∞ x i for i = 1 , m .Proof. The proof is by induction over dim M ( a , ..., a m ).Let m = 1. The sequence { x ( k ) } k = { x ( k )1 a } k is convergent (conv.), therefore it is fundamental(fund.) Assume that { x ( k )1 } k is not conv., then it isn’t fund. due to completeness of R : ∃ ε > ∀ N ∈ N : ∃ k , k > N : | x ( k )1 − x ( k )1 | (cid:62) ε But then (cid:107) x ( k ) − x ( k ) (cid:107) = | x ( k )1 − x ( k )1 | · (cid:107) a (cid:107) (cid:62) ε (cid:107) a (cid:107) >
0, which contradicts the fund. of { x ( k ) } k . Hence ∃ lim k →∞ x ( k )1 = (cid:101) x . Let (cid:101) x = (cid:101) x a . (cid:107) x ( k ) − (cid:101) x (cid:107) = | x ( k )1 − (cid:101) x | · (cid:107) a (cid:107) −−−−→ k →∞ ⇒ x ( k ) → (cid:101) x as k → ∞ . This means that x = (cid:101) x ∈ M ( a ) and x ( k )1 → x as k → ∞ .Now suppose that the statement holds true for dim M (cid:0) { a i } (cid:1) = 1 , , ..., m −
1. Consider theconv. { x ( k ) } k = { m (cid:80) i =1 x ( k ) i a i } k , it is fund. Take any i = 1 , m , for instance i = m . Assume that { x ( k ) m } isn’t conv., then it isn’t fund., ∃ ε > ∀ N ∈ N : ∃ k , k > N : | x ( k ) m − x ( k ) m | (cid:62) ε , and3 x ( k ) − x ( k ) (cid:107) = (cid:13)(cid:13) m (cid:80) i =1 ( x ( k ) i − x ( k ) i ) a i (cid:13)(cid:13) == | x ( k ) m − x ( k ) m | · (cid:13)(cid:13) a m + m − (cid:80) i =1 x ( k i − x ( k i x ( k m − x ( k m a i (cid:13)(cid:13) = | x ( k ) m − x ( k ) m | · (cid:107) a m − z (cid:107) where z ∈ M ( a , ..., a m − ) = M m − . It follows from dim M m − = m − M m − is closed. a m / ∈ M m − due to LI, therefore (cid:107) a m − z (cid:107) (cid:62) ρ ( a m , M m − ) > (cid:107) x ( k ) − x ( k ) (cid:107) (cid:62) ε ρ ( a m , M m − ) >
0, which contradicts the fund. of { x ( k ) } k .Hence ∃ lim k →∞ x ( k ) m = (cid:101) x m , and similarly ∃ lim k →∞ x ( k ) i = (cid:101) x i for i = 1 , m −
1. Let (cid:101) x = m (cid:80) i =1 (cid:101) x i a i . (cid:107) x ( k ) − (cid:101) x (cid:107) (cid:54) m (cid:80) i =1 | x ( k ) i − (cid:101) x i |·(cid:107) a i (cid:107) −−−−→ k →∞
0, so x ( k ) −−−−→ k →∞ (cid:101) x . Consequently, x = (cid:101) x ∈ M ( a , ..., a m )and x ( k ) i −−−−→ k →∞ x i for i = 1 , m . By induction principle, the statement is true for ∀ m ∈ N . Theorem 2. (Lusternik-Schnirelmann-Borsuk (LSB), [19, II.5], [5]; see also [20]). Let the sphere S m = (cid:8) x ∈ R m : (cid:107) x (cid:107) m = (cid:115) m (cid:80) j =1 x j = r (cid:9) = m (cid:83) i =1 A i , where A i are closed.Then ∃ i , ∃ x ∈ S m : { x, − x } ⊆ A i , — one of A i contains the pair of antipodal points of S m . The immediate corollary of LSB theorem is this generalization for normed spaces:
Lemma 6.
Let dim X (cid:62) m ∈ N , that is, ∃ a , ..., a m ∈ X , which are linearly independent.If S ( θ, r ) = m (cid:83) i =1 A i , where A i are closed, then ∃ A i , ∃ x ∈ S ( θ, r ) : { x, − x } ⊆ A i . See [4, p. 119], and most likely it’s mentioned in [23]; more general form is in e.g. [2, p. 39].
Proof.
