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Axial Borel functions
Topology and its Applications 160 (2013) 2049–2052
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
Axial Borel functions Marcin Szyszkowski Gdansk University, Wita Stwosza 57, Gdansk, 80-952, Poland
a r t i c l e
i n f o
Article history: Received 30 May 2012 Received in revised form 17 August 2013 Accepted 21 August 2013
a b s t r a c t A function from the plane to the plane is axial if it does not change one coordinate. We show that every Borel permutation of the plane is a superposition of 11 Borel axial permutations. © 2013 Elsevier B.V. All rights reserved.
Keywords: Borel maps Borel sets Axial functions
Definition 1. Function f : X × Y → X × Y is axial if f (x, y) = x, g(x, y) for some g : X × Y → Y (f is vertical)
or f (x, y) = g(x, y), y for some g : X × Y → X (f is horizontal).
The consideration of axial functions dates back to Banach and Ulam (in [7]). We state some known facts. Theorem 2. ([3,4]) Every permutation of X × Y is a composition of four axial permutations. We will be interested only in functions on the plane R2 (for results for other sets X and Y see [3,4,6]). On the plane we define a subclass of axial functions – slides, f is a slide if f (x, y) = (x, y + g(x)) or f (x, y) = (x + g(y), y) for some g : R → R. Note that a slide is a permutation of the plane. It’s worth to mention a result from [1] which says that every permutation of R2 is a composition of 209(!) slides. In [2] Eggleston (answering a question of Ulam in [7]) proved that every homeomorphism of R2 is a pointwise limit of compositions of axial homeomorphisms, however, there is a homeomorphism that cannot be even approximated in supremum metric by finite compositions of axial homeomorphisms. He does state, however, the following E-mail address: [email protected]. 0166-8641/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.topol.2013.08.010
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M. Szyszkowski / Topology and its Applications 160 (2013) 2049–2052
Theorem 3. Let h: [0, 1]2 → [0, 1]2 be any homeomorphism of the square [0, 1]2 being identity on the boundary and let ε > 0. There is g – a composition of axial homeomorphisms such that ∀p∈[0,1]2 |h(p) − g(p)| < ε. When we take h to be a continuous function instead of homeomorphism then the above is not true – see [8]. The aim of this paper is to prove the following Theorem 4. Every Borel permutation f : R2 → R2 is a composition of eleven axial Borel permutations of R2 . This answers a question in [10] and [9]. Note that when we allow axial functions to be not permutations then the situation is easier. Theorem 5. ([9]) Any Borel function f : R2 → R2 is a composition of only three axial Borel functions from R2 to R2 . 1. Notation By C we denote standard ternary Cantor set i.e. C = {0, x1 x2 . . . ∈ [0, 1]: xi = 0, 2}, where 0, x1 x2 . . . is the ternary expansion of x ∈ [0, 1] (in case of ambiguous expansion we use the one without 1). We can partition C into continuum many ‘subcantors’ Ct by fixing the odd coordinates, Ct = Cy1 y2 ... = {0, x1 x2 . . . ∈ C: x2i−1 = yi }. The set of sequences (yi ) treated as numbers i.e. (yi ) → 0, y1 y2 . . . (in ternary expansion) form a Cantor set and can be labeled with real numbers in a Borel way i.e. there is a Borel function c : C → R with c−1 ({t}) = Ct . We may additionally assume that c is onto R and that C0 = {0, 0x2 0x4 . . . : x2i = 0, 2} (the sequence (yi ) = (0)∞ i=1 ). Note that all Ct are isometric that is ∀t ∃mt Ct −mt = C0 , where mt = min Ct , moreover function t → mt is Borel. The following facts are either well known or obvious Facts 6. (1) [5, Rem. 1, §1, Chp. 13] If f is a 1-1 Borel function then f −1 is also Borel. (2) (Borel isomorphism theorem) [5, Cor. 1, §1, Chp. 13] Any two Borel subsets of R or R2 of the same cardinality are Borel isomorphic. (3) For any Borel sets A, B ⊂ R with |A| = |B| and |R \ A| = |R \ B| there is a Borel permutation f : R → R with f (A) = B. (4) Composition of Borel functions is Borel. (5) Function f = (f1 , f2 ) : R2 → R2 , where f1 , f2 : R2 → R, is Borel if and only if both f1 and f2 are Borel. (6) If a function f is axial so is f −1 , if f is a composition of axial functions so is f −1 . Borel isomorphism (on R2 ) is a Borel function whose inverse is also Borel. Note that by Fact 6(1) every Borel permutation is a Borel isomorphism. 2. Proofs Lemma 7. There are three axial Borel isomorphisms F1 , F2 , F3 : R2 → R2 such that F3 ◦F2 ◦F1 (R2 \C×{0}) = C × {0} (thus F3 ◦ F2 ◦ F1 (C × {0}) = R2 \ C × {0}).
