The Dimension Spectrum Conjecture for Planar Lines
aa r X i v : . [ c s . CC ] F e b The Dimension Spectrum Conjecture for PlanarLines
D. M. StullDepartment of Computer ScienceIowa State University [email protected]
February 17, 2021
Abstract
Let L a,b be a line in the Euclidean plane with slope a and intercept b . The dimension spectrum sp( L a,b ) is the set of all effective dimensionsof individual points on L a,b . The dimension spectrum conjecture statesthat, for every line L a,b , the spectrum of L a,b contains a unit interval.In this paper we prove that the dimension spectrum conjecture is true.Let ( a, b ) be a slope-intercept pair, and let d = min { dim( a, b ) , } . Forevery s ∈ (0 , x such that dim( x, ax + b ) = d + s .Thus, we show that sp( L a,b ) contains the interval ( d, d ). Results ofTuretsky [16], and Lutz and Stull [11], show that sp( L a,b ) contain theendpoints d and 1 + d . Taken together,[ d, d ] ⊆ sp( L a,b ),for every planar line L a,b . The effective dimension, dim( x ), of a Euclidean point x ∈ R n gives a fine-grained measure of the algorithmic randomness of x . Moreover, due to its strongconnection to (classical) Hausdorff dimension, effective dimension has proven tobe geometrically meaningful[4, 13, 2, 10]. A natural question, therefore, is thebehavior of the effective dimensions of points on a given line.Let L a,b be a line in the Euclidean plane with slope a and intercept b .Given the point-wise nature of effective dimension, one can study the dimensionspectrum of L a,b . That is, the setsp( L a,b ) = { dim( x, ax + b ) | x ∈ R } of all effective dimensions of points on L a,b . Jack Lutz posed the dimensionspectrum conjecture for lines. He conjectured that the dimension spectrum ofevery line in the plane contains a unit interval.1he first progress on this conjecture was made by Turetsky. He showed [16]that 1 ∈ sp( L a,b ) for every line L a,b . In [11], Lutz and Stull showed that theeffective dimension of points on a line is intimately connected to problems infractal geometry (more on this below). Among other things, they proved that1 + d ∈ sp( L a,b ) for every line L a,b , where d = min { dim( a, b ) , } .In this paper, we prove that dimension spectrum conjecture for planar lines istrue. For every s ∈ (0 , x such that dim( x, ax + b ) = d + s ,where d = min { dim( a, b ) , } . This, combined with the results of Turetsky andLutz and Stull, prove that [ d, d ] ⊆ sp( L a,b ),for every planar line L a,b .Apart from its intrinsic interest, recent work has shown that the effectivedimensions of points has connections to deep problems in classical analysis [8, 10,11, 15]. Lutz and Lutz [7] proved the point-to-set principle, which characterizesthe Hausdorff dimension of a set by effective dimension of its individual points .Lutz and Stull [11], using the point-to-set principle, showed that lower bounds ondim( x, ax + b ) implies lower bounds on the Hausdorff dimension of Furstenbergsets. The conditional Kolmogorov complexity of a binary string σ ∈ { , } ∗ givenbinary string τ ∈ { , } ∗ is K ( σ | τ ) = min π ∈{ , } ∗ { ℓ ( π ) : U ( π, τ ) = σ } , where U is a fixed universal prefix-free Turing machine and ℓ ( π ) is the lengthof π . The Kolmogorov complexity of σ is K ( σ ) = K ( σ | λ ), where λ is the emptystring. We stress that the choice of universal machine effects the Kolmogorovcomplexity by at most an additive constant (which, especially for our purposes,can be safely ignored). See [5, 14, 3] for a more comprehensive overview ofKolmogorov complexity.We can extend these definitions to Euclidean spaces by introducing “preci-sion” parameters [9, 7]. Let x ∈ R m , and r, s ∈ N . The Kolmogorov complexityof x at precision r is K r ( x ) = min { K ( p ) : p ∈ B − r ( x ) ∩ Q m } . The conditional Kolmogorov complexity of x at precision r given q ∈ Q m isˆ K r ( x | q ) = min { K ( p ) : p ∈ B − r ( x ) ∩ Q m } . The conditional Kolmogorov complexity of x at precision r given y ∈ R n atprecision s is K r,s ( x | y ) = max (cid:8) ˆ K r ( x | q ) : q ∈ B − r ( y ) ∩ Q n (cid:9) .