Let L = M ( a , ..., a m ) be the subspace of X generated by { a i } (by Lemma 5, L is closed), C = S ( θ, r ) ∩ L , S m = (cid:8) y ∈ R m : (cid:107) y (cid:107) m = 1 (cid:9) . ∀ x ∈ L has the unique representation x =( x ; ... ; x m ) = m (cid:80) i =1 x i a i . Therefore the mapping s : C → S m : s ( x ) = (cid:0) x / (cid:107) x (cid:107) m ; ... ; x m / (cid:107) x (cid:107) m (cid:1) iswell defined. Moreover, we claim that s is a homeomorphism.1) s is injective. Indeed, if s ( x (cid:48) ) = s ( x (cid:48)(cid:48) ), where x (cid:48) , x (cid:48)(cid:48) ∈ C , then x (cid:48) i (cid:107) x (cid:48) (cid:107) m = x (cid:48)(cid:48) i (cid:107) x (cid:48)(cid:48) (cid:107) m for i = 1 , m ,thus x (cid:48) i = αx (cid:48)(cid:48) i for α = (cid:107) x (cid:48) (cid:107) m / (cid:107) x (cid:48)(cid:48) (cid:107) m >
0. So x (cid:48) = αx (cid:48)(cid:48) ⇒ r = (cid:107) x (cid:48) (cid:107) = | α | · (cid:107) x (cid:48)(cid:48) (cid:107) = rα ⇒ α = 1.2) s is surjective. ∀ y = ( y ; ... ; y m ) ∈ S m : s − ( y ) = r (cid:107) x (cid:107) x , where x = m (cid:80) i =1 y i a i .3) s is continuous. Let C (cid:51) x ( k ) −−−−→ k →∞ x . Using closedness of S ( θ, r ) and Lemma 5, we obtain: x ∈ S ( θ, r ) ∩ L = C and x ( k ) i −−−−→ k →∞ x i . Therefore (cid:107) x ( k ) (cid:107) m −−−−→ k →∞ (cid:107) x (cid:107) m , and s ( x ( k ) ) −−−−→ k →∞ s ( x ).4) s − is continuous too. For S m (cid:51) y ( k ) −−−−→ k →∞ y : S m is closed ⇒ y ∈ S m , and y ( k ) i −−−−→ k →∞ y i .Let x ( k ) = m (cid:80) i =1 y ( k ) i a i , x = m (cid:80) i =1 y i a i , then x ( k ) −−−−→ k →∞ x , (cid:107) x ( k ) (cid:107) −−−−→ k →∞ (cid:107) x (cid:107) , so s − ( y ( k ) ) −−−−→ k →∞ s − ( y ).Consider C i = A i ∩ C = A i ∩ S ( θ, r ) ∩ L , they are closed. Hence the image s ( C i ) ⊆ S m , underhomeomorphic mapping s , is closed too ([17, XII, § m (cid:83) i =1 C i = (cid:0) m (cid:83) i =1 A i (cid:1) ∩ C = S ( θ, r ) ∩ C = C ,so m (cid:83) i =1 s ( C i ) = S m . By LSB theorem, ∃ i , ∃ y ∈ S m : { y, − y } ∈ s ( C i ). Since s − ( − y ) = − s − ( y )(by (2)), we obtain x = s − ( y ) ∈ C : { x, − x } ∈ C i ⊆ A i .In the Main section, certain infinite-dimensional ball covering will be considered, where thefollowing Hilbert space-related lemmas are needed.4e denote by H = l the separable infinite-dimensional Hilbert space over R . Until the endof this section, (cid:107) · (cid:107) = (cid:107) · (cid:107) H denotes the norm in H . S = S ( θ,
1) is the unit sphere of H . (cid:104) x, y (cid:105) is the scalar/inner product of x, y ∈ H , ∠ ( x, y ) = arccos (cid:104) x,y (cid:105)(cid:107) x (cid:107)·(cid:107) y (cid:107) ∈ [0; π ] is the anglebetween x and y ( ∠ ( x, y ) = 0 if x = θ or y = θ ). x ⊥ y means (cid:104) x, y (cid:105) = 0. The “basic” propertiesof H and (cid:104)· , ·(cid:105) (like (cid:104) x, x (cid:105) = (cid:107) x (cid:107) ) are assumed to be known; see e.g. [6, II.3], [18, 6]. Lemma 7. ∃ D = { d i } i ∈ N ⊂ S such that ∀ β > , ∀ x ∈ S : ∃ d ∈ D : ∠ ( x, d ) < β . In other words, there’s a countable subset D of S , which is everywhere dense (ED) in “geodesic”metric ρ S ( x, y ) = ∠ ( x, y ) on S (see [9, 6.4, 17.4]). Such D is said to be geodesically dense in S . Proof. H is separable: ∃ R ⊂ H , countable and ED in H . Let D = { y (cid:107) y (cid:107) | y ∈ R, y (cid:54) = θ } ; D ⊂ S and D is countable. Take any x ∈ S . ∀ δ > ∃ y ∈ R : (cid:107) x − y (cid:107) < δ . Then, by triangle inequality,1 − δ < (cid:107) x (cid:107) − (cid:107) x − y (cid:107) (cid:54) (cid:107) y (cid:107) (cid:54) (cid:107) y − x (cid:107) + (cid:107) x (cid:107) < δ hence (cid:107) x − y (cid:107) y (cid:107) (cid:107) = (cid:107) y (cid:107) · (cid:13)(cid:13) x · (cid:107) y (cid:107) − y (cid:13)(cid:13) (cid:54) (cid:107) y (cid:107) (cid:104)(cid:13)(cid:13) ( (cid:107) y (cid:107) − x (cid:13)(cid:13) + (cid:107) x − y (cid:107) (cid:105) (cid:54) − δ (cid:2) δ (cid:107) x (cid:107) + δ (cid:3) = δ − δ .Since δ − δ → δ →
0, we obtain for ∀ ε > ∃ d = y (cid:107) y (cid:107) ∈ D : (cid:107) x − d (cid:107) < ε . Consider ε = n toget { d n } n ∈ N ⊂ D : d n −−−−→ n →∞ x . (So, D is ED on S in (cid:107) · (cid:107) -induced metric).It follows from continuity of (cid:107) · (cid:107) and (cid:104)· , ·(cid:105) that (cid:104) d n ,x (cid:105)(cid:107) d n (cid:107)·(cid:107) x (cid:107) −−−−→ n →∞ (cid:104) x,x (cid:105)(cid:107) x (cid:107) = 1. In turn, continuousarccos (cid:104) d n ,x (cid:105)(cid:107) d n (cid:107)·(cid:107) x (cid:107) −−−−→ n →∞
0, therefore ∃ d n β ∈ D : ∠ ( x, d n β ) < β . Remark.