M. Szyszkowski / Topology and its Applications 160 (2013) 2049–2052
2051
Proof. We utilize partition of Cantor set C into sets Ct from previous section. Let
F1 (x, y) =
(x, y + c(x)) (x, y)
if x ∈ C, otherwise
(equivalently, we may write F1 (x, y) = (x, y + t) for x ∈ Ct ). F1 is a slide thus a bijection, it is Borel since the function c : C → R is Borel. We want F2 to satisfy F2 (Ct × {t}) = R × {t} \ Ct × {t} (thus F2 (R × {t} \ Ct × {t}) = Ct × {t}), this way we obtain F2 ◦ F1 (Ct × {0}) = R × {t} \ Ct × {t}. To define F2 we use three auxiliary functions g1 , g2 , g3 . We define g1 : R2 → R2 as g1 (x, t) = (x − mt , t) where mt = min Ct (as in previous section). Note that g1 (Ct × {t}) = C0 × {t} (all sets g1 (Ct × {t}) are “one over other”). Clearly g1 is a Borel isomorphism (as a slide and by borelity of assignment t → mt ). Let p : R → R be a Borel isomorphism such that p(C0 ) = R \ C0 (so p(R \ C0 ) = C0 ). We define g2 : R2 → R2 as g2 (x, y) = (p(x), y). Note that g2 ◦ g1 (Ct × {0}) = R × {t} \ C0 × {t}. Finally we ‘draw aside’ sets Ct i.e. g3 : R2 → R2 is the function g3 (x, t) = (x + mt , t) (so g3 = g1−1 ). Function F2 is defined as F2 = g3 ◦ g2 ◦ g1 : R2 → R2 . That is, F2 (x, t) = (p(x − mt ) + mt , t) is an axial (horizontal) Borel isomorphism. The last function F3 = F1−1 is vertical
F3 (x, y) =
(x, y − c(x)) (x, y)
(we may also write F3 (x, y) = (x, y − t) when x ∈ Ct ).