2e abbreviate K r,r ( x | y ) by K r ( x | y ).The effective Hausdorff dimension and effective packing dimension of apoint x ∈ R n aredim( x ) = lim inf r →∞ K r ( x ) r and Dim( x ) = lim sup r →∞ K r ( x ) r . Intuitively, these dimensions measure the density of algorithmic information inthe point x .By letting the underlying fixed prefix-free Turing machine U be a universal oracle machine, we may relativize the definition in this section to an arbitraryoracle set A ⊆ N . The definitions of K Ar ( x ), dim A ( x ), Dim A ( x ), etc. are then allidentical to their unrelativized versions, except that U is given oracle access to A . Note that taking oracles as subsets of the naturals is quite general. We can,and frequently do, encode a point y into an oracle, and consider the complexityof a point relative to y . In these cases, we typically forgo explicitly referring tothis encoding, and write e.g. K yr ( x ).Among the most used results in algorithmic information theory is the sym-metry of information . In Euclidean spaces, this was first proved, in a slightlyweaker form in [7], and in the form presented below in [11]. Lemma 1.
For every m, n ∈ N , x ∈ R m , y ∈ R n , and r, s ∈ N with r ≥ s , (i) | K r ( x | y ) + K r ( y ) − K r ( x, y ) (cid:12)(cid:12) ≤ O (log r ) + O (log log k y k ) . (ii) | K r,s ( x | x ) + K s ( x ) − K r ( x ) | ≤ O (log r ) + O (log log k x k ) . The proof of our main theorem will use the tools and techniques introduced byLutz and Stull [11]. The first lemma intuitively states the following. Supposethat L a,b is the only line of low complexity which intersects ( x, ax + b ). Thenit is possible to compute ( a, b ) given ( x, ax + b ) by simply enumerating over alllow complexity lines. Lemma 2 (Lutz and Stull [11]) . Suppose that a, b, x ∈ R , m, r ∈ N , δ ∈ R + ,and ε, η ∈ Q + satisfy r ≥ log(2 | a | + | x | + 5) + 1 and the following conditions. (i) K r ( a, b ) ≤ ( η + ε ) r . (ii) For every ( u, v ) ∈ B − m ( a, b ) such that ux + v = ax + b , K r ( u, v ) ≥ ( η − ε ) r + δ · ( r − t ) , whenever t = − log k ( a, b ) − ( u, v ) k ∈ (0 , r ] . Although effective Hausdorff was originally defined by J. Lutz [6] using martingales, it waslater shown by Mayordomo [12] that the definition used here is equivalent. For more details onthe history of connections between Hausdorff dimension and Kolmogorov complexity, see [3,13]. hen for every oracle set A ⊆ N , K Ar ( a, b, x | x, ax + b ) ≤ K Am,r ( a, b | x, ax + b ) + 4 εδ r − K ( ε, η ) + O a,b,x (log r ) . Note that, by the symmetry of information, the conclusion of the abovelemma can be equivalently written as K Ar ( x, ax + b ) ≥ K Ar ( a, b, x ) − K Am,r ( a, b | x, ax + b ) + εδ r − O a,b,x (log r )The second lemma which will be important in proving our main theorem isthe following. Lemma 3 ([11]) . Let a, b, x ∈ R . For all u, v ∈ B ( a, b ) such that ux + v = ax + b , and for all r ≥ t := − log k ( a, b ) − ( u, v ) k , K r ( u, v ) ≥ K t ( a, b ) + K r − t,r ( x | a, b ) − O a,b,x (log r ) . This lemma shows that we can give a lower bound on the complexity of anyline intersecting ( x, ax + b ).Finally, we also need the following oracle construction of Lutz and Stull. Lemma 4 ([11]) . Let r ∈ N , z ∈ R , and η ∈ Q ∩ [0 , dim( z )] . Then there is anoracle D = D ( r, z, η ) satisfying (i) For every t ≤ r , K Dt ( z ) = min { ηr, K t ( z ) } + O (log r ) . (ii) For every m, t ∈ N and y ∈ R m , K Dt,r ( y | z ) = K t,r ( y | z ) + O (log r ) and K z,Dt ( y ) = K zt ( y ) + O (log r ) . In this section, we prove the spectrum conjecture for lines with dim( a, b ) ≤ Theorem 5.