Given such D , it is easy to see that { A i } i ∈ N = { B ( d i , ε ) ∩ S } i ∈ N for ε < S by closed subsets (moreover, A i ∼ = A j , see Lemma 10). dim H = ℵ = |{ A i }| ,however, diam A i (cid:54) diam B ( d i , ε ) = 2 ε <
2, thus no A i contains antipodal points of S , — the“straight” attempt of infinite-dimensional generalization of LSB theorem fails. Cf. [8]. Lemma 8. If g : H ↔ H is a non-shift motion, g ( θ ) = θ , then ∀ x, y ∈ H : (cid:10) g ( x ) , g ( y ) (cid:11) = (cid:104) x, y (cid:105) .Proof. (cid:10) g ( x ) , g ( y ) (cid:11) = (cid:10) g ( x ) − g ( y ) + g ( y ) , g ( y ) (cid:11) = (cid:10) g ( x ) − g ( y ) , g ( y ) (cid:11) + (cid:107) g ( y ) (cid:107) == (cid:10) g ( x ) − g ( y ) , g ( y ) − g ( x ) + g ( x ) (cid:11) + (cid:107) y (cid:107) == − (cid:10) g ( x − y ) , g ( x − y ) (cid:11) + (cid:107) g ( x ) (cid:107) − (cid:10) g ( x ) , g ( y ) (cid:11) + (cid:107) y (cid:107) = (cid:107) x (cid:107) + (cid:107) y (cid:107) − (cid:107) x − y (cid:107) − (cid:10) g ( x ) , g ( y ) (cid:11) ,therefore (cid:10) g ( x ) , g ( y ) (cid:11) = (cid:2) (cid:107) x (cid:107) + (cid:107) y (cid:107) − (cid:104) x − y, x − y (cid:105) (cid:3) = (cid:2) (cid:104) x, y (cid:105) (cid:3) = (cid:104) x, y (cid:105) . Definition.
Let H (cid:51) s (cid:54) = e ∈ H , γ ∈ [0; π ]. We call the set C ( s, e, γ ) = (cid:8) x ∈ H : (cid:107) x − s (cid:107) (cid:54) (cid:107) e − s (cid:107) and ∠ ( x − s, e − s ) (cid:54) γ (cid:9) ⊆ B ( s, (cid:107) e − s (cid:107) )the (closed) ommatidium , with origin at s , around [ s, e ], of angle γ and of radius (cid:107) e − s (cid:107) .It’s actually a “sector” of the ball B ( s, (cid:107) e − s (cid:107) ), and would be a usual disk sector in R . Lemma 9. If s (cid:54) = x ∈ C ( s, e, γ ) , then ∀ λ ∈ [0; (cid:107) e − s (cid:107)(cid:107) x − s (cid:107) ] : s + λ ( x − s ) ∈ C ( s, e, γ ) .Proof. It follows simply from the definition.
Lemma 10.