for x ∈ C, otherwise
2
Theorem 8. Let f : R2 → R2 be a Borel permutation satisfying f (C × {0}) = C × {0} (so f (R2 \ C × {0}) = R2 \ C × {0}) then f is a composition of eight axial Borel permutations of R2 . Proof. We will construct Fi : R2 → R2 for i = 1, . . . , 8 such that F8 ◦ · · · ◦ F1 = f . Functions F1 , F2 and F3 are from Lemma 7 above. We define F˜4 : C × {0} → C × {0} as follows F˜4 = F3 ◦ F2 ◦ F1 ◦ f ◦ (F3 ◦ F2 ◦ F1 |C×{0} )−1 . We verify that F˜4 is a permutation of C × {0}. Indeed, (F3 ◦ F2 ◦ F1 )−1 (C × {0}) = R2 \ C × {0} and f (R2 \ C × {0}) = R2 \ C × {0} (assumption on f ) and F3 ◦ F2 ◦ F1 (R2 \ C × {0}) = C × {0}. Since F1 , F2 , F3 , f are Borel bijections so is F˜4 . We extend F˜4 to F4 : R2 → R2 putting identity on R2 \ C × {0}. To finish we take the next three functions to be F5 = F3−1 , F6 = F2−1 and F7 = F3−1 so F7 ◦ F6 ◦ F5 = (F3 ◦ F2 ◦ F1 )−1 . We verify now that when (x, y) ∈ R2 \C ×{0} then F7 ◦· · ·◦F1 (x, y) = f (x, y). Indeed, F3 ◦F2 ◦F1 (x, y) ∈ C ×{0} (by definition of F1 , F2 , F3 ), F4 (F3 ◦F2 ◦F1 (x, y)) = F3 ◦F2 ◦F1 ◦f ◦(F3 ◦F2 ◦F1 )−1 (F3 ◦F2 ◦F1 (x, y)) = F3 ◦ F2 ◦ F1 ◦ f (x, y) (by definition of F4 ), and so F7 ◦ F6 ◦ F5 ◦ F3 ◦ F2 ◦ F1 ◦ f (x, y) = f (x, y). Similarly we verify that for (x, 0) ∈ C × {0} we have F7 ◦ · · · ◦ F1 (x, 0) = (x, 0). To obtain f (x, 0) on (x, 0) ∈ C × {0} we put the last function F8 : R2 → R2 defined as F8 (x, 0) = f (x, 0) for (x, 0) ∈ C × {0} and F8 (x, y) = (x, y) for other (x, y). 2 Remark 9. The function F8 is horizontal (we will use this fact in the proof of Theorem 4). Lemma 10. For every Borel permutation f : R2 → R2 there are four axial Borel permutations g3 , g2 , g1 , g0 : R2 → R2 such that g3 ◦ g2 ◦ g1 ◦ f ◦ g0 (C × {0}) = C × {0}.
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M. Szyszkowski / Topology and its Applications 160 (2013) 2049–2052
Proof. By perfect set we understand additionally a compact set. Since f |R×{0} is Borel there is a perfect set P ⊂ R × {0} such that f |P is continuous (so homeomorphism). The set f (P ) is a perfect set in R2 . At least one of projection of f (P ) onto x-axis or y-axis is of size continuum. If such is the projection onto x-axis we set g1 as identity on R2 , otherwise we put g1 (x, y) = (x + y, y) (so images of vertical lines are slant lines) and the projection onto x-axis Π1 (g1 ◦ f (P )) has cardinality continuum. By Kunugui–Novikov ˜ ⊂ g1 ◦ f (P ) such that B ˜ is the graph uniformization theorem [5, Thm. 4, §3, Chp. 14] there is a Borel set B 2 ˜ of a Borel function b : Π1 (g1 ◦ f (P )) → R i.e. B = {(x, b(x)): x ∈ Π1 (g1 ◦ f (P ))}. ˜ be a perfect set. Projection Π1 (B) onto x-axis (= R × {0}) is a perfect set too. We define Let B ⊂ B g2 : R2 → R2 by g2 (x, y) = (x, y − b(x)) for x ∈ Π1 (B) and g2 (x, y) = (x, y) for x ∈ / Π1 (B). Note that g2 (B) = Π1 (B).