Let ( a, b ) ∈ R be a slope-intercept pair with dim( a, b ) ≤ .. Thenfor every s ∈ [0 , , there is a point x ∈ R such that dim( x, ax + b ) = s + dim( a, b ) . We will use many of the results of this section for the case when dim( a, b ) > a, b ),and a real s ∈ (0 , d = dim( a, b ). Let y ∈ R be random relative to ( a, b ).Thus, for every r ∈ N , K a,br ( y ) ≥ r − O (log r ).4efine the sequence of natural numbers { h j } j ∈ N inductively as follows. Define h = 1. For every j >
0, let h j = min (cid:26) h ≥ h j − : K h ( a, b ) ≤ (cid:18) d + 1 j (cid:19) h (cid:27) . Note that h j always exists. For every r ∈ N , let x [ r ] = ( a [ r − ⌊ sh j ⌋ ] if r ∈ ( ⌊ sh j ⌋ , h j ] for some j ∈ N y [ r ] otherwisewhere x [ r ] is the r th bit of x . Define x ∈ R to be the real number with thisbinary expansion.One of the most important aspects of our construction is that we encode(a subset of) the information of a into our point x . This is formalized in thefollowing lemma. Lemma 6.
For every j ∈ N , and every r such that sh j < r ≤ h j , K r − sh j ,r ( a, b | x, ax + b ) ≤ O (log h j ) .Proof. By definition, the last r − sh j bits of x are equal to the first r − sh j bitsof a . That is, x [ sh j ] x [ sh j + 1] . . . x [ r ] = a [0] a [1] . . . a [ r − sh j ].Therefore, since additional information cannot increase Kolmogorov complexity, K r − sh j ,r ( a | x, ax + b ) ≤ K r − sh j ,r ( a | x ) ≤ O (log h j ) . Note that, given 2 − ( r − sh j ) -approximations of a , x , and ax + b , it is possible tocompute an approximation of b . That is, K r − sh j ( b | a, x, ax + b ) ≤ O (log h j ).Therefore, by Lemma 1, K r − sh j ,r ( a, b | x, ax + b ) = K r − sh j ,r ( a | x, ax + b )+ K r − sh j ,r ( b | a, x, ax + b ) + O (log r ) ≤ O (log h j ) + K r − sh j ,r ( b | a, x, ax + b ) + O (log r ) ≤ O (log h j ) . The other important property of our construction is that ( a, b ) gives noinformation about x , beyond the information specifically encoded into x . Lemma 7.
For every j ∈ N , the following hold. . K a,bt ( x ) ≥ t − O (log h j ) , for all t ≤ sh j .2. K a,br ( x ) ≥ sh j + r − h j − O (log h j ) , for all h j ≤ r ≤ sh j +1 .Proof. We first prove item (1). Let t ≤ sh j . Then, by our construction of x ,and choice of y , K a,bt ( x ) ≥ K a,bt ( y ) − h j − − O (log t ) ≥ t − O (log t ) − log h j − O (log t ) ≥ t − O (log h j ) . For item (2), let h j ≤ r ≤ sh j +1 . Then, by item (1), Lemma 1 and ourconstruction of x , K a,br ( x ) = K a,bh j ( x ) + K a,br,h j ( x ) − O (log r ) [Lemma 1] ≥ sh j + K a,br,h j ( x ) − O (log r ) [Item (1)] ≥ sh j + K a,br,h j ( y ) − O (log r ) ≥ sh j + r − h j − O (log r ) , and the proof is complete.We now prove bounds on the complexity of our constructed point. We breakthe proof into two parts. In the first, we give lower bounds on K r ( x, ax + b ) atprecisions sh j < r ≤ h j . Lemma 8.