Two ommatidiums of the same angle and radius are congruent in H .Proof. Evidently, a parallel shift h ( x ) = x + a transforms C ( s, e, γ ) onto C ( s + a, e + a, γ ). Thus weconsider, without loss of generality, C = C ( θ, e , γ ) and C = C ( θ, e , γ ), where (cid:107) e (cid:107) = (cid:107) e (cid:107) = r , e (cid:54) = e . We are going to find the non-shift motion g such that g ( C ) = C .It suffices to obtain g such that g ( e ) = e . Indeed, ∀ x ∈ C we have then (cid:107) g ( x ) (cid:107) = (cid:107) x (cid:107) (cid:54) r and ∠ ( g ( x ) , e ) = arccos (cid:104) g ( x ) ,g ( e ) (cid:105)(cid:107) g ( x ) (cid:107)·(cid:107) g ( e ) (cid:107) Lemma 8 = arccos (cid:104) x,e (cid:105)(cid:107) x (cid:107)·(cid:107) e (cid:107) = ∠ ( x, e ) (cid:54) γ , so g ( x ) ∈ C .Conversely, ∀ x ∈ C : g − ( x ) ∈ C , because g − is a non-shift motion as well, and g − ( e ) = e .We apply the “coordinate” approach to define such g .Let e (cid:48) = e /r , e (cid:48) = e /r . They generate the 2-dimensional subspace M = M ( e (cid:48) , e (cid:48) ) of H . ∃ u ∈ M such that (cid:107) u (cid:107) = 1 and u ⊥ e (cid:48) , hence M = M ( e (cid:48) , u ) and ∀ z ∈ M : z = z e (cid:48) + z u , (cid:107) z (cid:107) = z + z . Then e (cid:48) = (cos α ) e (cid:48) + (sin α ) u for some α ∈ (0; 2 π ).5 = M ⊕ L , where L is the orthogonal complement of M . It follows that ∀ x ∈ H has uniquerepresentation x = x e (cid:48) + x u + w x , where w x ∈ L , and (cid:107) x (cid:107) = x + x + (cid:107) w x (cid:107) . In particular, e = re (cid:48) and e = ( r cos α ) e (cid:48) + ( r sin α ) u .Let g ( x ) = ( x cos α − x sin α ) e (cid:48) + ( x sin α + x cos α ) u + w x . It has the required properties:1) g is isometric: (cid:107) g ( x ) − g ( y ) (cid:107) == (cid:2) ( x − y ) cos α − ( x − y ) sin α (cid:3) + (cid:2) ( x − y ) sin α + ( x − y ) cos α (cid:3) + (cid:107) w x − w y (cid:107) == ( x − y ) + ( x − y ) + (cid:107) w x − w y (cid:107) = (cid:107) x − y (cid:107) g is surjective: g − ( x ) = ( x cos α + x sin α ) e (cid:48) + ( − x sin α + x cos α ) u + w x .3) g ( θ ) = θ + θ + θ = θ , and 4) g ( e ) = ( r cos α ) e (cid:48) + ( r sin α ) u + θ = e . Lemma 11. If D = { d i } i ∈ N ⊂ S is geodesically dense in S , then ∀ β > : B ( θ,
1) = (cid:83) i ∈ N C ( θ, d i , β ) .Proof. C ( θ, d i , β ) ⊆ B ( θ,
1) is obvious. θ ∈ C ( θ, d i , β ) for any i . Take ∀ x ∈ B ( θ, \{ θ } , then x (cid:48) = x/ (cid:107) x (cid:107) ∈ S and x = (cid:107) x (cid:107) · x (cid:48) . By definition of D , ∃ d ∈ D : ∠ ( x (cid:48) , d ) < β , thus x (cid:48) ∈ C ( θ, d, β ).By Lemma 9, x ∈ C ( θ, d, β ) too. Lemma 12.
Let d ∈ S and γ (cid:54) arccos . Then C = C ( − d, d, γ ) ⊂ B ( θ, .Proof. Due to convexity of B ( θ, ∀ x ∈ S ( − d, ∩ C : (cid:107) x (cid:107) (cid:54) ∀ y ∈ C \{− d } : y = λx + (1 − λ )( − d ) for x = − d + y + d (cid:107) y + d (cid:107) ∈ S ( − d, ∩ C ( ∈ C follows from Lemma 9) and λ = (cid:107) y + d (cid:107) ∈ [0; 1].Let H = M ( d ) ⊕ T , then ∀ y ∈ H : y = y d + y u , where d ⊥ u ∈ T , (cid:107) u (cid:107) = 1, and (cid:107) y (cid:107) = y + y .In particular, for y = x + d : (cid:107) y (cid:107) = 1, hence we can represent y = cos β (cid:62) y = sin β for β ∈ [0; 2 π ). At that y = (cid:104) y, d (cid:105) = (cid:104) y,d (cid:105)(cid:107) y (cid:107)·(cid:107) d (cid:107) = cos ∠ ( x + d, d ). x ∈ C , so y = cos β (cid:62) cos γ (cid:62) , (cid:107) x (cid:107) = (cid:107) y − d (cid:107) = (cid:107) (cos β − ) d + sin β · u (cid:107) = (cos β − ) + sin β = − cos β (cid:54) Lemma 13. If γ < π , then s / ∈ Int C ( s, e, γ ) .Proof. Let v = e − s , then ∀ ε > ∠ ( − εv, e − s ) = arccos (cid:104)− εv,v (cid:105)(cid:107)− εv (cid:107)·(cid:107) v (cid:107) = arccos( −
1) = π > γ , hence B ( s, ε (cid:107) e − s (cid:107) ) (cid:51) s − εv / ∈ C ( s, e, γ ), and B ( s, ε (cid:107) e − s (cid:107) ) (cid:42) C ( s, e, γ ). Lemma 14. If γ > , then ( s + e ) ∈ Int C ( s, e, γ ) .Proof. Without loss of generality, assume that s = θ , (cid:107) e (cid:107) = 1, and γ (cid:54) π (otherwise move theommatidium so that its origin becomes θ by Lemma 10, scale it to attain (cid:107) e (cid:107) = 1 ( x ↔ x/ (cid:107) e (cid:107) ),and consider C ( θ, e, π ) ⊆ C ( θ, e, γ )). We need to show that ∃ ε > B ( e, ε ) ⊆ C ( θ, e, γ ) ⇔∀ x ∈ B ( e, ε ): (cid:107) x (cid:107) (cid:54) ∠ ( x, e ) (cid:54) γ ; the latter inequality is equivalent to cos ∠ ( x, e ) (cid:62) cos γ .For arbitrary ε > ∀ x ∈ B ( e, ε ): x = e + b , where (cid:107) b (cid:107) < ε .Then (cid:107) x (cid:107) (cid:54) (cid:107) e (cid:107) + (cid:107) b (cid:107) < + ε ; the constraint ε < ensures (cid:107) x (cid:107) < ∠ ( x, e ) = (cid:104) x,e (cid:105)(cid:107) x (cid:107)·(cid:107) e (cid:107) = (cid:107) e + b (cid:107) (cid:2) (cid:104) e, e (cid:105) + (cid:104) b, e (cid:105) (cid:3) = (cid:107) e +2 b (cid:107) + (cid:107) e +2 b (cid:107) (cid:104) b, e (cid:105) (cid:107) e + 2 b (cid:107) (cid:54) (cid:107) e (cid:107) + 2 (cid:107) b (cid:107) < ε , hence (cid:107) e +2 b (cid:107) > ε .2) On the other hand, (cid:107) e + 2 b (cid:107) (cid:62) (cid:107) e (cid:107) − (cid:107) − b (cid:107) > − ε ⇒ (cid:107) e +2 b (cid:107) < − ε , and Cauchy-Bunyakowsky-Schwartz inequality implies (cid:12)(cid:12) (cid:104) b, e (cid:105) (cid:12)(cid:12) (cid:54) (cid:107) b (cid:107) · (cid:107) e (cid:107) < ε , therefore (cid:107) e +2 b (cid:107) (cid:104) b, e (cid:105) > − ε − ε .Consequently (for sufficiently small ε ) cos ∠ ( x, e ) > ε − ε − ε → ε →
0, thus for some ε ∈ (0; ) we obtain: cos ∠ ( x, e ) (cid:62) cos γ for each x ∈ B ( e, ε ). Lemma 15. If γ (cid:54) π , then C ( s, e, γ ) is convex.Proof. Again, we assume s = θ and (cid:107) e (cid:107) = 1 without loss of generality.Let x, y ∈ C ( θ, e, γ ). We claim that ∀ λ ∈ [0; 1]: z = λx + (1 − λ ) y ∈ C ( θ, e, γ ).If x = θ or y = θ , then z ∈ C ( θ, e, γ ) by Lemma 9. If not: clearly θ ∈ C ( θ, e, γ ), suppose z (cid:54) = θ .1) (cid:107) z (cid:107) (cid:54) λ (cid:107) x (cid:107) + (1 − λ ) (cid:107) y (cid:107) (cid:54) λ · − λ ) · ∠ ( z, e ) = (cid:104) z,e (cid:105)(cid:107) z (cid:107)·(cid:107) e (cid:107) = (cid:2) λ (cid:104) x,e (cid:105)(cid:107) z (cid:107) + (1 − λ ) (cid:104) y,e (cid:105)(cid:107) z (cid:107) (cid:3) = (cid:2) λ (cid:104) x,e (cid:105)(cid:107) x (cid:107) · (cid:107) x (cid:107)(cid:107) z (cid:107) + (1 − λ ) (cid:104) y,e (cid:105)(cid:107) y (cid:107) · (cid:107) y (cid:107)(cid:107) z (cid:107) (cid:3) == (cid:2) λ (cid:107) x (cid:107)(cid:107) z (cid:107) cos ∠ ( x, e ) + (1 − λ ) (cid:107) y (cid:107)(cid:107) z (cid:107) cos ∠ ( y, e ) (cid:3) (cid:62) λ (cid:107) x (cid:107) +(1 − λ ) (cid:107) y (cid:107)(cid:107) λx +(1 − λ ) y (cid:107) cos γ cos γ (cid:62) (cid:62) cos γ ⇔ ∠ ( z, e ) (cid:54) γ Main
Proposition 1.