6
f (P ) g1 ◦
@ B @
Π1 (B) = g2 (B)
-
There is g3 : R×{0} → R×{0} – a Borel permutation satisfying g3 (Π1 (B)) = C ×{0}, this way g3 ◦g2 (B) = C × {0}. Putting identity on R2 \ R × {0} we extend g3 to the Borel permutation on R2 . Finally we take function g0 : R × {0} → R × {0} as a Borel permutation with g0 (C × {0}) = (g1 ◦ f )−1 (B) ⊂ P ⊂ R × {0} and extend it to R2 using identity on R2 \ R × {0}. We obtain g3 ◦ g2 ◦ g1 ◦ f ◦ g0 (C × {0}) = g3 ◦ g2 ◦ g1 ◦ f ((g1 ◦ f )−1 (B)) = g3 ◦ g2 (B) = C × {0}. 2 Remark 11. The function g3 is horizontal (this fact will be used in the proof of Theorem 4). Proof of Theorem 4. Let f : R2 → R2 be a Borel permutation. Combining Lemma 10 and Theorem 8 (and using their notations) we obtain g3 ◦ g2 ◦ g1 ◦ f ◦ g0 = F8 ◦ · · · ◦ F1 thus f = (g3 ◦ g2 ◦ g1 )−1 ◦ F8 ◦ · · · ◦ F1 ◦ g0−1 . Since g3−1 and F8 are both horizontal (Remarks 9 and 11) we treat g3−1 ◦ F8 as one (horizontal) permutation and conclude the proof. 2 Although some axial functions in Theorem 4 are slides we cannot expect to obtain any Borel permutation as a composition of Borel slides. Since Borel slides preserve measure (see [1]) their composition is a measure preserving permutation. We do not know if any measure preserving Borel permutation is a composition of Borel slides or at least measure preserving axial Borel permutations. Acknowledgements The author thanks the referees for their remarks. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
M. Abert, T. Keleti, Shuffle the plane, Proc. Am. Math. Soc. 130 (2) (2002) 549–553. H.G. Eggleston, A property of plane homeomophisms, Fundam. Math. 42 (1955) 61–74. A. Ehrenfeucht, E. Grzegorek, On axial maps of direct products I, Colloq. Math. 32 (1974) 1–11. E. Grzegorek, On axial maps of direct products II, Colloq. Math. 34 (1976) 145–164. K. Kuratowski, A. Mostowski, Set Theory, PWN, Warszawa, 1976. K. Płotka, Composition of axial functions of products of finite sets, Colloq. Math. 107 (2007) 15–20. R. Mauldin (Ed.), The Scottish Book, Birkhäuser, Boston, 1981. M. Szyszkowski, Axial continuous functions, Topol. Appl. 157 (2010) 559–562. M. Szyszkowski, A note on axial functions on the plane, Tatra Mt. Math. Publ. 40 (2008) 59–62. S. Ulam, A Collection of Mathematical Problems, Interscience, New York, 1960.
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
Axial Borel functions Marcin Szyszkowski Gdansk University, Wita Stwosza 57, Gdansk, 80-952, Poland
a r t i c l e
i n f o
Article history: Received 30 May 2012 Received in revised form 17 August 2013 Accepted 21 August 2013
a b s t r a c t A function from the plane to the plane is axial if it does not change one coordinate. We show that every Borel permutation of the plane is a superposition of 11 Borel axial permutations. © 2013 Elsevier B.V. All rights reserved.
Keywords: Borel maps Borel sets Axial functions
Definition 1. Function f : X × Y → X × Y is axial if f (x, y) = x, g(x, y) for some g : X × Y → Y (f is vertical)
or f (x, y) = g(x, y), y for some g : X × Y → X (f is horizontal).