For every γ > and all sufficiently large j ∈ N , K r ( x, ax + b ) ≥ ( s + min { dim( a, b ) , } ) r − γr ,for every r ∈ ( sh j , h j ] .Proof. Let d = min { dim( a, b ) , } . Let η ∈ Q such that d − γ/ < η < d − γ .Let ε ∈ Q such that ε < γ ( d − η ) / D = D ( r, ( a, b ) , η ) be the oracle of Lemma 4.Let ( u, v ) be a line such that t := k ( a, b ) − ( u, v ) k ≥ r − sh j , and ux + v = ax + b . Note that r − t ≤ sh j . Then, by Lemma 3, Lemma 4 and Lemma 7(1), K Dr ( u, v ) ≥ K Dt ( a, b ) + K Dr − t,r ( x | a, b ) − O (log r ) [Lemma 3] ≥ K Dt ( a, b ) + K r − t,r ( x | a, b ) − O (log r ) [Lemma 4] ≥ dt + r − t − O (log r ) [Lemma 7(1)]= ηr + (1 − η ) r − t (1 − d ) − O (log r ) ≥ ηr + (1 − η )( r − t ) − O (log r ) ≥ ( η − ε ) r + (1 − η )( r − t ) . (1)6herefore we may apply Lemma 2, K r ( x, ax + b ) ≥ K Dr ( a, b, x ) − K Dr − sh j ,r ( a, b | , x, ax + b ) [Lemma 2] − ε − η r − O a,b,x (log r ) ≥ K Dr ( a, b, x ) − K Dr − sh j ,r ( a, b | x, ax + b ) − γ ( d − η )4(1 − η ) r − γ r = K Dr ( a, b, x ) − K Dr − sh j ,r ( a, b | x, ax + b ) − γ r. (2)By Lemma 7(1), our construction of oracle D , and the symmetry of information, K Dr ( a, b, x ) = K Dr ( a, b ) + K Dr ( x | a, b ) − O (log r ) [Lemma 1]= K Dr ( a, b ) + K r ( x | a, b ) − O (log r ) [Lemma 4(ii)] ≥ ηr + K r ( x | a, b ) − O (log r ) [Lemma 4(i)] ≥ ηr + sh j − γ r. (3)Finally, by Lemma 6, K Dr − sh j ,r ( a, b | , x, ax + b ) ≤ γ r. (4)Together, inequalities (2), (3) and (4) imply that K r ( x, ax + b ) ≥ K Dr ( a, b, x ) − K Dr − sh j ,r ( a, b | , x, ax + b ) − γ r ≥ ηr + sh j − γ r − γ r − γ r ≥ dr − γ r + sh j − γ r ≥ dr + sh j − γr ≥ ( s + d ) r − γr, and the proof is complete.We now give lower bounds on the complexity of our point, K r ( x, ax + b ), for h j < r ≤ sh j +1 . Lemma 9.