Let dim X (cid:62) m ∈ N , B ( θ, ⊆ E ⊆ B ( θ, , and E = m (cid:83) i =1 A i , where A i ∼ = A j .Then either θ ∈ m (cid:84) i =1 Int A i , or θ / ∈ m (cid:83) i =1 Int A i .Proof. Suppose m (cid:62)
2. Let K = B ( θ, S = S ( θ, K = B ( θ, ⊆ E ⊆ B ( θ,
1) = K ⇔ E = K .Let f ij be the motion transforming A i to A j , so that f ij ( A i ) = A j , and f ji = f − ij ( f ii = I ).Consider S i = A i ∩ S . They are closed and m (cid:83) i =1 S i = (cid:0) m (cid:83) i =1 A i (cid:1) ∩ S = m (cid:83) i =1 A i ∩ S = K ∩ S = S .By Lemma 6, ∃ S k , ∃ d ∈ S : { d, − d } ⊆ S k .Take any i (cid:54) = k . Let A (cid:48) k = A k ∪ S k ∪ f − ki ( S i ) and A (cid:48) i = A i ∪ S i ∪ f ki ( S k ).1) A (cid:48) i ⊆ K . Indeed, a) A i ⊆ E ⊆ K , b) S i ⊆ S ⊂ K , c) ∀ x ∈ S k ⊆ A k ∃{ x l } ∞ l =1 , x l ∈ A k : x l −−−→ l →∞ x , then continuous f ki ( x l ) −−−→ l →∞ f ki ( x ). f ki ( x l ) ∈ A i , hence f ki ( x ) ∈ A i ⊆ E = K .2) f ki ( A (cid:48) k ) = f ki ( A k ) ∪ f ki ( S k ) ∪ f ki (cid:0) f − ki ( S i ) (cid:1) = A i ∪ S i ∪ f ki ( S k ) = A (cid:48) i .Since { d, − d } ⊆ S k ⊆ A (cid:48) k , we obtain f ki (cid:0) { d, − d } (cid:1) ⊆ A (cid:48) i ⊆ K , so (cid:107) f ki ( d ) (cid:107) (cid:54) (cid:107) d (cid:107) and (cid:107) f ki ( − d ) (cid:107) (cid:54) (cid:107) d (cid:107) . By Lemma 4, the shift component h ki of f ki = h ki ◦ g ki is trivial. Then theshift component h ik of f ik = f − ki is also trivial.There are 2 possible cases: either ∃ i : θ ∈ Int A i , or ∀ i : θ / ∈ Int A i ⇔ θ / ∈ m (cid:83) i =1 Int A i .Consider the former case, then ∃ B ( θ, ε ) ⊆ A i . Take ∀ j (cid:54) = i . By Lemma 2, f ik (cid:0) B ( θ, ε ) (cid:1) = B (cid:0) f ik ( θ ) , ε (cid:1) = B (cid:0) g ik ( θ ) , ε (cid:1) = B ( θ, ε ), hence B ( θ, ε ) ⊆ A k . Apply Lemma 2 again: f kj (cid:0) B ( θ, ε ) (cid:1) = B (cid:0) f kj ( θ ) , ε (cid:1) = B ( θ, ε ) ⊆ A j , and θ ∈ Int A j . Therefore θ ∈ m (cid:84) i =1 Int A i . Corollary.
If dim X (cid:62) ℵ , then the statement of Prop. 1 holds true for ∀ m ∈ N : a ball insuch X cannot be covered by any finite number of congruent subsets so that its centre belongs tothe interiors of certain of them and doesn’t belong to the interiors of the others.As for infinite coverings, see Ex. 3 and Ex. 4 below. Remark 1.
One may ask why we do not generalize the approach from [10] instead.The reasonings there essentially make use of the inequality ∀ x, y, z ∈ X : (cid:107) x − y (cid:107) + (cid:107) z (cid:107) (cid:54) (cid:107) x (cid:107) + (cid:107) y (cid:107) + (cid:107) x − z (cid:107) + (cid:107) y − z (cid:107) which is the implication of the inequality [10, p. 184, (c)] (for p ← θ , q ← z = σ A ( θ )), establishedfor Euclidean/pre-Hilbert X . Unfortunately, it is not true for arbitrary NCS X : consider X = R / with (cid:107) x (cid:107) = (cid:13)(cid:13) ( x ; x ) (cid:13)(cid:13) / = (cid:0) | x | / + | x | / (cid:1) / and let x = (1; 0), y = (0; 1), z = (1; 1). Then (cid:107) x − y (cid:107) + (cid:107) z (cid:107) = 2 · = 4 · > (cid:107) x (cid:107) + (cid:107) y (cid:107) + (cid:107) x − z (cid:107) + (cid:107) y − z (cid:107) (Maybe some subtler form of the inequality would work.) Remark 2.
On the other hand, LSB theorem is applied here too, being a “foundation stone”of the inference; another pebble is that the motions transforming the subsets onto each other don’tinclude parallel shift, otherwise one of antipodal points moves outside of the ball.Antipodal/“diametral” points and the constraints they impose are exploited, — without resortto LSB theorem, — in [11, § Remark 3.
