aa r X i v : . [ m a t h . L O ] O c t Intersection points of planar curves can be computed
Klaus WeihrauchFernUniversit¨at in Hagen
Abstract
Consider two paths φ, ψ : [0; 1] → [0; 1] in the unit square such that φ (0) = (0 , φ (1) = (1 , ψ (0) = (0 ,
1) and ψ (1) = (1 , φ and ψ there is a point of intersection. We provethat from φ and ψ we can compute closed intervals S φ , S ψ ⊆ [0; 1] such that φ ( S φ ) = ψ ( S ψ ). A path in the Euclidean plane is a continuous function f : [0; 1] → R , a curve is the range of a path.The following is known about planar curves. Theorem 1.1
Let φ, ψ : [0; 1] → [0; 1] be two paths in the unit square such that φ (0) = (0 , , φ (1) = (1 , , ψ (0) = (0 , and ψ (1) = (1 , . (1) Then the two curves range( φ ) and range( ψ ) intersect. Figure 1 visualizes the theorem. In Markov-style computable analysis [4] Manukyan [7] has proved arelated theorem (in Russian), cited in [5, Page 279] as follows:
Theorem 1.2 (Manukyan)
There are two constructive (and therefore continuous) planar curves ϕ and ϕ such that ϕ (0) = (0 , , ϕ (1) = (1 , , ϕ (0) = (0 , , ϕ (1) = (1 , , (2) for every < t < both ϕ ( t ) and ϕ ( t ) belong to the open unit square, (3) the paths of ϕ and ϕ do not intersect. (4)While in (Grzegorczyk-Lacombe- [2, 6]) computable analysis the following has been proved [10]: Theorem 1.3 (Weihrauch) If φ and ψ in Theorem 1.1 are computable then there is a computable point x ∈ range( φ ) ∩ range( ψ ) . This is not contradictory. In Markov’s approach only functions on the computable real numberswhich are encoded by G¨odel numbers are considered and computations transform G¨odel numbers toG¨odel numbers. While in the Grzegorczyk-Lacombe-approach all real numbers are considered, wherereal numbers are encoded by (fast converging) Cauchy sequences of rational numbers and computationstransform infinite Cauchy sequences to infinite Cauchy sequences.In this article we prove Claim 6.2 from [10]: 1 ✻ t x x φψgf fgφ (0) ψ (0) ψ (1) φ (1) f ( − g ( − g (2) f (2)Figure 1: Intersecting curves φ and ψ with extensions f and g . Theorem 1.4
Let T be the multi-valued operator mapping every pair φ, ψ : [0; 1] → [0; 1] of paths inthe unit square such that φ (0) = (0 , , φ (1) = (1 , , ψ (0) = (0 , and ψ (1) = (1 ,
0) (5) to some pair ( S φ , S ψ ) of closed intervals such that φ ( S φ ) = ψ ( S ψ ) . Then the operator T is computable. Theorem 1.3 follows straightforwardly from Theorem 1.4. In the proof from φ and ψ we computesequences I ⊇ I ⊇ I ⊇ . . . and J ⊇ J ⊇ J ⊇ . . . of closed intervals with rational endpoints such that φ ( T i I i ) = ψ ( T i J i ).Curves (even computable ones) can be much more complicated than the examples shown in Figure 1.Consider, for example, space-filling curves or curves with infinitely many spirals, each of which containinginfinitely many sub-spirals etc. infinitely often or curves with “completely” chaotic behavior.This article is a contribution to computable analysis. There are various non equivalent definitions ofcomputability in analysis. One of these is Markov’s constructive analysis [4, 5]. Theorem 1.2 is a resultin this theory. We use ”TTE”, an approach which is based on ideas from [2, 3, 6]. In TTE computabilityon { , } ∗ and Cantor space { , } ω (the finite and infinite 0-1-sequences) is defined explicitly (e.g. byTuring machines with finite or infinite one-way input and output tapes) and computability on other sets X is induced via representations δ : ⊆ { , } ∗ → X or δ : ⊆ { , } ω → X (partial surjective) where finite orinfinite 0-1-sequences are interpreted as names and computations are performed on names. We considercanonical representations of the real numbers, open subsets, closed subsets, compact subsets and realfunctions. Equivalently any finite alphabet Σ (with at least two elements) can be used instead of { , } .We assume that the reader is familiar with the basic concepts of TTE. Details can be found in [9, 1, 11].For technical reasons we extend φ and ψ trivially to continuous functions f, g : [ −
1; 2] → R whichintersect in the same way as φ and ψ , that is, φ ( s ) = ψ ( t ) ⇐⇒ f ( s ) = g ( t ) (see Figure 1): f ( t ) := ( t,
0) if − ≤ t ≤ φ ( t ) if 0 ≤ t ≤ t,
1) if 1 ≤ t ≤ , (6) g ( t ) := ( t.
1) if − ≤≤ ψ ( t ) if 0 ≤ t ≤ t,
0) if 1 ≤ t ≤ , (7)2e will consider closed rational sub-intervals I = [ a I ; b I ] and J = [ a J ; b J ] of the real interval [ −
1; 2]such that for the restrictions f | I of f to I and g | J of g to J ,the end-points f | I ( a I ) and f | I ( b I ) are not in g ( J ) andthe end-points g | J ( a J ) and g | J ( b J ) are not in f ( I ) . Since f ( I ) and g ( J ) are compact this means α IJ := min( d s ( { f ( a I ) , f ( b I ) } , g ( J )) , d s ( { g ( a J ) , g ( b J ) } , f ( I ))) > . (8)where d s ( A , A ) := inf {k z − z k | z ∈ A , z ∈ A } .We will approximate f | I and g | J by rational polygon paths h and h ′ , respectively, and consider theintersections ( s, t ), that is, pairs such that h ( s ) = h ′ ( t ). In order to keep this number finite we consideronly pairs ( h, h ′ ) such that range( h ) ∩ range( h ′ ) contains no straight line segment. For such pairs everyintersection ( s, t ) is either a crossing or tangent. As a central lemma we will prove that the parity (evenor odd) of the number of crossings does not depend on h and h ′ (it is an invariant of ( f | I , g | J )). We callit the crossing parity of the pair ( f | I , g | J ). In the proof we will apply transformations of polygon pathswhich may change the number of crossings but do not change the parity (even or odd) of the number ofcrossings. For points x, y ∈ R such that x = y let xy ⊆ R be the straight line segment from x to y . In this article xx is not a straight line segment. Definition 2.1
1. A track is a sequence p = (( s , x ) , ( s , x ) , . . . , ( s k , x k )) such that s i < s i +1 and x i = x i +1 for ≤ i < k . The points x , x , . . . , x k are the vertices of p .2. The track p spans a (polygon) path h p : [ s ; s k ] → R by h p ( s ) = x i + s − s i s i +1 − s i ( x i +1 − x i ) if s i ≤ s ≤ s i +1 . (9)By (9), h p ( s i ) = x i and h p [ s i ; s i +1 ] = x i x i +1 .For tracks p, q we want to count the number of crossings of the paths h p and h q . In order to keep thisnumber finite we consider only pairs ( p, q ) such that range( h p ) and range( h q ) have no common straightline segment. Furthermore, we will not count all intersections of h p and h q but only “proper” crossings. Definition 2.2