For every γ > and all sufficiently large j ∈ N , K r ( x, ax + b ) ≥ ( s + min { dim( a, b ) , } ) r − γr ,for every r ∈ ( h j , sh j +1 ] .Proof. Let d = min { dim( a, b ) , } , and r ∈ ( h j , sh j +1 ]. We consider two cases,when s ≤ dim( a, b ) and when s > dim( a, b ).First assume that s ≤ dim( a, b ). Define7 = s ( r − h j )+dim( a,b ) h j r .Let η ∈ Q ∩ (0 , α ) such that α − η < ( d − s ) γ/ ε ∈ Q such that ε < ( α − η ) γ/ D = D ( r, ( a, b ) , η ) be the oracle of Lemma 4.Let ( u, v ) be a line such that t := k ( a, b ) − ( u, v ) k ≥ h j , and ux + v = ax + b .Then, by Lemmas 3, 4 and 7, K Dr ( u, v ) ≥ K Dt ( a, b ) + K Dr − t,r ( x | a, b ) − O (log r ) [Lemma 3] ≥ K t ( a, b ) + K r − t,r ( x | a, b ) − O (log r ) [Lemma 4] ≥ dim( a, b ) t − o ( t ) + s ( r − t ) − O (log r ) [Lemma 7(1)]= dim( a, b ) h j + dim( a, b )( t − h j ) + s ( r − t ) − o ( r )= dim( a, b ) h j + dim( a, b )( t − h j ) + s ( r − h j ) − s ( t − h j ) − o ( r )= αr + (dim( a, b ) − s )( t − h j ) − o ( r )= ηr + ( α − η ) r + (dim( a, b ) − s )( t − h j ) − o ( r ) ≥ ηr + ( α − η )( r − t ) − o ( r ) ≥ ( η − ε ) r + ( α − η )( r − t ) . Therefore we may apply Lemma 2, which yields K Dr ( a, b, x | x, ax + b ) ≤ K Dh j ,r ( a, b, x | x, ax + b ) [Lemma 2]+ 4 εα − η r + K ( ε, η ) + O a,b,x (log r ) ≤ K Dh j ,r ( a, b, x | x, ax + b ) + γ ( α − η )4( α − η ) r + γ r = K Dh j ,r ( a, b, x | x, ax + b ) + 3 γ r. (5)By Lemma 7, and our construction of oracle D , K Dr ( a, b, x ) = K Dr ( a, b ) + K Dr ( x | a, b ) − O (log r ) [Lemma 1]= ηr + K r ( x | a, b ) − O (log r ) [Lemma 4] ≥ ηr + sh j + r − h j − O (log r ) [Lemma 7(2)] ≥ αr − γ r + sh j + r − h j − O (log r ) ≥ s ( r − h j ) + dim( a, b ) h j − γ r + sh j + r − h j − O (log r ) ≥ (1 + s ) r − (1 − dim( a, b )) h j − γ r. (6)8y Lemmas 8, and 1, and the fact that additional information cannot in-crease Kolmogorov complexity K h j ,r ( a, b, x | x, ax + b ) ≤ K h j ,h j ( a, b, x | x, ax + b )= K h j ( a, b, x ) − K h j ( x, ax + b ) [Lemma 1] ≤ K h j ( a, b ) − dh j + γ h j [Lemma 8] ≤ dim( a, b ) h j + h j /j − dh j + γ r ≤ dim( a, b ) h j − dh j + γ r (7)Combining inequalities (5), (6) and (7) , we see that K Dr ( x, ax + b ) ≥ K Dr ( a, b, x ) − (dim( a, b ) − d ) h j − γ r − γ r ≥ (1 + s ) r − (1 − dim( a, b )) h j − γ r − (dim( a, b ) − d ) h j − γ r = (1 + s ) r − h j (1 − d ) − γr. Note that, since d ≤
