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Strange Subsets of the Euclidean Plane
Chapter 6 Strange su bsets of the Euclidean plane
We begin this chapter with consideration of non Lebesgue-measurable subsets of the euclidean plane R 2, which have small linear sections. We will investigate these sets from the point of view of measurability with respect to appropriate invariant extensions of the Lebesgue measure on R 2. Our interest in such subsets of the plane is inspired by a classical Sierpifiski partition of R 2 whose properties were thoroughly discussed in Chapter 4. In the literature there are many examples of paradoxical subsets of a finite-dimensional euclidean space, having small sections by certain affine hyperplanes of this space. One of the earliest examples is due again to Sierpifiski who constructed a function
f :R~R such that its graph is a A2-thick (or A2-massive) subset of the plane R 2. This construction was presented in Chapter 2. Here A2 denotes, as usual, the standard two-dimensional Lebesgue measure on R 2, and we recall that a subset X of R 2 is A2-thick (or A2-massive) in R 2 if the inner A2-measure of the set R 2 \ X is equal to zero (see the corresponding definition in [62]). In particular, the A2-thickness of the graph F / o f f implies that F / i s nonmeasurable with respect to A2 and: therefore, f is not measurable in the Lebesgue sense (see Theorem 4 from Chapter 2). At the same time, any straight line in R 2 parallel to the line {0} • R meets F / at exactly one point. Another interesting example was given by Mazurkiewicz who constructed a subset Y of R 2 having the property that for each straight line I in R 2, the set 1 N Y consists of exactly two points. In the sequel, any such subset of the plane will be called a Mazurkiewicz set. 102
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The descriptive structure of a Mazurkiewicz set turned out to be rather complicated. In general, one cannot assert that a Mazurkiewicz set is necessarily nonmeasurable with respect to &2. Indeed, there are Mazurkiewicz subsets of the plane which have &2-measure zero (see Exercise 1 of this chapter). Moreover, by using the classical Fubini theorem, one can easily conclude that if a Mazurkiewicz set is )~2-measurable, then it must be of ~2-measure zero. Slightly changing the argument of Mazurkiewicz, we shall demonstrate below the existence of a Mazurkiewicz set which is )~2-thick (consequently, it is nonmeasurable with respect to &2). Actually, the same argument works not only for the plane R 2 and the measure &2 on it, but also for the euclidean n-dimensional space R n, where n __ 2, and for the standard n-dimensional Lebesgue measure &n on this space. In fact, various analogous constructions of sets in R n which have small sections but are large in some sense can be carried out by utilizing the method of Mazurkiewicz (compare Exercise 3 of this chapter). T h e o r e m 1. Let n > 2. There exists a subset X of R n such that: 1) each affine hyperplane in R n intersects X in exactly n points; consequently, X is a set of points in general position; 2) X is thick with respect to )~n. P r o o f . Let (~ be the least ordinal number of cardinality continuum (denoted by c), let {L~ 9~ < c~} be the family of all affine hyperplanes in R n and let {B~ 9~ < a} be the family of all those Borel subsets of R n which have strictly positive An-measure. We are going to construct, using the transfinite recursion method, a family {X~ "~ < a} of subsets of R n satisfying the following conditions" (1) if r < ~ < c~, then Xr C X~ (in other words, this family is increasing by inclusion); (2) card(X~) <_ card(~) + w for any ordinal ~ < c~; (3) for each ~ < c~, the set X~ is a set of points in general position in the space R n ; (4) if ~ < a, then card(X~ N L~) - n; (5) if ~ < c~, then X~ N B~ % 0. Suppose that, for an ordinal ~ < c~, the partial family of sets <
has already been constructed. Consider the set Bf. Let B~ be a subset of B~ of cardinality continuum, whose all points are in general position.
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Note that the existence of such a subset of B~ can easily be established by utilizing Lemma 1 of this chapter, which is presented below. Further, let us denote
Yr = u { x r
r < ,1}.
Evidently, Y~ is a set of points in general position and
card(Y~) < card(~) + w. Consequently, we may write
card(Y~
n
L~) < n.
Now, by using ordinary recursion, it is not hard to define a finite subset Z~ of L~ such that: (i) card((Y~ U Z~) n L~) = n; (ii) Y~ U Z~ is a set of points in general position. Note that in order to define Z~, it suffices to apply a simple geometric fact stating that no affine linear manifold L in R n can be covered by a family of affine linear manifolds whose cardinality is strictly less than c and whose members all have dimension strictly less than dim(L) (compare Lemma 1 below in which a more general result is formulated). Further, since
card(Y~ U Z~) < c,
card(B~) - c
and B~ is a set of points in general position, there exists a point x~ E B~ for which the union {x~ } U Y~ U Z~ is also a set of points in general position. So, we may put
x~ = {x~} u Y~ u z~. Proceeding in this way, we are able to construct the required family of sets
{ x ~ : ,1 < ~}. Finally, we define
x = u { x ~ : ~ < o~}. It can easily be verified, in view of our construction, that the set X satisfies relations 1) and 2) of the theorem. This finishes the proof. In particular, for n = 2, the theorem proved above yields a Mazurkiewicz subset of the euclidean plane R 2, which is thick with respect to the Lebesgue
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measure A2. Hence, this subset turns out to be nonmeasurable with respect to ~2. Now, according to the standard construction of extending the Lebesgue measure (see, for instance, Exercise 10 of Chapter 2), any Mazurkiewicz set can be included in the domain of a translation-invariant extension of ~2, and we can assert that any Mazurkiewicz set necessarily becomes a set of measure zero with respect to such an extension (compare Exercise 4 of this chapter). The same result remains valid for those extensions of ~2 which are invariant under the group of all isometric transformations (motions) of R 2. Actually, we are interested in subsets of R 2 having small linear sections in connection with the following natural question: how small are these subsets from the point of view of invariant extensions of the Lebesgue measure We are going to demonstrate in our further considerations that there are invariant extensions of ~2, concentrated on sets with small linear sections in all directions. In fact, a more general result will be obtained stating t h a t the corresponding sets can have small sections by all analytic curves lying in R 2. Note t h a t our further argument is applicable to the space R n, where n >_ 2, and to the s t a n d a r d Lebesgue measure An on this space. In the sequel, we need the following simple auxiliary statement which is useful in many analogous situations. L e m m a 1. Let Z be a An-measurable subset of the Euclidean space R n with An(Z) > 0 and let {Mi : i E I} be a family of analytic manifolds in R n, such that: 1) card(I) is strictly less than c; 2) for each index i E I, the dimension of Mi is strictly less than n. Then the relation
Z\u{M
: i E I } #O
is satisfied. In particular, Lemma 1 states that the space R ~ cannot be covered by a small (in the sense of cardinality) family of analytic submanifolds of R n whose dimensions are strictly less than n. The proof of this lemma is not difficult and can be carried out by induction on n. Here the classical Fubini theorem plays an essential role. We leave the corresponding purely technical details to the reader.