Let ( p, q ) where p = (( s , x ) , s , x ) , . . . , ( s k , x k )) and q = (( t , y ) , t , y ) , . . . , ( t l , y l )) be a pair of tracks such that range( h p ) and range( h q ) have no common straight line segment.1. An intersection of p and q is a pair ( s, t ) such that s < s < s k , t < t < t l and h p ( s ) = h q ( t ) . Wecall x:= h p ( s ) = h q ( t ) the corresponding intersection point. . For an intersection ( s, t ) of p and q with intersection point h p ( s ) = h q ( t ) = x let δ st > be anumber such that B ( x, δ st ) \ { x } contains no vertex of p and no vertex of q . Let s < := inf { s ′ < s | h p [ s ′ ; s ] ⊆ B ( x, δ st ) } , x < := h p ( s < ) ,s > := sup { s ′ > s | h p [ s ; s ′ ] ⊆ B ( x, δ st ) } , x > := h p ( s > ) ,t < := inf { t ′ < t | h q [ t ′ ; t ] ⊆ B ( x, δ st ) } , y < := h q ( t < ) ,t > := sup { t ′ > t | h q [ t ; t ′ ] ⊆ B ( x, δ st ) } , y > := h q ( t > ) . If on the boundary of B ( x, δ st ) the four points x < , x > , y < and y > occur in the order ( x < , y < , x > , y > ) or in the order ( x < , y > , x > , y < ) , we call ( s, t ) a crossing and x the corresponding crossing point,else x is a touch point.3. Let CN( p, q ) the number of crossings of p and q and let π ( p, q ) := CN( p, q ) mod 2 be its parity ( even and odd). Obviously, s < < s i < s for no number i , x < = h p ( s < ) ∈ ∂B ( x, δ st ) and h p [ s < ; s ] = x < x where ∂A denotes the boundary of A ⊆ R . This is true correspondingly for s > , t < and t > .Figure 2 shows several kinds of intersection of p and q (thin lines for h q and thick lines for h p ) thefirst two of which are crossings. In (a) possibly the center x is no vertex of p or no vertex of q . r r r r x > x < y < y > ✟✟✟(cid:0)(cid:0)(cid:0) x < x > y < y > ✟✟✟ y > y < x > x < y < = y > x < x > (a) (b) (c) (d)Figure 2: h p and h q in the ball B ( x, δ st ) for x = h p ( s ) = h q ( t ).The definition of a crossing ( s, t ) depends on the number δ st only formally. Lemma 2.3
The definition of a crossing does not depend on the choice of δ st . Proof
Let ( s, t ) be a crossing of p and q defined via some δ st . Let 0 < δ < δ st .Let s < := inf { s ′ < s | h p [ s ′ ; s ] ⊆ B ( x, δ ) } and x < := h p ( s < ). Then x < = x < x ∩ δB ( x, δ ). This is truecorrespondingly for the other three cases. Obviously the four points on B ( x, δ ) alternate in the same wayas the four corresponding points on B ( x, δ st ). ✷ Notice that { x < , x > } ∩ { y < , y > } = ∅ since range( h p ) and range( h q ) have no common straight linesegment. In the applications below the endpoints of h p are not in range( h q ) and the endpoints of h q are not in range( h p ). Therefore it suffices to consider s < s < s k and t < t < t l in the definition ofintersections.We introduce a separation concept for tracks p and q which induces that range( h p ) and range( h q )have no common straight line segment. that is, on the boundary of B ( x, δ st ) the four points alternate in x and y efinition 2.4 Let p = (( s , x ) , ( s , x ) , . . . , ( s k , x k )) be a track.1. Define V ( p ) := { x , x , . . . , x k } .2. For x, y ∈ R with x = y let l ( x, y ) be the straight line through x and y .3. Define L ( p ) := [ { l ( x i − , x i ) | ≤ i ≤ k } , (10) L ( p ) := [ { l ( x i , x j ) | ≤ i < j ≤ k, x i = x j } . (11)
4. We call tracks p and q weakly separated, p ⊲⊳ q , iff V ( p ) ∩ L ( q ) = ∅ and (12) V ( q ) ∩ L ( p ) = ∅ . (13)Figure 3 shows on the left the set L ( p ) of a track p and a straight line through a point x
6∈ L ( p ) andon the right the set L ( q ) of a track q and a straight line through a point y
6∈ L ( q ). ✡✡✡✡✡✡✡(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)❏❏❏❏❏❏❏❏❏❏❏ s s ss ❳❳❳❳❳❳❳❳❳❳❳❳❳❳ ❝ x ✡✡✡✡✡✡✡(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)❏❏❏❏❏❏❏❏❏❏❏ s s ss ❳❳❳❳❳❳❳❳❳❳❳❳❳❳ ❝ y Figure 3: L ( p ) and L ( q ) Lemma 2.5
1. If y
6∈ L ( p ) then every straight line through the point y intersects every straight line from L ( p ) atmost once.2. If V ( p ) ∩ L ( q ) = ∅ or V ( q ) ∩ L ( p ) = ∅ then then range( h p ) and range( h q ) have no common straightline segment. Proof
If a straight line through y intersects a straight line from L ( p ) twice then y ∈ L ( p ). Contradiction.Let q = (( t , y ) , . . . , ( t m , y m )). Suppose for some i and j , x i x i +1 and y j y j +1 have a common straightline segment. Then x i ∈ l ( y j , y j +1 ) ⊆ L ( q ), but V ( p ) ∩ L ( q ) = ∅ . Correspondingly, V ( q ) ∩ L ( p ) = ∅ . ✷ As an essential tool we will use local transformations of tracks which leave the crossing parity invariant.The following lemma justifies these transformations.5 emma 2.6
Let p = (( s , x ) , ( s , x ) , . . . , ( s k , x k )) ,q = (( r, y ) , ( t , z ) , ( r ′ , y ′ )) and q = (( r, y ) , ( t , z ) , ( r ′ , y ′ )) be tracks and let B be a ball such that { y, y ′ } ∩ L ( p ) = ∅ , (14) { y, y ′ , z , z } ⊆ B , (15) { x , x k } ∩ B = ∅ . (16) Then π ( p, q ) = π ( p, q ) . Proof
Remember that by Definition 2.1, r < t < r ′ , r < t < r ′ , z
6∈ { y, y ′ } and z
6∈ { y, y ′ } . By (14), y, y ′ range( h p ).If range( h p ) and range( h q ) have a common straight line segment then y ∈ L ( p ) or y ′ ∈ L ( p ),but { y, y ′ } 6∈ L ( p ). Therefore, CN( p, q ) is well-defined. Correspondingly, CN( p, q ) is well-defined. InFigure 4 the “open ended” line segments are parts of range( h p ). We distinguish several cases, see Figure 4) ❇❇❇❇❇❇❇❇❇❇❇❇❇❇ s rr ( t , z )( r, y ) ( r ′ , y ′ ) ✟✟❍❍❳❳✑✑PPP (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0) s s s s ( t , z )( t , z )( r, y )( r ′ , y ′ ) ❳❳❳ ❏❏❏ ◗◗ s s s (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)❍❍❍❍❍❍❍❍❍ ( r, y ) ( r ′ , y ′ )( t , z )( t , z ) r b σ r σ r✚✙✛✘ a σ ✘✘✘ s Case (a) Case (b) Case (c)Figure 4: Illustration for Lemma 2.6.
Case (a) y = y ′ : Obviously, CN( p, q ) and CN( p, q ) are even, hence π ( p, q ) = 0 = π ( p, q ). Case (b) y = y ′ , z ∈ yy ′ and z ∈ l ( y, y ′ ): In Figure 4(b) the track q is drawn in thick lines.If z ∈ y, y ′ then h q = h q , hence CN( p, q ) = CN( p, q ). If z y, y ′ then CN( p, q ) and CN( p, q ) differby an even number. In both cases π ( p, q ) = 0 = π ( p, q ). Case (c) y = y ′ , z ∈ yy ′ and z l ( y, y ′ ): In Figure 4(c) the track q is drawn in thick lines.Let ∆ be the closed triangle with boundary ∂ ∆ := range( h q ) ∪ range( h q ) (= yz ∪ z y ′ ∪ yy ′ ) and let∆ ◦ be its interior.If yz and range( h p ) have a common straight line segment then y ∈ L ( p ), a contradiction by (14).Therefore yz and range( h p ) have no common straight line segment. This is true correspondingly for z y ′ and yy ′ . Therefore ∂ ∆ and range( h p ) have no common straight line segment. (17)Let s < σ < s k such that h p ( σ ) ∈ ∆ ◦ . Let [ a σ ; b σ ] be the longest interval such that s ≤ a σ < σ < b σ ≤ s k and h p [ a σ ; b σ ] ⊆ ∆. Figure 4(c) shows an example for σ with a σ and b σ positioned at the images under h p h p ( a σ ) ∈ range( h q ). Obviously, h p ( a σ ) , h p ( b σ ) ∈ ∆. We show that h p ( a σ ) and h p ( b σ ) are crossingpoints of p and q or of p and q .Suppose h p ( a σ ) ∈ range( h q )Since range( h p ) is a chain of straight line segments there is some a σ < s ′ < σ such that λ := h p ([ a σ ; s ′ ])is a straight line segment with h p ( a σ ) ∈ λ ∩ ∂ ∆ and by (17) λ \ { h p ( a σ ) } ⊆ ∆ ◦ .Since range( h p ) is a chain of straight line segments and a σ is the smallest number a with h p [ a ; σ ] ∈ ∆,there is some s ′′ < a σ such that λ ′′ := h p [ s ′′ ; a σ ] is straight line segment with h p ( a σ ) ∈ λ ′′ ∩ ∂ ∆ and( λ ′′ \ { h p ( a σ ) } ) ∩ ∆ = ∅ by (17).If we draw a sufficiently small circle around h p ( a σ ) then (with the terminology from Definition 2.2))the intersections x < and x > of it with range( h p ) and the intersections y < and y > of it with range( h q )alternate on this circle in x and y . Therefore, for every σ such that h p ( σ ) ∈ ∆ ◦ , h p ( a σ ) is a crossing point of p and q if h p ( a σ ) ∈ range( h q )and correspondingly, h p ( a σ ) is a crossing point of p and q if h p ( a σ ) ∈ range( h q ), h p ( b σ ) is a crossing point of p and q if h p ( b σ ) ∈ range( h q ) and h p ( b σ ) is a crossing point of p and q if h p ( b σ ) ∈ range( h q ).On the other hand, every crossing point of p and q or q is equal to h p ( a σ ) or h p ( b σ ) for some σ with h p ( σ ) ∈ ∆ ◦ . Therefore, the number N of crossings of p with q or q is even. Since by (14) { y, y ′ } ∩ range( h p ) = ∅ , N = CN( p, q ) + CN( p, q ) is an even number, hence π ( p, q ) = π ( p, q ). Case (d) y = y ′ , z ∈ l ( y, y ′ ) \ yy ′ and z l ( y, y ′ ): Let q := ( r, y )( t , z )( r ′ , y ′ )) for z := ( z + z ) /
2. By Case (b), π ( p, q ) = π ( p, q ) and by Case (c), π ( p, q ) = π ( p, q ). Case (e) y = y ′ , z , z l ( y, y ′ ): Let q := ( r, y )( t , z )( r ′ , y ′ )) for z := ( z + z ) /
2. By Case (c), π ( p, q ) = π ( p, q ) and π ( p, q ) = π ( p, q ). Case (f ) y = y ′ and z , z ∈ l ( y, y ′ ) \ yy ′ : Proof via q as in (d) and (e). ✷ For tracks p, q and q ′ , by tiny shifts of the vertices of p we can obtain a track p such that p ⊲⊳ q and p ⊲⊳ q ′ . Lemma 2.7
1. Let p = (( s , x ) , ( s , x ) . . . . , ( s k , x k )) and let q and q ′ be tracks. Then for every δ > there is sometrack p = (( s , y ) , ( s , y ) , . . . , ( s k , y k )) such that p ⊲⊳ q , p ⊲⊳ q ′ and k x i − y i k < δ for ≤ i ≤ k .2. If p = (( s , x ) , ( s , x ) . . . . , ( s k , x k )) and p = (( s , y ) , ( s , y ) , . . . , ( s k , y k )) are tracks such that k x i − y i k < δ for ≤ i ≤ k then k h p ( s ) − h p ( s ) k < δ for all s ≤ s ≤ s k . Proof
1. We must find (prove the existence of) points y , y , . . . , y k such that k x i − y i k < δ and for p :=( s , y ) , ( s , y ) , . . . , ( s k , y k )), V ( p ) ∩ ( L ( q ) ∪ L ( q ′ )) = ∅ and (18) L ( p ) ∩ ( V ( q ) ∪ V ( q ′ )) = ∅ (19)The following statements are true since V ( q ) ∪ V ( q ′ ) is a finite set of points and L ( q ) ∪ L ( q ′ ) is a finiteset of straight lines. 7 There is some y ∈ B ( x , δ ) such that y
6∈ L ( q ) ∪ L ( q ′ ).– Suppose y i has been determined for some 0 ≤ i < k . There is some y i +1 ∈ B ( x i +1 , δ ) such that y i +1 = y i , y i +1 ( L ( q ) ∪ L ( q ′ )) and l ( y i , y i +1 ) ∩ ( V ( q ) ∪ V ( q ′ )) = ∅ , Then p := (( s , y ) , ( s , y ) . . . . , ( s k , y k )) is a track such that p ⊲⊳ q , p ⊲⊳ q ′ and k x i − y i k < δ for0 ≤ i ≤ k .2. For s i ≤ s ≤ s i +1 by (9), k h p ( s ) − h p ( s ) k = k h p ( s i ) + s − s i s i +1 − s i ( h p ( s i +1 ) − h p ( s i )) − h p ( s i ) − s − s i s i +1 − s i ( h p ( s i +1 ) − h p ( s i )) k = k x i + s − s i s i +1 − s i ( x i +1 − x i ) − y i − s − s i s i +1 − s i ( y i +1 − y i ) k = k (1 − s − s i s i +1 − s i ) · ( x i − y i )+ s − s i s i +1 − s i ) · ( x i +1 − y i +1 ) k < δ ✷ In the following let f, g : [ −
1; 2] → R , let I = [ a I ; b I ] and J = [ a J ; b J ] be intervals withrational endpoints such that I, J ⊆ [ −
1; 2] and let α IJ be the number from (8). Since f and g are continuous, there is a modulus of uniform continuity md : N → N for f and g , thatis, md is increasing and for all t, t ′ ∈ [ −
1; 2], k f ( t ) − f ( t ′ ) k < − n and k g ( t ) − g ( t ′ ) k < − n if | t − t ′ | < − md( n ) . (20)We approximate f | I and g | J by tracks. Definition 2.8
A track p = (( s , x ) , ( s , x ) , . . . , ( s k , x k )) is an approximation with precision − n (shortlyan n -approximation) of f | I if s = a I and s k = b I , (21) s i +1 − s i < − md( n ) and (22) k f ( s i ) − x i k < − n . (23)Remember that by Definition 2.1, x i = h p ( s i ). Approximations of g | J are defined accordingly. Lemma 2.9
In Definition 2.8, for all meaningful i and s , k x i − x i +1 k < · − n , (24) k f ( s ) − h p ( s ) k < · − n . (25)8 roof k x i − x i +1 k ≤ k x i − f ( s i ) k + k f ( s i ) − f ( s i +1 ) k + k f ( s i +1 ) − x i +1 ) k < · − n .For s i ≤ s ≤ s i +1 , k f ( s ) − h p ( s ) k ≤ k f ( s ) − f ( s i ) k + k f ( s i ) − h p ( s i ) k + k h p ( s i ) − h p ( s ) k < − n + 2 − n + k x i − x i +1 k < · − n by (9) and (24). ✷ After these technical preparations prove the following central lemma.
Lemma 2.10
Let − n < α IJ / . Let p and p ′ be n -approximations of f | I and let q and q ′ be n -approximations of g | J such that p ⊲⊳ q and p ′ ⊲⊳ q ′ . Then π ( p, q ) = π ( p ′ , q ′ ) . Proof
Let p = (( s , x ) , ( s , x ) . . . . , ( s k , x k ))where s = a I and s k = b I . Notice that possibly p and q ′ as well as p ′ and q have common straight linesegments such that π ( p, q ′ ) and π ( p ′ , q ) are not defined. By Lemma 2.7 there is some track p = (( s , y ) , ( s , y ) , . . . , ( s k , y k ))such that p ⊲⊳ q, p ⊲⊳ q ′ and k x i − y i k < − n for 0 ≤ i ≤ k . (26)Figure 5 shows the ⊲⊳ -relation (indicated by lines) between the tracks p, q, p, p ′ and q ′ . It suffices to prove ✟✟✟✟✟ ❍❍❍❍❍ qp p p ′ q ′ Figure 5: The ⊲⊳ -relation between the tracks p, q, p, p ′ and q ′ . π ( p, q ) = π ( p, q ) = π ( p, q ′ ) = π ( p ′ , q ′ ) . (27)We use the facts that by the definitions the points f ( t ), h p ( t ), h p ( t ) and h p ′ ( t ) are close together andthat the points g ( t ), h q ( t ) and h q ′ ( t ) are close together.In the following we prove the second equation π ( p, q ) = π ( p, q ′ ) of (27) in detail. Let q = (( r , u ) , ( r , u ) , . . . , ( r l − , u l − ) , ( r l , u l )) , (28) q ′ = (( r ′ , u ′ ) , ( r ′ , u ′ ) , . . . , ( r ′ m − , u ′ m − ) , ( r ′ m , u ′ m )) (29)where r = r ′ = a J and r l = r ′ m = b J . (30)be n -approximations of g | J . As an example, Figure 6 shows the arguments r , . . . , r of a track q on thereal line and and the arguments r ′ , . . . , r ′ of a track q ′ on the real line (thick black dots).From (26) we know that V ( q ) ∩ L ( p ) = ∅ and V ( q ′ ) ∩ L ( p ) = ∅ . In a first step we add “redundant”vertices to q and q ′ such that the resulting tracks q and q ′ have the same number of vertices and suchthat h q = h q , h q ′ = h q ′ , V ( q ) ∩ L ( p ) = ∅ and V ( q ′ ) ∩ L ( p ) = ∅ .Let i be a number such that r i