If we replace the condition “ A i ∼ = A j ” by “Int A i ∼ = Int A j ”, then “ θ ∈ Int A and θ / ∈ Int A ” becomes possible, evidently; for example, in R take z ∈ K : (cid:107) z (cid:107) = , and A = B ( θ, ) ∪ { ( x ; y ) ∈ K | x ∈ Q and y ∈ Q } , A = B ( z, ) ∪ { ( x ; y ) ∈ K | x / ∈ Q or y / ∈ Q } then A ∪ A = K , Int A = B ( θ, ) (cid:51) θ , Int A = B ( z, ), so Int A ∼ = Int A , A ∩ A = ∅ .Same happens if we replace “congruence” by “homotheticity”: take A = K and let A , A ,... be the balls of sufficiently small radius ρ so that all of them can be placed within K and don’tcontain its centre (in other words, A i = ρK + c i , ρ < (cid:107) c i (cid:107) < − ρ for i (cid:62) xample 1. (cid:67) Without NCS, the Prop. 1 statement can become false. Consider non-NCS l ∞ = (cid:8) x = ( x ; x ; ... ) : (cid:107) x (cid:107) ∞ = sup i ∈ N | x i | < ∞ (cid:9) and its unit ball B ( θ,
1) = (cid:8) x ∈ l ∞ : sup i | x i | (cid:54) (cid:9) .For any odd n (cid:62) A i = (cid:8) x ∈ B ( θ,
1) : x ∈ [ − i − n ; − in ] (cid:9) , i = 1 , n , arecongruent (motion f ij ( x ) = ( x +2 j − in ; x ; x ; ... ) transforms A i to A j ), θ ∈ B ( θ, n ) ⊂ Int A (cid:98) n (cid:99) ,while θ / ∈ A i for i (cid:54) = 1 + (cid:98) n (cid:99) , and A ∪ ... ∪ A n = B ( θ, l ∞ , we can take R m ∞ if m (cid:62) n .Note that this decomposition of B ( θ,
1) is a dissection and a covering (but not a partition). (cid:66)
Remark 4.
Consider R p , (cid:107) x (cid:107) p = (cid:0) | x | p + | x | p (cid:1) p . For p = 2, usual Euclidean metric, theoriginal dissection problem posed in [7, C6] remains unsolved. For p = 1 or p = ∞ , — non-NCScase, — B ( θ,
1) is a square (sides being parallel to Ox i for p = ∞ , rotated by π for p = 1), triviallydissectable into 3 (5, 7, ...) congruent rectangles such that the centre θ is within one of them.It is shown in [11, §
2] that B ( θ,
1) in non-NCS c , C [0;1] is partitionable into n congruent subsetsfor ∀ n (cid:54) ℵ , while in NCS Banach X there’s no such partition if 2 (cid:54) n < min { dim X, ℵ } + 1. Example 2. (cid:67)
Obviously, as Fig. 1 illustrates, the ball/disk in R can be covered by n (cid:62) n = 4 n = 5 n = 17Figure 1: Covering a disk by n (cid:62) n = 3 is slightly different: the sets are not convex and not 1-connected, each one hasa circular hole in one of two symmetric segments it consists of. At Fig. 2, ∠ AOB = 150 ◦ (forinstance). We do not know is there any such covering by three 1-connected congruent subsets.Figure 2: Covering a disk by n = 3 congruent subsets (resembles “Biohazard” symbol)In fact, the case n = k can be extended to all n > k (which makes Fig. 1 redundant), becausecovering allows A i = A j (not so for dissection): take A , ..., A k − , and A k = A k +1 = ... = A n .Similar constructions can be used in R m . In particular, when n = m +2, note that the “hollow”around the centre at Fig. 1, case n = 4, is an equilateral triangle and a 2-simplex in R . (cid:66) emark 5. Convexity of parts implies the negative answer not only to the original dissectionproblem, but also to its generalization: the closed disk in R cannot be dissected into n (cid:62) Proof.
Let K = B ( θ, S = S ( θ,
1) in R , and let { A i } ni =1 be the parts, so that K = n (cid:83) i =1 A i , A i ∼ A j , Int A i ∩ Int A j = ∅ for i (cid:54) = j . Also, let ∂A i = A i ∩ R \ A i ⊂ A i be the boundary of A i .1) Claim: if ∂A i contains 2 n +4 different points x , ..., x n +4 that belong to some circle S ( a, r ),then S ( a, r ) = S . (“The strictly convex section of ∂A i has to be on ∂K = S , not inside K .”)To show that this claim is true, assume the contrary: ∃ x j / ∈ S . Let N + be the number of x j ∈ S , and N − = (cid:12)(cid:12) { x j : x j / ∈ S } (cid:12)(cid:12) . N + + N − = 2 n + 4. If N + (cid:62)