The consideration of axial functions dates back to Banach and Ulam (in [7]). We state some known facts. Theorem 2. ([3,4]) Every permutation of X × Y is a composition of four axial permutations. We will be interested only in functions on the plane R2 (for results for other sets X and Y see [3,4,6]). On the plane we define a subclass of axial functions – slides, f is a slide if f (x, y) = (x, y + g(x)) or f (x, y) = (x + g(y), y) for some g : R → R. Note that a slide is a permutation of the plane. It’s worth to mention a result from [1] which says that every permutation of R2 is a composition of 209(!) slides. In [2] Eggleston (answering a question of Ulam in [7]) proved that every homeomorphism of R2 is a pointwise limit of compositions of axial homeomorphisms, however, there is a homeomorphism that cannot be even approximated in supremum metric by finite compositions of axial homeomorphisms. He does state, however, the following E-mail address: [email protected]. 0166-8641/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.topol.2013.08.010
2050
M. Szyszkowski / Topology and its Applications 160 (2013) 2049–2052
Theorem 3. Let h: [0, 1]2 → [0, 1]2 be any homeomorphism of the square [0, 1]2 being identity on the boundary and let ε > 0. There is g – a composition of axial homeomorphisms such that ∀p∈[0,1]2 |h(p) − g(p)| < ε. When we take h to be a continuous function instead of homeomorphism then the above is not true – see [8]. The aim of this paper is to prove the following Theorem 4. Every Borel permutation f : R2 → R2 is a composition of eleven axial Borel permutations of R2 . This answers a question in [10] and [9]. Note that when we allow axial functions to be not permutations then the situation is easier. Theorem 5. ([9]) Any Borel function f : R2 → R2 is a composition of only three axial Borel functions from R2 to R2 . 1. Notation By C we denote standard ternary Cantor set i.e. C = {0, x1 x2 . . . ∈ [0, 1]: xi = 0, 2}, where 0, x1 x2 . . . is the ternary expansion of x ∈ [0, 1] (in case of ambiguous expansion we use the one without 1). We can partition C into continuum many ‘subcantors’ Ct by fixing the odd coordinates, Ct = Cy1 y2 ... = {0, x1 x2 . . . ∈ C: x2i−1 = yi }. The set of sequences (yi ) treated as numbers i.e. (yi ) → 0, y1 y2 . . . (in ternary expansion) form a Cantor set and can be labeled with real numbers in a Borel way i.e. there is a Borel function c : C → R with c−1 ({t}) = Ct . We may additionally assume that c is onto R and that C0 = {0, 0x2 0x4 . . . : x2i = 0, 2} (the sequence (yi ) = (0)∞ i=1 ). Note that all Ct are isometric that is ∀t ∃mt Ct −mt = C0 , where mt = min Ct , moreover function t → mt is Borel. The following facts are either well known or obvious Facts 6. (1) [5, Rem. 1, §1, Chp. 13] If f is a 1-1 Borel function then f −1 is also Borel. (2) (Borel isomorphism theorem) [5, Cor. 1, §1, Chp. 13] Any two Borel subsets of R or R2 of the same cardinality are Borel isomorphic. (3) For any Borel sets A, B ⊂ R with |A| = |B| and |R \ A| = |R \ B| there is a Borel permutation f : R → R with f (A) = B. (4) Composition of Borel functions is Borel. (5) Function f = (f1 , f2 ) : R2 → R2 , where f1 , f2 : R2 → R, is Borel if and only if both f1 and f2 are Borel. (6) If a function f is axial so is f −1 , if f is a composition of axial functions so is f −1 . Borel isomorphism (on R2 ) is a Borel function whose inverse is also Borel. Note that by Fact 6(1) every Borel permutation is a Borel isomorphism. 2. Proofs Lemma 7. There are three axial Borel isomorphisms F1 , F2 , F3 : R2 → R2 such that F3 ◦F2 ◦F1 (R2 \C×{0}) = C × {0} (thus F3 ◦ F2 ◦ F1 (C × {0}) = R2 \ C × {0}).
M. Szyszkowski / Topology and its Applications 160 (2013) 2049–2052
2051
Proof. We utilize partition of Cantor set C into sets Ct from previous section. Let
F1 (x, y) =
(x, y + c(x)) (x, y)
if x ∈ C, otherwise
(equivalently, we may write F1 (x, y) = (x, y + t) for x ∈ Ct ). F1 is a slide thus a bijection, it is Borel since the function c : C → R is Borel. We want F2 to satisfy F2 (Ct × {t}) = R × {t} \ Ct × {t} (thus F2 (R × {t} \ Ct × {t}) = Ct × {t}), this way we obtain F2 ◦ F1 (Ct × {0}) = R × {t} \ Ct × {t}. To define F2 we use three auxiliary functions g1 , g2 , g3 . We define g1 : R2 → R2 as g1 (x, t) = (x − mt , t) where mt = min Ct (as in previous section). Note that g1 (Ct × {t}) = C0 × {t} (all sets g1 (Ct × {t}) are “one over other”). Clearly g1 is a Borel isomorphism (as a slide and by borelity of assignment t → mt ). Let p : R → R be a Borel isomorphism such that p(C0 ) = R \ C0 (so p(R \ C0 ) = C0 ). We define g2 : R2 → R2 as g2 (x, y) = (p(x), y). Note that g2 ◦ g1 (Ct × {0}) = R × {t} \ C0 × {t}. Finally we ‘draw aside’ sets Ct i.e. g3 : R2 → R2 is the function g3 (x, t) = (x + mt , t) (so g3 = g1−1 ). Function F2 is defined as F2 = g3 ◦ g2 ◦ g1 : R2 → R2 . That is, F2 (x, t) = (p(x − mt ) + mt , t) is an axial (horizontal) Borel isomorphism. The last function F3 = F1−1 is vertical
F3 (x, y) =
(x, y − c(x)) (x, y)
(we may also write F3 (x, y) = (x, y − t) when x ∈ Ct ).