6∈ { r ′ , . . . , r ′ m } . There is a unique j such that r ′ j < r i < r ′ j +1 (e.g. i = 3 and j = 1 in Figure 6). 9 ✲ RR t t t t t t t t❣r ❣r ❣r ❣r ❣r r ′ r ′ r ′ r ′ r ′ r ′ r ′ r ′ t t t t t t t❣ ❣ ❣ ❣r r r r r r r r r r r Figure 6: Insertion of redundant vertices into q and q ′ .The point h q ′ ( r i ) is an element of the straight line segment u ′ j u ′ j +1 and different from u ′ j and u ′ j +1 .We would like to add the “redundant” pair ( r i , h q ′ ( r i )) to q ′ with result(( r ′ , u ′ ) , . . . , ( r ′ j , u ′ j ) , ( r i , h q ′ ( r i )( r ′ j +1 , u ′ j +1 ) , . . . , ( r ′ m , u ′ m ))But possibly h q ′ ( r i ) ∈ L ( p ). Instead, we add a pair ( t, v ) with v = h q ′ ( t )
6∈ L ( p ) to q ′ which is very closeto ( r i , h q ′ ( r i )). Let γ := min {| c − d | | c, d ∈ { r , . . . , r l , r ′ , . . . , r ′ m } and c = d } . (31)Since u ′ j
6∈ L ( p ), by Lemma 2.5 l ( u ′ j u ′ j +1 ) intersects every straight line from L ( p ) at most once. Therefore,the straight line segment u ′ j u ′ j +1 contains only finitely many points of L ( p ). Thus there is some t suchthat | t − r i | < γ/ v := h q ′ ( t )
6∈ L ( p ). Then v ∈ u ′ j u ′ j +1 (see (9)). We add the pair ( t, v ) to q ′ withresult q := (( r ′ , u ′ ) , . . . , ( r ′ j , u ′ j ) , ( t, h q ′ ( t )( r ′ j +1 , u ′ j +1 ) , . . . , ( r ′ m , u ′ m )) . Then h q ′ = h q . Let q ′ be the track obtained from q ′ by adding a pair ( t, v ) in this way for every i suchthat r i
6∈ { r ′ , . . . , r ′ m } in turn. Then h q ′ = h q ′ and V ( q ′ ) ∩ L ( p ) = ∅ .Correspondingly let q be the track obtained from q in the same way. Then h q = h q and V ( q ) ∩L ( p ) = ∅ . In summary, h q = h q , h q ′ = h q ′ , V ( q ) ∩ L ( p ) = ∅ and V ( q ′ ) ∩ L ( p ) = ∅ . (32)In Figure 6 the inserted arguments t are within the the circles. The new tracks can be written as q = (( t , v ) , ( t , v ) , . . . , ( t µ , v µ )) , (33) q ′ = (( t ′ , v ′ ) , ( t ′ , v ′ ) , . . . , ( t ′ µ , v ′ µ )) (34)where a J = r = t = t ′ and b J = r l = t µ = t ′ µ . By the condition | t − r i | < γ/ t, v )above, | t ν − t ′ ν | < γ/ { t ν , t ′ ν } < { t ν +1 , t ′ ν +1 } . (35)The next Proposition prepares the proof or Proposition 2.12. Proposition 2.11
For the tracks q and q ′ , for all < ν < µ , { v ν − , v ν , v ν +1 , v ′ ν − , v ′ ν , v ′ ν +1 } ⊆ B ( g ( t ν ) , · − n ) and (36) { y , y k } ∩ B ( g ( t ν ) , · − n ) = ∅ , (37) B ( g ( t ) , · − n ) ∩ range( h p ) = ∅ , (38) B ( g ( t µ − ) , · − n ) ∩ range( h p ) = ∅ . (39)10 roof (Proposition 2.11) Proof of (36):
Consider (( t ν − , v ν − ) , ( t ν , v ν ) , ( t ν +1 , v ν +1 )) as a part of q . Then ( t ν , v ν ) has alreadybeen in q , that is, ( t ν , v ν ) = ( r i , u i ) for some i or it has been inserted via some j such that r ′ j
6∈ { r , . . . , r l } .Therefore, there is some i such that (see Figure 6) r i − ≤ t i − < r i < t ν +1 ≤ r i +1 or (40) r i ≤ t ν − < t ν < t ν +1 ≤ r i +1 . (41)Consider (40). Then k g ( t ν ) − v ν k = k g ( r i ) − u i k < − n by (23). Furthermore, k g ( t ν ) − v ν − k ≤k g ( t ν ) − v ν k + k v ν − v ν − k < − n + k u i − v ν − k . Since h q = h q , v ν − ∈ u i − u i , hence k u i − v ν − k ≤k u i − − u i k < · − n by (24). Therefore, k g ( t ν ) − v ν − k < · − n . By symmetry k g ( t ν ) − v ν +1 k < · − n .Therefore, { v ν − , v ν , v ν +1 } ⊆ B ( g ( t ν ) , · − n ).Consider (41).Then k g ( t ν ) − v ν k = k g ( t ν ) − h q ( t ν ) k < · − n by(25). Furthermore, k g ( t ν ) − v ν − k ≤ k g ( t ν ) − v ν k + k v ν − v ν − k < − n + k u i +1 − u i k < − n + 3 · − n = 4 · − n (since v ν , v ν − ∈ u i u i +1 and by (24)). By symmetry k g ( t ν ) − v ν +1 k < · − n . Therefore, { v ν − , v ν , v ν +1 } ⊆ B ( g ( t ν ) , · − n ).As a summary, in both cases (40) and (41), { v ν − , v ν , v ν +1 } ⊆ B ( g ( t ν ) , · − n ). By symmetry, { v ′ ν − , v ′ ν , v ′ ν +1 } ⊆ B ( g ( t ′ ν ) , · − n ). Since | t ν − t ′ ν | < γ/ < − md( n ) , k g ( t ν ) − g ( t ′ ν k < − n by (35), (31),(22) and (20). Therefore, { v ν − , v ν , v ν +1 , v ′ ν − , v ′ ν , v ′ ν +1 } ⊆ B ( g ( t ν ) , · − n ). Proof of (37):
By (8), (23) and Lemma 2.7 for all ν · − n < α IJ ≤ k f ( s ) − g ( t ν ) k ≤ k f ( s ) − x k + k x − y k + k y − g ( t ν ) k < − n + 2 − n + k y − g ( t ν ) k ,hence k y − g ( t ν ) k > · − n . By symmetry, k y k − g ( t ν ) k > · − n . Proof of (38):
For every s ≤ s ≤ s k , k f ( s ) − g ( r ) k ≥ α IJ > · − n . By (25) and Lemma 2.7, k h p ( s ) − g ( r ) k > · − n . Since | r − t | ≤ | s − s | < md ( n ), k h p ( s ) − g ( t ) k > · − n , hence B ( g ( t ) , · − n ) ∩ range( h p ) = ∅ . Proof of (39):
From (38) by symmetry. ✷ (Proposition 2.11) Proposition 2.12 π ( p, q ) = π ( p, q ′ ) . Proof (Proposition 2.12): We transform the track q ′ in two phases to the track q preserving the crossingparity in each step. Remember (32) and Proposition 2.11 for q = (( t , v ) , ( t , v ) , ( t , v ) , . . . , ( t µ , v µ )) and q ′ = (( t ′ , v ′ ) , ( t ′ , v ′ ) , ( t ′ , v ′ ) , . . . , ( t ′ µ , v ′ µ )) . In the first phase for every 0 ≤ i < µ we replace every v ′ i by some w i without changing q ′ i . For0 ≤ i < µ let Q ( i ) be the following property:There are points w , . . . , w i such that for all 1 ≤ j ≤ i k w j − v ′ j k < − n / , w j
6∈ L ( p ) , w j
6∈ { v j − , w j − , v ′ j +1 } ) and (42) π ( p, q ′ ) = π ( p, q i ) (43)where q i is the track q i := (( t ′ , w ) , . . . , ( t ′ i , w i ) , ( t ′ i +1 , v ′ i +1 ) , . . . , ( t ′ µ , v ′ µ )) . Let w := v ′ . Then q ′ = q , hence Q (0) is true. Suppose for some 0 ≤ i ≤ µ − Q ( i ).There is some w i +1 such that k w i +1 − v ′ i +1 k < − n / , w i +1
6∈ L ( p ) , w i +1