1, and h j ≤ r ,(1 + s ) r − h j (1 − d ) − ( s + d ) r = r (1 − d ) − h j (1 − d )= ( r − h j )(1 − d ) ≥ . Thus, since oracles cannot increase Kolmogorov complexity K r ( x, ax + b ) ≥ K Dr ( x, ax + b ) ≥ (1 + s ) r − h j (1 − d ) − γr ≥ ( s + d ) r − γr, and the proof is complete for the case s ≤ dim( a, b ).The proof when s > dim( a, b ) is essentially that of Lutz and Stull [11]. Inthis case, let η ∈ Q ∩ (0 , dim( a, b )) such thatdim( a, b ) − η < γ .Let ε ∈ Q such that ε < ( s − η ) γ/ D = D ( r, ( a, b ) , η ) be the oracle of Lemma 4.9et ( u, v ) be a line such that t := k ( a, b ) − ( u, v ) k ≥
1, and ux + v = ax + b .Then, by Lemmas 3, 4 and 7, K Dr ( u, v ) ≥ K Dt ( a, b ) + K Dr − t,r ( x | a, b ) − O (log r ) [Lemma 3] ≥ ηt + K r − t,r ( x | a, b ) − O (log r ) [Lemma 4] ≥ ηt + s ( r − t ) − O (log r ) [Lemma 7(1)]= ηt + η ( r − t ) + ( s − η )( r − t ) − O (log r )= ηr + ( s − η )( r − t ) + (1 − η ) t − O (log r ) ≥ ( η − ε ) r + ( s − η )( r − t ) . Therefore we may apply Lemma 2, which yields K Dr ( a, b, x | x, ax + b ) ≤ K D ,r ( a, b, x | x, ax + b ) [Lemma 2]+ 4 εs − η r + K ( ε, η ) + O a,b,x (log r ) ≤ γ ( α − η )4( s − η ) r + γ r = 3 γ r. (8)By Lemma 7, and our construction of oracle D , K Dr ( a, b, x ) = K Dr ( a, b ) + K Dr ( x | a, b ) − O (log r ) [Lemma 1]= ηr + K r ( x | a, b ) − O (log r ) [Lemma 4] ≥ ηr + sh j + r − h j − O (log r ) [Lemma 7(2)] ≥ dr − γ r + sh j + r − h j − O (log r ) ≥ dr + sh j + r − h j − γ r. (9)Finally, by combining inequalities (8), (9) and (7) , we see that K Dr ( x, ax + b ) ≥ K Dr ( a, b, x ) − γ r ≥ dr + sh j + r − h j − γ r − γ r ≥ dr + sr − γr, and the proof is complete.We are now able to prove our main theorem. Proof of Theorem 5.
Let ( a, b ) ∈ R be a slope-intercept pair with d = dim( a, b ) ≤
1. Let s ∈ [0 , ∈ sp( L a,b ), hence, if s = 0, the conclu-sion holds.If s = 1, then by [11], for any point x which is random relative to ( a, b ),10im( x, ax + b ) = 1 + d ,and the claim follows.Therefore, we may assume that s ∈ (0 , x be the point constructedin this section. Let γ >
0. Let j be large enough so that the conclusions ofLemmas 8 and 9 hold for these choices of ( a, b ), x , s and γ . Then, by Lemmas8 and 9, dim( x, ax + b ) = lim inf r →∞ K r ( x, ax + b ) r ≥ lim inf r →∞ ( s + d ) r − γrr = s + d − γ. Since we chose γ arbitrarily, we see thatdim( x, ax + b ) ≥ s + d .For the upper bound, let j ∈ N be sufficiently large. Then K h j ( x, ax + b ) ≤ K h j ( x, a, b )= K h j ( a, b ) + K h j ( x | a, b ) ≤ dh j + sh j = ( d + s ) h j . Therefore, dim( x, ax + b ) ≤ s + d ,and the proof is complete. Theorem 10.