for x ∈ C, otherwise
2
Theorem 8. Let f : R2 → R2 be a Borel permutation satisfying f (C × {0}) = C × {0} (so f (R2 \ C × {0}) = R2 \ C × {0}) then f is a composition of eight axial Borel permutations of R2 . Proof. We will construct Fi : R2 → R2 for i = 1, . . . , 8 such that F8 ◦ · · · ◦ F1 = f . Functions F1 , F2 and F3 are from Lemma 7 above. We define F˜4 : C × {0} → C × {0} as follows F˜4 = F3 ◦ F2 ◦ F1 ◦ f ◦ (F3 ◦ F2 ◦ F1 |C×{0} )−1 . We verify that F˜4 is a permutation of C × {0}. Indeed, (F3 ◦ F2 ◦ F1 )−1 (C × {0}) = R2 \ C × {0} and f (R2 \ C × {0}) = R2 \ C × {0} (assumption on f ) and F3 ◦ F2 ◦ F1 (R2 \ C × {0}) = C × {0}. Since F1 , F2 , F3 , f are Borel bijections so is F˜4 . We extend F˜4 to F4 : R2 → R2 putting identity on R2 \ C × {0}. To finish we take the next three functions to be F5 = F3−1 , F6 = F2−1 and F7 = F3−1 so F7 ◦ F6 ◦ F5 = (F3 ◦ F2 ◦ F1 )−1 . We verify now that when (x, y) ∈ R2 \C ×{0} then F7 ◦· · ·◦F1 (x, y) = f (x, y). Indeed, F3 ◦F2 ◦F1 (x, y) ∈ C ×{0} (by definition of F1 , F2 , F3 ), F4 (F3 ◦F2 ◦F1 (x, y)) = F3 ◦F2 ◦F1 ◦f ◦(F3 ◦F2 ◦F1 )−1 (F3 ◦F2 ◦F1 (x, y)) = F3 ◦ F2 ◦ F1 ◦ f (x, y) (by definition of F4 ), and so F7 ◦ F6 ◦ F5 ◦ F3 ◦ F2 ◦ F1 ◦ f (x, y) = f (x, y). Similarly we verify that for (x, 0) ∈ C × {0} we have F7 ◦ · · · ◦ F1 (x, 0) = (x, 0). To obtain f (x, 0) on (x, 0) ∈ C × {0} we put the last function F8 : R2 → R2 defined as F8 (x, 0) = f (x, 0) for (x, 0) ∈ C × {0} and F8 (x, y) = (x, y) for other (x, y). 2 Remark 9. The function F8 is horizontal (we will use this fact in the proof of Theorem 4). Lemma 10. For every Borel permutation f : R2 → R2 there are four axial Borel permutations g3 , g2 , g1 , g0 : R2 → R2 such that g3 ◦ g2 ◦ g1 ◦ f ◦ g0 (C × {0}) = C × {0}.