6∈ { v i , w i , v ′ i +2 } . (44)11y definition, q i = (( t ′ , w ) , . . . , ( t ′ i , w i ) , ( t ′ i +1 , v ′ i +1 ) , ( t ′ i +2 , v ′ i +2 ) , . . . , ( t ′ µ , v ′ µ )) and q i +1 = (( t ′ , w ) , . . . , ( t ′ i , w i ) , ( t ′ i +1 , w i +1 ) , ( t ′ i +2 , v ′ i +2 ) , . . . , ( t ′ µ , v ′ µ )) . Since q i is a track, by (44), q i +1 is a track (neighboring vertices must be different). Also by (44),(42) is true also for i + 1. By Proposition 2.11 Lemma 2.6 can be applied to the boldface sub-tracks τ := (( t ′ i , w i ) , ( t ′ i +1 , v ′ i +1 ) , ( t ′ i +2 , v ′ i +2 )) and τ := (( t ′ i , w i ) , ( t ′ i +1 , w i +1 ) , ( t ′ i +2 , v ′ i +2 )) such that π ( p, τ ) = π ( p, τ ). Since the vertices of q i and q i +1 are not in L ( p ), hence not in range( h p ), π ( p, q ′ ) = π ( p, q i ) = π ( p, q i +1 ). Therefore, we have proved Q ( i + 1).By induction Q ( µ −
1) is true, hence π ( p, q ′ ) = π ( p, q µ − ) . (45)Figure 7 shows the =-relation for the vertices of q = (( t , v ) , ( t , v ) , ( t , v ) , . . . , ( t µ , v µ )) (top) and q µ − = (( t ′ , w ) , ( t ′ , w ) , . . . , ( t ′ µ − , w µ − ) , ( t ′ µ , v ′ µ )) (bottom). A line between x and y means x = y . ❅❅❅ ❅❅❅ ❅❅❅ ❅❅❅ ❅❅❅ ❅❅❅ ❅❅❅ q q q q q w w w w w µ − w µ − v ′ µ v v v v v µ − v µ − v µ Figure 7: The =-relation for the vertices of q (top) and q µ − (bottom).In the second phase two we transform q µ − step by step to q without changing the crossing paritywith p . For 0 ≤ i ≤ µ − q ′′ i := (( t , v ) , . . . , ( t i , v i )( t ′ i +1 , w i +1 ) , . . . , ( t ′ µ − , w µ − ) , ( t ′ µ , v ′ µ )) . Since q and q µ − are tracks and v i = w i +1 , q ′′ i is a track.By (36) and by k w − v ′ k < − n /
16 (42), v w ∪ w w ⊆ B ( g ( t ) , · − n ). By (38), v w and w w do not intersect h p . Therefore, π ( p, q µ − ) = π ( p, q ′′ ) (46)Suppose for 0 ≤ i ≤ µ − π ( p, q µ − ) = π ( p, q ′′ i ) . By definition, q ′′ i = (( t , v ) , . . . , ( t i , v i ) , ( t ′ i +1 , w i +1 ) , ( t ′ i +1 , w i +2 ) , . . . , ( t ′ µ , v ′ µ )) and q ′′ i +1 = (( t , v ) , . . . , ( t i , v i ) , ( t i +1 , v i +1 ) , ( t ′ i +1 , w i +2 ) , . . . , ( t ′ µ , v ′ µ )) , By Proposition 2.11 Lemma 2.6 can be applied to the boldface sub-tracks τ := (( t i , v i ) , ( t ′ i +1 , w i +1 ) , ( t ′ i +1 , w i +2 ) and τ := (( t i , v i )( t i +1 , v i +1 ) , ( t ′ i +1 , w i +2 )) 12uch that π ( p, τ ) = π ( p, τ ).Since the vertices of q ′′ i and q ′′ i +1 are not in L ( p ), hence not in range( h p ), π ( p, q µ − ) = π ( p, q ′′ i ) = π ( p, q ′′ i +1 ). By induction, π ( p, q µ − ) = π ( p, q ′′ µ − )Finally we compare q = (( t , v ) , , . . . , ( t µ − , v µ − ) , ( t µ − , v µ − ) , ( t µ , v µ )) and q ′′ µ − = (( t , v ) , , . . . , ( t µ − , v µ − ) , ( t ′ µ − , w µ − ) , ( t ′ µ , v ′ µ )) . Since k w µ − − v ′ µ − k < − n by Proposition 2.11 and( v µ − v µ − ∪ v µ − v µ ∪ v µ − w µ − ∪ w µ − v ′ µ ) ∩ range( h p ) = ∅ ,π ( p, q ′′ µ − ) = π ( p, q ), hence π ( p, q ′ ) = π ( p, q ). ✷ (Proposition 2.12)We continue the proof of Lemma 2.10. Since h q = h q and h q ′ = h q ′ by (32), π ( p, q ′ ) = π ( p, q ), which isthe second equation from (27). The first and the third equation can be poved accordingly. We omit thedetails. ✷ (Lemma 2.10)Lemma 2.10 allows to define the parity of the pair ( f | I , g | J ). Definition 2.13 π ( f | I , g | J ) := π ( p, q ) for an arbitrary n -approximation p of f | I and an arbitrary n -approximation q of g | J such that p ⊲⊳ q and − n < α IJ / . Lemma 2.14
Let f, g, J and I = [ a I ; b I ] be as before such that (8). Let I := [ a I ; c ] and I := [ c ; b I ] where a I < c < b I and f ( c ) g ( J ) . Then π ( f | I , g | J ) = ( π ( f | I , g | J ) + π ( f | I , g | J )) mod 2 . (47) Proof
There are numbers α > n ∈ N such that α < d s ( { f ( a I ) , f ( c ) , f ( b I ) } , g ( J )), α < d s ( { g ( a J ) , g ( b J ) } , f ( I )) and 16 · − n < α . There are n -approximations p = (( s , x ) , . . . , ( s k , x x )) and q of f | I and g | q , respectively, such that c = s i for some i and p ⊲⊳ q .Let p := (( s , x ) , . . . , ( s i , x i )) and p := (( s i , x i ) , . . . , ( s k , x k )).Then p ⊲⊳ q and p ⊲⊳ q . Therefore CN( p , q ) and CN( p , q ) are well-defined such that (see Defini-tion 2.2) CN( p, q ) = CN( p , q ) + CN( p , q ) , hence π ( p, q ) = CN( p, q ) mod 2 = (CN( p , q ) mod 2 + CN( p , q ) mod 2) mod 2 = ( π ( p , q ) + π ( p , q )) mod 2. (47) follows by Definition 2.13. ✷ In TTE computability is defined on represented sets (
X, δ ). A representation is a function δ : ⊆ Σ ∗ → X (if X is finite) or δ : ⊆ Σ ω → X (if X has at most continuum cardinality). If δ ( w ) = x then w is consideredas a “name” (or a “ δ -name”) of x . Every x ∈ X must have a name (and may have many names) but notevery w ∈ Σ ∗ must be a name of some x ∈ X , hence δ is a partial surjective function.13or represented sets ( X i , δ i ) where δ i : ⊆ A i → X i , A i ∈ { Σ ∗ , Σ ω } , a partial function f : ⊆ X → X iscomputable, if there is a computable function h : ⊆ A → A which realizes f , that is, if w is a δ -nameof x ∈ dom( f ) then h ( w ) is a δ -name of f ( x ) (accordingly functions on Cartesian products). For mostsets in Analysis there are canonical (or standard or effective or obvious) representations which we usehere. We will say “computable” without mentioning the (standard) representations.We need the concept of computable multi-functions on represented sets. As an example, for thestandard represented sets ( R , ρ ), ( Q , ν Q ) and ( N , ν N ) there is no computable function f : R × N → Q suchthat | x − f ( x, n ) | < − n . But there is a computable function h : Σ ω × Σ ∗ → Σ ∗ which from every name of x and every name of n computes a name of some a ∈ Q such that | x − a | < − n . The computed number a depends on the names of x and n and not only on ( x, n ). The function h realizes the multi-function f : R × N ⇒ Q such that f ( x, n ) = { a ∈ Q | | x − q | < − n } . Informally we say: f maps every ( x, n )to some a ∈ Q such that | x − a | < − n , more formally, f : ( x, n ) | ⇒ a such that | x − a | < − n . In thefollowing we apply results about computability on sets with standard representations from [9, 1, 11, 8].In Lemma 3.1 we consider n -approximations p and q of f | I and g | J , respectively such that p ⊲⊳ q . Weprove that p and q can be computed from ( f, g, I, J, n ). We use from TTE, that ( f, s, n ) | ⇒ y ( s ∈ Q , y ∈ Q ) such that k f ( s ) − y k < − n ) is computable. Lemma 3.1
Let R : ( f, g, I, J, n ) | ⇒ ( p, q ) be the multi-function mapping continuous functions f, g :[ −
1; 2] → R , closed rational intervals I, J ⊆ [ −
1; 2] and a number n ∈ N to a pair ( p, q ) , p =(( s , x ) , . . . , ( s k , x k )) and q = (( t , y ) , . . . , ( t l , y l )) , of rational tracks such that p is an n -approximation of f | I , (48) q is an n -approximation of g | J and (49) p ⊲⊳ q . (50) Then R is computable. Since rational tracks are “discrete” objects there is no compuble funtion but only a computable multi-function (known from TTE).