Let ( a, b ) ∈ R be a slope-intercept pair with dim( a, b ) > ..Then for every s ∈ [0 , , there is a point x ∈ R such that dim( x, ax + b ) = s + 1 . We will build on the construction, and results, of the previous section. Wefirst slightly modify the construction. Let y ∈ R be random relative to ( a, b ).Define the sequence of natural numbers { h j } j ∈ N inductively as follows. Define h = 1. For every j >
0, let h j = min (cid:26) h ≥ h j − : K h ( a, b ) ≤ (cid:18) d + 1 j (cid:19) h (cid:27) . Note that h j always exists. For every r ∈ N , let x [ r ] = ( r ∈ ( ⌊ sh j ⌋ , h j ] for some j ∈ N y [ r ] otherwise11here x [ r ] is the r th bit of x . Define x ∈ R to be the real number with thisbinary expansion.Fix a j ∈ N which is sufficiently large. Let m = h j − sh j + 1. We now definethe set of points x , . . . , x m by x n = x + 2 − h j + n /a .A useful fact about each point x n is that ax n + b = ax + b + 2 − h j + n (10) Lemma 11.
For every n, r such that ≤ n ≤ m and sh j ≤ r ≤ h j the followinghold.1. K n,h j ( a | x n ) ≤ log( n ) .2. For every n ′ > n , | K r ( x ′ n , ax ′ n + b ) − K r ( x n , ax n + b ) | < n ′ − n + log( r ) + log( r ) .Proof. By our construction of x and x n , the x n [ h j − n ] . . . x n [ h j ]are the first n bits of 1 /a . Since we can compute a 2 − n -approximation of a givena 2 − n -approximation of 1 /a , the first claim follows.For item (2), first assume that r < h j − n ′ . Then K r ( x ′ n , ax ′ n + b ) = K r ( x n , ax n + b ),and the claim follows. Now assume that r ≥ h j − n ′ . By property (10), K r ( x ′ n , ax ′ n + b ) = K r ( a , x n , ax n + b ) + O (log n ),where a is binary string of length n ′ − n encoding the bits a [ r − ( h j − n )] . . . a [ r − ( h j − n ′ )].Therefore, K r ( x ′ n , ax ′ n + b ) = K r ( a , x n , ax n + b ) + O (log n )= K r ( x n , ax n + b ) + K r ( a | x n , ax n + b ) + O (log n ) . Since K r ( a | x n , ax n + b ) ≤ n ′ − n ,the proof of the second item is complete.We will use (a discrete, approximate, version of) the mean value theorem.For each n , define M n = min { K r ( x n ,ax n + b ) r | sh j ≤ r ≤ h j } .12he proof of our theorem essentially reduces to showing that, for some n ,1 + s − ǫ ≤ M n ≤ s + ǫ. (11)We first note, by Lemma 8, that K r ( x m , ax m + b ) & (1 + s ) r − γr ,for every sh j ≤ r ≤ h j , and so M m & (1 + s ). We also have the fact that K h j ( x, ax + b ) ≤ (1 + s ) h j .That is, we have M ≤ s ≤ M m .If either inequality is not strict then we have shown that (11) holds and weare done. We may therefore assume that M < s − ǫ , and 1 + s + ǫ < M m .Define L = { n | M n < s − ǫ } G = { n | M n > s + ǫ } . By our assumption, L and G are non-empty. Suppose that L and G partitionthe naturals. Then there is a n such that, without loss of generality, n ∈ L and n + 1 ∈ G . However, by Lemma 11, | K r ( x n +1 , ax n +1 + b ) − K r ( x n , ax n + b ) | < r ) + log( r ),for every r . Let r be a precision testifying to x n ∈ L . Then1 + log( r ) + log( r ) > | K r ( x n , ax n + b ) − K r ( x n , ax n + b ) | > K r ( x n ′ , ax n ′ + b ) − (1 + s − ǫ ) r> (1 + s + ǫ ) r − (1 + s − ǫ ) r = 2 ǫr, which is false for all sufficiently large r . Therefore, we can conclude that (11)holds.We now fix an n such that K r ( x n , ax n + b ) ≥ (1 + s ) r − ǫr (12)for every sh j ≤ r ≤ h j .We will modify x n on the bits larger than h j as follows. For every h j ≤ r ≤ sh j +1 , let x n [ r ] = y [ r ].13o avoid notational inconvenience, we will now refer to this point as x .To complete the proof, let r ≥ h j . Then, by essentially the argument ofLemma 9, we can prove the following. Lemma 12.