3, then 3 points that ∈ S among x , ..., x n +4 determine the circle S ( a, r ) uniquely (see e.g. [1, 2.3, Cor. 7]), so S ( a, r ) = S , whichcontradicts the assumption. Thus N + (cid:54) ⇒ N − (cid:62) n + 2: we can take 2 n + 2 points on S ( a, r )in Int K = B ( θ, x (cid:48) , ..., x (cid:48) n +2 .Let x (cid:48) x (cid:48) ...x (cid:48) n +1 be the convex polygon, with interior, inscribed into S ( a, r ); it follows fromconvexity of A i and n + 1 (cid:62) x (cid:48) x (cid:48) ...x (cid:48) n +1 ⊆ A i and ∅ (cid:54) = Int x (cid:48) x (cid:48) ...x (cid:48) n +1 ⊆ Int A i . Considerthe rest of points: x (cid:48) , x (cid:48) , ..., x (cid:48) n +2 . Since x (cid:48) j ∈ ∂A i ∩ Int K , each of these n + 1 points belongs tothe boundary ∂A k j of at least one other part, k j (cid:54) = i . There are n − x (cid:48) (1) and x (cid:48) (2) , belong to the boundary of the same A k , k (cid:54) = i . Then [ x (cid:48) (1) , x (cid:48) (2) ] ⊂ A k .It is easy to see that [ x (cid:48) (1) , x (cid:48) (2) ] intersects Int x (cid:48) x (cid:48) ...x (cid:48) n +1 , hence Int A i ∩ Int A k (cid:54) = ∅ (because ∀ y ∈ [ x (cid:48) (1) , x (cid:48) (2) ], ∀ B ( y, ε ): B ( y, ε ) ∩ Int A k (cid:54) = ∅ ), — a contradiction.2) n (cid:83) i =1 ( A i ∩ S ) = S and | S | > ℵ imply ∃ A i : ∂A i ⊇ A i ∩ S ⊇ { x , ..., x n +4 } . Consequently, ∂A k = f ik ( ∂A i ) of any other A k contains x ( k ) j = f ik ( x j ), j = 1 , n + 4, which belong to some circle S ( a, r ) = f ik ( S ); here f ik : X ↔ X , (cid:107) f ik ( x ) − f ik ( y ) (cid:107) = α ik (cid:107) x − y (cid:107) is the homothety transforming A i to A k . By (1), x ( k ) j ∈ S for any j and k .3) Now assume that there’s exactly one part A i such that θ ∈ Int A i : B ( θ, δ ) ⊆ A i . { x ( i ) j } n +4 j =1 ⊆ S ∩ ∂A i , and the point θ ∈ A i is at distance 1, equidistant, from each x ( i ) j .Take any k (cid:54) = i . As above, { y ( k ) j = f i k ( x ( i ) j ) } n +4 j =1 ⊆ S ∩ A k . And there must be z ∈ A k ,which is equidistant from each y ( k ) j ; clearly, z = θ ( y ( k )1 , y ( k )2 , y ( k )3 determine it uniquely). Also, B ( θ, δ ) ⊆ A k (apply similar arguments to ∀ x ∈ B ( θ, δ ) ⊆ A i ), so θ ∈ Int A k . A contradiction.(1st step shortens if we assume that one of ∂A i contains the arc ˘ a , which has to be on S then,otherwise 2 internal points of ˘ a are in ∂A k , k (cid:54) = i , too, implying a contradiction.) Example 3. (cid:67)
Without the upper bound for the cardinal number of covering, there is a“universal” covering of B ( θ,
1) such that the interior of exactly one subset contains the centre: let C = { A θ } ∪ (cid:8) A y (cid:9) y ∈ S ( θ, ) , where A θ = B ( θ, ) and A y = B ( y, ). Indeed, θ ∈ Int A θ = B ( θ, ),while ∀ y ∈ S ( θ, ): θ / ∈ Int A y = B ( y, ), and for ∀ x ∈ B ( θ, \{ θ } we take y x = (cid:107) x (cid:107) x ∈ S ( θ, ),then (cid:107) x − y x (cid:107) = | − (cid:107) x (cid:107) | · (cid:107) x (cid:107) = (cid:12)(cid:12) (cid:107) x (cid:107) − (cid:12)(cid:12) (cid:54) , thus x ∈ A y x . Certainly, A i ∼ = A j .This covering doesn’t require NCS or dim X >
1; meanwhile dim
X > |C| > ℵ . (cid:66) xample 4. (cid:67) Consider the Hilbert space over R , X = H = l , and its closed unit ball B H = B ( θ, S H = S ( θ, B H by itscongruent and convex subsets { A i } such that the interior of exactly one set contains the centre. Proof.
It’s a direct corollary of Lemmas 7, 10–15:1) Lemma 7 and Lemma 11 along with Lemma 10 provide the countable covering of B H bycongruent ommatidiums A i = C ( θ, d i , π ), i ∈ N , where d i ∈ S H . By Lemma 13, θ / ∈ Int A i .2) Then Lemma 12 allows to add the ommatidium A = C ( − d , d , π ), which is containedin B H since π < arccos and congruent with A i by Lemma 10.3) Finally, by Lemma 14, θ = (cid:0) − d + d (cid:1) ∈ Int A .By Lemma 15, A i are convex. |{ A i } i ∈ N ∪{ } | = ℵ = dim H .This covering somewhat resembles those from Fig. 1, except that a) the sets intersect “a lot”,b) there’s no “hollow” at the centre (corrigible by erasing sufficiently small neighborhood of tem-plate ommatidium’s origin), and c) it’s infinite-dimensional. (cid:66) The covering problem turns out to be easier about “positive” results than the problems ofdissection and partition types.
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