Proof
It is known [9, 1, 11] that from continuous functions f, g : [ −
1; 2] → R we can compute a modulusof continuity md f of f and a modulus of continuity md g of g . We can compute md := max(md f , md g ),which is a modulus of continuity of f and of g . Let I = [ a I ; b I ] and J = [ a J ; b J ]. Consider Definition 2.8.First we compute p .Choose some k and rational numbers s := a I , s , . . . , s k − , s k := b I such that 0 < s i +1 − s i < − md( n ) for 0 ≤ i < k . Choose some x ∈ Q such that k f ( s ) − x k < − n . Suppose, x i has been chosen for i < k . Then choose some x i +1 ∈ Q such that k f ( s i +1 ) − x i +1 k < − n and x i +1 = x i . Since x i = x i +1 for i < k , p := (( s , x ) , . . . , ( s k , x k )) is a rational track. It is an n -approximation of f | I .Next we compute q .Choose some l and rational numbers t := a J , t , . . . , t l − , t l := b J such that 0 < t i +1 − t i < − md( n ) for0 ≤ i < l . Choose some y ∈ Q such that k g ( t ) − y k < − n and y
6∈ L ( p ). Suppose, y , . . . , y i havebeen chosen for some i < l . Then choose some y i +1 ∈ Q such that y i +1 = y i , k g ( t i +1 ) − y i +1 k < − n , y i +1
6∈ L ( p ) } and l ( y i , y i +1 )) ∩ V ( p ) = ∅ .Since y i = y i +1 for 0 i < l , q = (( t , y ) , . . . , ( t l , y l )) is a rational track. It is an n -approximation of g | J .We have V ( q ) ∩ L ( p ) = ∅ since y i
6∈ L ( p ) for all i . We have L ( q ) ∩ V ( p ) = ∅ since l ( y i , y i +1 )) ∩ V ( p ) = ∅ for all i < l . Therefore, p ⊲⊳ q .All of this can be computed from continuous functions f, g : [ −
1; 2] → R , rational intervals I, J ⊆ [ −
1; 2] and n . ✷ orollary 3.2 Let f, g : [ −
1; 2] → R be continuous functions and let I, J ⊆ [ −
1; 2] be closed rationalintervals such that f ( I ) ∩ g ( J ) = ∅ . Then π ( f | I , g | J ) = 0 . Proof
Since f ( I ) and g ( J ) are compact, d s ( f ( I ) , g ( J )) > α IJ >
0. There is some number n such that 11 · − n < d s ( f ( I ) , g ( J )) and 2 − n < α IJ / n -approximations p and q of ( f | I and g | J , respectively, such that p ⊲⊳ p .By definition 2.13, π ( f | I , g | J ) = π ( p, q ). By (25), range( h p ) ⊆ U (range( f I ) , · − n ) and range( h q ) ⊆ U (range( g J ) , · − n ) . Since 11 · − n < d s ( f ( I ) , g ( J )), U (range( f I ) , · − n ) ∩ U (range( g J ) , · − n ) = ∅ ,hence range( h p ) ∩ range( h q ) = ∅ . We obtain CN( h p , h q ) = 0 and π ( p, q ) = 0. By Definition 2.13, π ( f | I , g | J ) = 0. ✷ Lemma 3.3
1. From continuous functions f, g : [ −
1; 2] → R and closed rational intervals I, J ⊆ [ −
1; 2] we cancompute α IJ .2. From continuous functions f, g : [ −
1; 2] → R and closed rational intervals I, J ⊆ [ −
1; 2] such that α IJ > we can compute π ( f | I , g | J ) . Remember that f ( a I ) , f ( b I ) g ( J ) and g ( a J ) , g ( b J ) f ( I ) is equivalent to α IJ > Proof (1) From f , a I and b I we can compute the compact set { a I , b I } . from g and J we can computethe compact set g ( J ). The function ( A, B ) d s ( A, B ) on (non-empty) compact set is computable.Therefore, from f, g, I, J we can compute d s ( { f ( a I ) , f ( b I ) } , g ( J )). Correspondingly we can compute d s ( { g ( a J ) , g ( b J ) } , f ( I )) and hence their minimum α IJ .(2) Compute α IJ . Find some n such that 2 − n < α/
16. By Lemma 3.1 we can compute n -approximations p and q of f I and g J , respectively such that p ⊲⊳ q . Compute the number of crossingsCN( p, q ) and its parity π ( p, q ). By Definition 2.13, π ( f | I , g | J ) = π ( p, q ) ✷ f and g . The crossing parity π ( f, g ) does not depend on f and g as long as φ, ψ : [0; 1] → [0; 1] are continuousand (1) is true. For showing π ( p, q ) = 1 we construct special tracks p and q . For tracks p define L ( p ) := S { l ( x, y ) | x, y ∈ V ( p ) , x = y } ( (11) and Figure 3). Lemma 4.1
There are -approximations p = (( s , x ) , . . . , ( s k , x k )) and q = (( t , y ) , . . . , ( t l , y l )) of f and g , respectively, such that card V ( p ) = k + 1 and card V ( q ) = l + 1 and (51) V ( p ) ∩ L ( q ) = ∅ and V ( q ) ∩ L ( p ) = ∅ . (52) Proof
There is a modulus of uniform continuity md of f and g (cf. (20)). There are some k andrational numbers s := − , s , . . . , s k − , s k := 2 such that 0 < s i +1 − s i < − md(5) for 0 ≤ i < k .There are points x , . . . , x k ∈ Q such that k f ( s ) − x k < − and x i = x j for all i = j . Then p = (( s , x ) , . . . , ( s k , x k )) is a 5-approximation of f such that card V ( p ) = k + 1. There are some l and where U ( A, δ ) := { x ∈ R | ( ∃ y ∈ A ) k x − y k < δ } t := − , t , . . . , t k − , t k := 2 such that 0 < t i +1 − t i < − md(5) for 0 ≤ i < l . There issome y ∈ Q such that k g ( t ) − y k < − k and y
6∈ L ( p ). Suppose y , . . . , y i ( i < l ) have been found.Then there is some y i +1 ∈ Q such that k g ( s i +1 ) − y i +1 k < − , y i +1
6∈ { y , . . . , y i } , y i +1
6∈ L ( p ) and( l ( y , y i +1 ) ∪ . . . ∪ l ( y i , y i +1 )) ∩ V ( p ) = ∅ . By induction, we can define y , . . . , y l . The obviously (51) and(52) are true for p and q . ✷ Lemma 4.2 π ( f, g ) = 1 . Proof
For I := J := [ −
1; 2] we have f | I = f , g | J = g and α I J = 1 ((8) and Figure 1). Since2 − < α IJ /
16 by Definition 2.13, π ( f, g ) = π ( p, q ) for (arbitrary) 5-approximations p and q of f and g such that (51) and (52). Notice that (52) implies p ⊲⊳ q . In the following we prove π ( p, q ) = 1. We define z := (0 . , . , (53) R f := [ − − − ; − . · − ] × [ − − ; 2 − ] and (54) R f := [1 . − · − ; 2 + 2 − ] × [ − − ; 2 − ] . (55)Figure 8 shows f and g , the ball B ( z, .
31) and the two boxes R f and R f . By the definitions of f and g in (6) and (7), f ( s ) = ( s,
0) for − ≤ s ≤
0. First we prove h p ( s ) ∈ R f for s < − . − and (56) k h p ( s ) − z k < .
31 for − . − − < s ≤ . − and (57) h p ( s ) ∈ R f for s > . − − . (58) ✲✻ s z z φψg f f gφ (0) ψ (0) ψ (1) φ (1) f ( − g ( − g (2) f (2) rr rr r z ✏✏✏✏✶ ( − . , q q ✂✂✂✍ ✂✂✂✌ x β x γ Figure 8: Intersecting curves φ and ψ with extensions f and g We show (56). Let s < − . − . Then for some i , s i ≤ s < s i +1 < − . − . Then x i ∈ R f since k f ( s i ) − x i k − − . Correspondingly, x i +1 ∈ R f . Since h p ( s ) ∈ x i x i +1 and R f is convex, h p ( s ) = ∈ R f .(58) is true by symmetry.We show (57). If − . − − < s ≤ − . k h p ( s ) − z k ≤ k h p ( s ) − f ( s ) k + k f ( s ) − f ( − . k + k f ( − . − z k < · − + 2 − + k f ( − . − z k ≤ · − + p / < .
19 + 1 .
12 = 1 .
31. If − . ≤ s ≤ k h p ( s ) − z k ≤ k h p ( s ) − f ( s ) k + k f ( s ) − z k < · − + k f ( − . − z k < .
31 (from above). Hence k h p ( s ) − z k < .
31 for − . − − < s ≤
0. By symmetry, k h p ( s ) − z k < .
31 also for 1 ≤ s < . − .For 0 ≤ s ≤ k h p ( s ) − z k ≤ k h p ( s ) − f ( s ) k + k f ( s ) − z k ≤ · − + √ / < .
31. Therefore (57) istrue. Since | s i +1 − s i | < − md(5) ≤ − there are indices β < γ such that − . − − < s β < − . − and 1 . − − < s γ < . − . (59)By (56) - (58), x i ∈ R f if i ≤ β , (60) x i ∈ B (( z, .
31) if β ≤ i ≤ γ , (61) x i ∈ R f if γ ≤ i . (62)We simplify the track p step by step by means of Lemma 2.6. For β < i ≤ γ define p i := (( s , x ) , . . . , ( s β , x β ) , ( s i , x i ) , . . . , ( s k , x k ))Since the x i are pairwise different (by card( V ( p )) = k + 1), x β = x i . Therefore, p i is a track, seeDefinition 2.1. We prove by induction π ( p i , q ) = π ( p, q ) for β < i ≤ γ . (63)Since p β +1 = p , π ( p β +1 , q ) = π ( p, q ).Suppose π ( p i , q ) = π ( p, q ) for some β < i < γ . We define p a := (( s , x ) , . . . , ( s β , x β )) , (64) p b := (( s β , x β ) , ( s i , x i ) , ( s i +1 , x i +1 )) , (65) p c := (( s β , x β ) , ( s i , z ) , ( s i +1 , x i +1 )) (66)for some z ∈ x β x i +1 \ { x β , x i +1 } ,p d := (( s β , x β ) , ( s i +1 , x i +1 )) (67) p e := (( s i +1 , x i +1 ) , . . . , ( s k , x k )) (68)We apply Lemma 2.6 to p b and p c . We have (cf. (14 - 16))– { x β , x i +1 } ∩ L ( q ) = ∅ , (since V ( p ) ∩ L ( q ) = ∅ ),– { x β , x i , x i +1 , z } ∈ B ( z, .