For every γ > and all sufficiently large j ∈ N , K r ( x, ax + b ) ≥ ( s + 1) r − γr ,for every r ∈ ( h j , sh j +1 ] .Proof. Define α = s ( r − h j )+dim( a,b ) h j r .Let η ∈ Q ∩ (0 , α ) such that α − η < (1 − s ) γ/ ε ∈ Q such that ε < ( α − η ) γ/ D = D ( r, ( a, b ) , η ) be the oracle of Lemma 4.Let ( u, v ) be a line such that t := k ( a, b ) − ( u, v ) k ≥ h j , and ux + v = ax + b .Then, by Lemmas 3, 4 and 7, K Dr ( u, v ) ≥ K Dt ( a, b ) + K Dr − t,r ( x | a, b ) − O (log r ) [Lemma 3] ≥ K t ( a, b ) + K r − t,r ( x | a, b ) − O (log r ) [Lemma 4] ≥ dim( a, b ) t − o ( t ) + s ( r − t ) − O (log r ) [Lemma 7(1)]= dim( a, b ) h j + dim( a, b )( t − h j ) + s ( r − t ) − o ( r )= dim( a, b ) h j + dim( a, b )( t − h j ) + s ( r − h j ) − s ( t − h j ) − o ( r )= αr + (dim( a, b ) − s )( t − h j ) − o ( r )= ηr + ( α − η ) r + (dim( a, b ) − s )( t − h j ) − o ( r ) ≥ ηr + ( α − η )( r − t ) − o ( r ) ≥ ( η − ε ) r + ( α − η )( r − t ) . Therefore we may apply Lemma 2, which yields K Dr ( a, b, x | x, ax + b ) ≤ K Dh j ,r ( a, b, x | x, ax + b ) [Lemma 2]+ 4 εα − η r + K ( ε, η ) + O a,b,x (log r ) ≤ K Dh j ,r ( a, b, x | x, ax + b ) + γ ( α − η )4( α − η ) r + γ r = K Dh j ,r ( a, b, x | x, ax + b ) + 3 γ r. (13)14y Lemma 7, and our construction of oracle D , K Dr ( a, b, x ) = K Dr ( a, b ) + K Dr ( x | a, b ) − O (log r ) [Lemma 1]= ηr + K r ( x | a, b ) − O (log r ) [Lemma 4] ≥ ηr + sh j + r − h j − O (log r ) [Lemma 7(2)] ≥ αr − γ r + sh j + r − h j − O (log r ) ≥ s ( r − h j ) + dim( a, b ) h j − γ r + sh j + r − h j − O (log r ) ≥ (1 + s ) r − (1 − dim( a, b )) h j − γ r (14) ≥ (1 + s ) r − γ r. (15)By Lemmas 1, inequality (12), and the fact that additional informationcannot increase Kolmogorov complexity K h j ,r ( a, b, x | x, ax + b ) ≤ K h j ,h j ( a, b, x | x, ax + b )= K h j ( a, b, x ) − K h j ( x, ax + b ) [Lemma 1] ≤ K h j ( a, b, x ) − (1 + s ) h j + γ h j [In. (12)] ≤ dim( a, b ) h j + h j /j + sh j − (1 + s ) h j + γ h j ≤ dim( a, b ) h j + h j /j − h j + γ r ≤ dim( a, b ) h j − h j + γ r (16)Combining inequalities (13), (15) and (16) , we see that K Dr ( x, ax + b ) ≥ K Dr ( a, b, x ) − (dim( a, b ) − h j − γ r ≥ (1 + s ) r − γ r − (dim( a, b ) − h j − γ r = (1 + s ) r − γr. Thus, since oracles cannot increase Kolmogorov complexity K r ( x, ax + b ) ≥ K Dr ( x, ax + b ) ≥ (1 + s ) r − γr, and the proof is complete References [1] Adam Case and Jack H. Lutz. Mutual dimension.
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