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M. Szyszkowski / Topology and its Applications 160 (2013) 2049–2052
Proof. By perfect set we understand additionally a compact set. Since f |R×{0} is Borel there is a perfect set P ⊂ R × {0} such that f |P is continuous (so homeomorphism). The set f (P ) is a perfect set in R2 . At least one of projection of f (P ) onto x-axis or y-axis is of size continuum. If such is the projection onto x-axis we set g1 as identity on R2 , otherwise we put g1 (x, y) = (x + y, y) (so images of vertical lines are slant lines) and the projection onto x-axis Π1 (g1 ◦ f (P )) has cardinality continuum. By Kunugui–Novikov ˜ ⊂ g1 ◦ f (P ) such that B ˜ is the graph uniformization theorem [5, Thm. 4, §3, Chp. 14] there is a Borel set B 2 ˜ of a Borel function b : Π1 (g1 ◦ f (P )) → R i.e. B = {(x, b(x)): x ∈ Π1 (g1 ◦ f (P ))}. ˜ be a perfect set. Projection Π1 (B) onto x-axis (= R × {0}) is a perfect set too. We define Let B ⊂ B g2 : R2 → R2 by g2 (x, y) = (x, y − b(x)) for x ∈ Π1 (B) and g2 (x, y) = (x, y) for x ∈ / Π1 (B). Note that g2 (B) = Π1 (B).
6
f (P ) g1 ◦
@ B @
Π1 (B) = g2 (B)
-
There is g3 : R×{0} → R×{0} – a Borel permutation satisfying g3 (Π1 (B)) = C ×{0}, this way g3 ◦g2 (B) = C × {0}. Putting identity on R2 \ R × {0} we extend g3 to the Borel permutation on R2 . Finally we take function g0 : R × {0} → R × {0} as a Borel permutation with g0 (C × {0}) = (g1 ◦ f )−1 (B) ⊂ P ⊂ R × {0} and extend it to R2 using identity on R2 \ R × {0}. We obtain g3 ◦ g2 ◦ g1 ◦ f ◦ g0 (C × {0}) = g3 ◦ g2 ◦ g1 ◦ f ((g1 ◦ f )−1 (B)) = g3 ◦ g2 (B) = C × {0}. 2 Remark 11. The function g3 is horizontal (this fact will be used in the proof of Theorem 4). Proof of Theorem 4. Let f : R2 → R2 be a Borel permutation. Combining Lemma 10 and Theorem 8 (and using their notations) we obtain g3 ◦ g2 ◦ g1 ◦ f ◦ g0 = F8 ◦ · · · ◦ F1 thus f = (g3 ◦ g2 ◦ g1 )−1 ◦ F8 ◦ · · · ◦ F1 ◦ g0−1 . Since g3−1 and F8 are both horizontal (Remarks 9 and 11) we treat g3−1 ◦ F8 as one (horizontal) permutation and conclude the proof. 2 Although some axial functions in Theorem 4 are slides we cannot expect to obtain any Borel permutation as a composition of Borel slides. Since Borel slides preserve measure (see [1]) their composition is a measure preserving permutation. We do not know if any measure preserving Borel permutation is a composition of Borel slides or at least measure preserving axial Borel permutations. Acknowledgements The author thanks the referees for their remarks. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
M. Abert, T. Keleti, Shuffle the plane, Proc. Am. Math. Soc. 130 (2) (2002) 549–553. H.G. Eggleston, A property of plane homeomophisms, Fundam. Math. 42 (1955) 61–74. A. Ehrenfeucht, E. Grzegorek, On axial maps of direct products I, Colloq. Math. 32 (1974) 1–11. E. Grzegorek, On axial maps of direct products II, Colloq. Math. 34 (1976) 145–164. K. Kuratowski, A. Mostowski, Set Theory, PWN, Warszawa, 1976. K. Płotka, Composition of axial functions of products of finite sets, Colloq. Math. 107 (2007) 15–20. R. Mauldin (Ed.), The Scottish Book, Birkhäuser, Boston, 1981. M. Szyszkowski, Axial continuous functions, Topol. Appl. 157 (2010) 559–562. M. Szyszkowski, A note on axial functions on the plane, Tatra Mt. Math. Publ. 40 (2008) 59–62. S. Ulam, A Collection of Mathematical Problems, Interscience, New York, 1960.