31) (by (61)) and– { y , y l } ∩ B ( z, .
31) = ∅ .(For the last line: 1 . < √ . + 0 . = k g ( − − z k ≤ k g ( − − y k + k y − z k < k y − z k + 2 − < k y − z k + 0 .
04, hence k y − z k > .
31. Accordingly k y l − z k > . π ( p b , q ) = π ( p c , q ).Since z ∈ x β x i +1 \ { x β , x i +1 } , π ( p c , q ) = π ( p d , q ), hence π ( p b , q ) = π ( p d , q ). Therefore, replacing p b by p d in p β,i does not change the crossing parity. We show this in detail.Since V ( p ) ∩ L ( q ) = ∅ , x i range( h q ) for all i , henceCN( p i , q ) = CN( p a , q ) + CN( p b , q ) + CN( p e , q ) andCN( p i +1 , q ) = CN( p a , q ) + CN( p d , q ) + CN( p e , q ) . Since ( m + n ) mod 2 = ( m mod 2 + n mod 2) mod 2, π ( p i , q ) = ( π ( p a , q ) + π ( p b , q ) + π ( p e , q )) mod 2= ( π ( p a , q ) + π ( p d , q ) + π ( p e , q )) mod 2= π ( p i +1 , q ) .
17y induction π ( p, q ) = π ( p γ , q ) for p γ = (( s , x ) , . . . , ( s β , x β ) , ( s γ , x γ ) , . . . , ( s k , x k )) . (69)By (56) and (58), h p γ runs in the box R f for − ≤ s ≤ s β , then its graph is the straight line segment x β x γ and finally h p γ remains in the box R f for s γ < s , see Figure 8.Since V ( p γ ) ⊆ V ( p ), V ( q ) ∩ L ( p γ ) = ∅ . Keeping p γ fixed we can simplify the track q accordingly.Therefore there are indices µ < ν such that for q ν := (( t , y ) , . . . , ( t µ , y µ )( t ν , y ν ) , . . . , ( t l , y l )), π ( p γ , q ) = π ( p γ , q ν ), where h q ν runs in the box R g for − ≤ t µ , then its graph is the straight line segment y µ y ν and finally h p γ remains in the box R g for t ν < t as shown in Figure 9. ✲✻ s z z rr rrq q ✂✂✂✍ ✂✂✂✌ x β x γ q q ❇❇❇◆ ❇❇❇▼ y µ y ν R f R g R g R f Figure 9: The tracks p γ and q ν .Since ( R f ∪ R f ) ∩ range( h q ν ) = ∅ and ( R g ∪ R g ) ∩ range( h p γ ) = ∅ (we omit the straightforwardformal verifications), CN( p γ , q ν ) = 1. We obtain π ( p, q ) = π ( p γ , q ) = π ( p γ , q ν ) = 1. ✷ (Lemma 4.2) For A ∈ R and a number δ > U ( A, δ ) := { x ∈ R | ( ∃ y ∈ A ) k x − y k < δ } be the δ -neighborhoodof A . From f and g we will compute sequences I ⊇ I ⊇ I ⊇ . . . and J ⊇ J ⊇ J ⊇ . . . of rationalclosed intervals such that f ( T i I i ) = g ( T i J i ). First we prove the following lemma. Lemma 5.1
From f, g , rational intervals
I, J ⊆ [ −
1; 2] and n ⊆ N such that such that − n < α IJ and π ( f | I , g | J ) = 1 we can compute a rational interval K ⊆ I such that α KJ > , f ( K ) ⊆ U ( g ( J ) , − n ) and π ( f | K , g | J ) = 1 . (70) Proof
Perform the following computations:– From J and g compute the compact set g ( J ).– Compute some sequence ( a I = s , s , . . . , s k − , s k = b I ) of rational numbers such that 0 < s i +1 − s i < − md ( n +4) .For every index i do the following:– From f and s i compute x i := f ( s i ) ∈ R .– From x i and g ( J ) compute c i := d s ( { x i } , g ( J )) ∈ R .18 ✻ − n − n d ss ✉ r s ✉r r r rr r rr s ✉rr rr s ✉r rr rr rr r r r r r s ✉ rrr r r r r s ✉ r rr r s ✉ Figure 10: d i ≈ d s ( f ( s i ) , g ( J ))– From c i ∈ R and n compute some d i ∈ Q such that | c i − d i | < − n / d i ≈ d s ( f ( s i ) , g ( J )).Suppose | s i − s | < − md ( n +4) and d i ≥ − n /
2. Then for every z ∈ g ( J ),2 − n / ≤ d i < c i + 2 − n / ≤ k x i − z k + 2 − n / ≤ k f ( s i ) − f ( s ) k + k f ( s ) − z k + 2 − n / < − n /
16 + k f ( s ) − z k + 2 − n /
16, hence k f ( s ) − z k > / · − n . Therefore,if | s i − s | < − md ( n +4) and d i ≥ − n / ∀ z ∈ f ( J )) k f ( s ) − z k > / · − n . (71)Suppose | s i − s | < − md ( n +4) and d i < − n /
2. Then c i < d i + 2 − n / ≤ / · − n , hence for some z ∈ g ( J ), k f ( s i ) − z k < / · − n and k f ( s ) − z k ≤ k f ( s ) − f ( s i ) k + k f ( s i ) − z k ≤ / · − n . Therefore,if | s i − s | < − md ( n +4) and d i < − n / ∃ z ∈ f ( J )) k f ( s ) − z k < / · − n . (72)Since 2 − n < α IJ and s = a I , 2 − n < d s ( f ( a I ) , g ( J )) = c < d + 2 − n /
16, hence d > − n / , correspondingly, d k > − n / . (73)Therefore, we can find a sub-sequence 0 = i < i < . . . < i m = k of 0 , , . . . , k such that( ∀ j ) d i j ≥ − n / ∀ i j < l < i j +1 ) d l ≥ − n / ∀ i j < l < i j +1 ) d l < − n / . (76)Let K j := [ s i j ; s i j +1 ]. (In Figure 10, m = 6 and ( i , i , . . . , i ) = (0 , , , , , , , ∀ i j < l < i j +1 ) d l ≥ − n /
2. (In Figure 47 this is true for j = 0 , , z ∈ g ( J ) and s ∈ K j , k f ( s ) − z k > / · − n , hence f ( K j ) ∩ g ( J ) = ∅ , therefore, π ( f | K j , g | J ) = 0 . (77)Suppose (76), that is, ( ∀ i j < l < i j +1 ) d l < − n /
2. (In Figure 47 this is true for j = 1 , , s ∈ K j there is some z ∈ g ( J ) such that k f ( s ) − z k < / · − n , hence f ( K j ) ⊆ U ( g ( J ) , − n ).Furthermore by(74), 2 − n / − − n / ≤ d i j − − n / < c i j = d s ( { f ( s i j ) } , g ( J )). The same is true for( j + 1) instead of j , therefore 0 < d s ( { f ( s i j ) , f ( s i j +1 ) } , g ( J )).19ince K j ⊆ I and α IJ > d s ( f ( K j ) , { g ( a J ) , g ( b J ) } ≥ d s ( f ( I ) , { g ( a J ) , g ( b J ) } ≥ α IJ . In summary,0 < α K j J and f ( K j ) ⊆ U ( g ( J ) , − n ) . (78)By Lemma 2.14, (cid:0) π ( f | K , g | J ) + · · · + ( π ( f | K m − , g | J ) (cid:1) mod 2 = π ( f | I , g | J ) = 1 . Therefore, there is some j such that π ( f | K j , g | J ) = 1. By (77), (75) cannot be true for j , hence (76) istrue for j , hence (78) is true for K j .For computing K from f, g, I, J and n , first compute ( d , . . . , d k ), then compute the sub-sequence0 = i < i <, . . . , < i m = k of 0 , , . . . , k such that (74–76). The find some j such that π ( f | K j , g | J ) = 1.Let K := K j . Then (70) is true by (78). ✷ Lemma 5.2
From functions f and g , closed rational intervals I, J ⊆ [ −
1; 2] and m ∈ N such that α IJ > and π ( f | I , g | J ) = 1 we can compute closed rational intervals I ′ , J ′ such that α I ′ J ′ > and π ( f | I ′ , g | J ′ ) = 1 , (79) I ′ ⊆ I and J ′ ⊆ J , (80) f ( I ′ ) ⊆ U ( g ( J ) , − m ) and g ( J ′ ) ⊆ U ( f ( I ′ ) , − m ) . (81) Proof
Compute α IJ and some number n > m such that 2 − n < α IJ .By Lemma 5.1 we can compute a rational interval I ′ ⊆ I ( I m = K in the lemma) such that by (70), α I ′ J > , f ( I ′ ) ∈ U ( g ( J ) , − n ) and π ( f | I ′ , g | J ) = 1 . (82)Again by Lemma 5.1 with ( f and g exchanged) first compute α I ′ J and some number n ′ > m such that2 − n ′ < α I ′ J , then compute some J ′ ⊆ J such that α I ′ J ′ > , g ( J ′ ) ∈ U ( f ( I ′ ) , − n ′ ) and π ( f | I ′ , g | J ′ ) = 1 . (83)(79) follows from (83), (80) is true by the construction and (81) follows from (82) and (83) by m < n and m < n ′ . ✷ Lemma 5.3
From functions f and g (as defined in (6) and (7)) we can compute sequences I ⊇ I , ⊇ I , . . . and J ⊇ J ⊇ J , . . . of rational closed intervals such that f ( T m I m ) = g ( T m J m ) . Proof
By Lemma 5.2 starting with I := J := [ −
1; 2] we can compute sequences I ⊇ I , ⊇ I , . . . and J ⊇ J ⊇ J , . . . of rational closed intervals such that for all m ≥ f ( I m ) ⊆ U ( g ( J m − ) , − ( m − ) and g ( J m ) ⊆ U ( f ( I m )) , − m ) . (84)The following equations follow from continuity of f and g . f ( \ m I m ) = \ m f ( I m ) = \ m U ( f ( I m ) , − m ) , (85) g ( \ m J m ) = \ m g ( J m ) = \ m U ( g ( J m ) , − m ) . (86)20e give elementary proofs. Since f ( T k I k ) ⊆ f ( I m ) ⊆ U ( f ( I m ) , − m ) for all m , f ( T k I k ) ⊆ T m f ( I m ) ⊆ T m U ( f ( I m ) , − m ).Suppose x ∈ T m f ( I m ). Then there is a sequence a , a , a , . . . such that for all m , a m ∈ I m and f ( a m ) = x . Since I is compact there is a subsequence a m , a m , a m , . . . converging to some a ∈ I .Since f is continuous, x = lim i f ( a m i ) = f (lim i a m i ) = f ( a ). Assume a I m k for some k . Then for some j > k , a m j I m k , hence a m j I m j , a contradiction. Therefore a ∈ I m k for all k , hence a ∈ T m I m and x = f ( a ) ∈ f ( T m I m ).Suppose x ∈ T m U ( f ( I m ) , − m ). Assume x f ( I k ) for some k . Since f ( I k ) is compact x U ( f ( I k ) , − j ) for some j > k . Therefore, x U ( f ( I j ) , − j ), hence x T m U ( f ( I m ) , − m ), a contra-diction. Therefore x ∈ f ( I k ) for all k , hence x ∈ T m f ( I m ).Therefore (85) is true. By symmetry also (86) is true. By (84–86), f ( T m I m ) = T m f ( I m ) = T m ≥ f ( I m ) ⊆ T m ≥ U ( g ( J m − ) , − ( m − ) = T m U ( g ( J m ) , − m ) = g ( T m J m ).Accordingly, g ( T m J m ) ⊆ f ( T m I m ). ✷ After these preparations Theorem 1.4 can be proved straightforwardly. For the canonical representa-tion (most conveniently the canonical multi-representation [11]) of functions h : ⊆ R → R the function f can be computed from φ and the function g can be computed from ψ . By the outer representation of theset I of the closed real intervals, a name of S is a sequence I ⊇ I ⊇ . . . of closed rational intervals suchthat S = T n I n . Therefore, by Lemma 5.3 the multi-function ( φ, ψ ) | ⇒ ( S φ , S ψ ) such that φ ( S φ ) = ψ ( S ψ )is computable. Thus we have proved Theorem 1.4.The main theorem from [10] follows straightforwardly from Theorem 1.4.21 orollary 5.4
1. If φ and ψ in Theorem 1.1 are computable then there are computable numbers a and b such that φ ( a ) ∈ range( ψ ) (hence x := φ ( a ) ∈ range( φ ) ∩ range( ψ ) ) and ψ ( b ) ∈ range( φ ) (hence y := ψ ( b ) ∈ range( φ ) ∩ range( ψ ) ).2. Restricted to the pairs ( φ, ψ ) which have a unique intersection point, the point x such that { x } =range( φ ) ∩ range( ψ ) can be computed from φ and ψ .3. Restricted to the pairs ( φ, ψ ) such that φ ( a ) = ψ ( b ) for a unique pair ( a, b ) ∈ [0; 1] the function ( φ, ψ ) ( a, b ) is computable. Proof
1. Suppose S φ has length >
0. Then a ∈ S φ for some a ∈ Q , which is a computable real number suchthat φ ( a ) ∈ ψ ( S β ) ⊆ range( ψ ).Suppose S φ has length 0, that is, S φ = { a } for some a . Since { a } = S φ = T n I n for a computablesequence I ⊇ I ⊇ . . . of rational intervals, a is a computable real number such that φ ( a ) ∈ ψ ( S β ) ⊆ range( ψ ).By symmetry there is a computable number b such that ψ ( b ) ∈ range( φ ).2. Notice that possibly S φ and S ψ have positive lengths. Suppose φ and ψ have a single intersectionpoint x . Then { x } = φ ( S φ ) = ψ ( S ψ ). For every m > ∅ = { x } \ B ( x, − m ) = T n ∈ N φ ( I n ) \ B ( x, − m ) = T n ∈ N ( φ ( I n ) \ B ( x, − m )) by (85). The countable intersection of closed subsets of the compact set [0; 1] is empty. Therefore finitely many suffice: there is some N such that ∅ = T n ≤ N ( φ ( I n ) \ B ( x, − m )) = T n ≤ N φ ( I n ) \ B ( x, − m ) = φ ( I N ) \ B ( x, − m ), hence φ ( I N ) ⊆ B ( x, − m ). Therefore, for every m > N ∈ N and some z ∈ Q such that x ∈ φ ( I N ) ⊆ B ( z, · − m ). Since ( φ, N ) φ ( I N ) iscomputable and the set ( K, z, j ) such that K is compact and K ⊆ B ( z, j ) is c.e. [11], from φ and thelist I , I , . . . we can compute a sequence of rational balls contracting to x . Therefore we can computethe single intersection point of φ and ψ .3. We can compute a sequence I ⊇ I ⊇ . . . converging to a . Therefore we can compute a . Corre-spondingly we can compute b . ✷ In Corollary 5.4.1 in general φ ( a ) = ψ ( b ). From φ and ψ we cannot compute some a such that φ ( a ) ∈ range( ψ ) or some x ∈ range( φ ) ∩ range( ψ ). We do not even know whether for the computablenumber a there is a computable number c such that φ ( a ) = ψ ( c ). In Corollary 5.4.2 from φ and ψ wecannot compute some a such that φ ( a ) = x . References [1] Vasco Brattka, Peter Hertling, and Klaus Weihrauch. A tutorial on computable analysis. In S. BarryCooper, Benedikt L¨owe, and Andrea Sorbi, editors,
New Computational Paradigms: Changing Con-ceptions of What is Computable , pages 425–491. Springer, New York, 2008.[2] Andrzej Grzegorczyk. Computable functionals.
Fundamenta Mathematicae , 42:168–202, 1955. computably enumerable or recursively enumerable FundamentaMathematicae , 44:61–71, 1957.[4] Boris Abramovich Kuˇsner.
Lectures on Constructive Mathematical Analysis , volume 60 of
Transla-tions of Mathematical Monographs . American Mathematical Society, Providence, 1984.[5] Boris Abramovich Kuˇsner. Markov’s constructive analysis; a participant’s view.
Theoretical Com-puter Science , 219:267–285, 1999.[6] Daniel Lacombe. Extension de la notion de fonction r´ecursive aux fonctions d’une ou plusieursvariables r´eelles I.
Comptes Rendus Acad´emie des Sciences Paris , 240:2478–2480, June 1955. Th´eoriedes fonctions.[7] S. N. Manukyan. O nekotorykh topologicheskikh osobennostyakh konstruktivnykh prostykh dug. (inRussian). In B.A. Kushner and A.A. Markov, editors,
Issledovaniya po teorii algorifmov i matem-aticheskoy logike , volume 2, pages 122–129. Vychislitel’ny Tsentr AN SSSR, Moscow, 1976.[8] Nazanin R. Tavana and Klaus Weihrauch. Turing machines on represented sets, a model of compu-tation for analysis.
Logical Methods in Computer Science , 7(2):2:19, 21, 2011.[9] Klaus Weihrauch.
Computable Analysis . Springer, Berlin, 2000.[10] Klaus Weihrauch. Computable planar curves intersect in a computable point.
Computability , 8(3,4):399–415, 2019.[11] Klaus Weihrauch and Tanja Grubba. Elementary computable topology.