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Some applications of Lemma 1 to certain questions of the geometry of euclidean spaces may be found in [87]. Note that the lemma does not hold true for topological manifolds in R n. For example, there are models of set theory in which the plane R 2 can be covered by a family of Jordan curves, whose cardinality is strictly less than c. (Here a Jordan curve is any homeomorphic image of the unit circumference Sl C R2.) Starting with the above-mentioned lemma, we are able to establish the following result. T h e o r e m 2. Let G be the group of all analytic diffeomorphisms of the space R n (n > 1) onto itself. Then there exists a subset X of R n such that: 1) X is almost G-invariant; in other words, we have (Vg E G ) ( c a r d ( g ( X ) A X ) < c),
where the symbol A denotes the operation of symmetric difference of sets; 2) card(X) = c and X is )~n-thick in Rn; 3) for any analytic manifold M in R n with dim(M) < n, we have card(M 0 X ) < c.
P r o o f . We use the method developed in [75], [82] and [100]. Let a denote the least ordinal number of cardinality continuum. Since the equality card(G) = c holds, we may write a=u{a
:
where {G~ 9 ~
card(G~) <_ card(~) + w; (b) the family {G~ : ~ < a} is increasing by inclusion. Also, we introduce the family {Y~ : ~ < a} consisting of all those Borel subsets of R n which are of strictly positive An-measure. Finally, we denote by {M~ : ~ < a} the family of all analytic manifolds in R n whose dimensions are strictly less than n.
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Let us now define, by the method of transfinite recursion, an injective c~-sequence {x( : ( < c~} of points in R n, satisfying the following relations: (i) for any ( < c~, the point x( belongs to Y(; (ii) for any ~ < c~, we have
9~ r G~((u{Mr
: ~ < ( } ) u {~r : ( < (}).
Note that Lemma 1 guarantees, at each ~-step of our recursion, the existence of a point x~. So the recursion can be continued up to c~. Proceeding in this way, we are able to construct the required c~-sequence of points {x~ : ~ < c~}. Now, putting
x = {ar
: ~ < ~},
we can easily check that the set X is the desired one. (All details are left to the reader.) This finishes the proof of Theorem 2. The next statement is a trivial consequence of the theorem just proved. T h e o r e m 3. Let G denote the group of all isometric transformations (motions) of the space R n. Then there exists a subset X of R n such that: 1) X is almost G-invariant; 2) card(X) = c and X is An-thick in Rn; 3) for each analytic manifold M in R n with dim(M) < n, we have
card(X n M) < c. In particular, for any affine hyperplane L in R n, the inequality card(X n L) < c is true. It immediately follows from Theorem 3 that if the Continuum Hypothesis holds, then, for each analytic manifold M in R n with dim(M) < n, we have card(X n M) < w, where X is the set from Theorem 3. Let us point out one straightforward application of Theorem 3.
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Various statements related to the classical Fubini theorem lead to the following intriguing question: does there exist a measure #n on R n extending the Lebesgue measure An, invariant under the group of all isometric transformations of R n and concentrated on some subset X of R n with small sections by all hyperplanes? Here the smallness of sections of X means that, for any hyperplane L in R n, the cardinality of L N X is strictly less than the cardinality of the continuum. Theorem 3 immediately yields a positive answer to this question. Indeed, let G be the group of all isometric transformations of R n and consider the G-invariant a-ideal of subsets of R n, generated by the set R n \ X, where X is the set from Theorem 3. We denote this a-ideal by 3". Then, for any set Z E if, the inner An-measure of Z is equal to zero (since X is almost Ginvariant and An-thick in Rn). Taking this fact into account and applying the standard methods of extending invariant measures (see, for instance, [82], [100], [212] and [214]), we infer that there exists a measure #n on R n satisfying the relations: 1) ~Un is complete and extends )~n; 2) #n is invariant under the group G; 3) ff C dom(#n); 4) for each set Z E ,7, we have # n ( Z ) - O . Relation 4) implies at once that
u . ( R n \ X ) =0. In other words, our measure/~n is concentrated on the set X. At the same time, we know that X has small sections by all hyperplanes in R n and, moreover, by all analytic manifolds in R n whose dimensions are strictly less than n. In addition to this, the measure #n being complete and metrically transitive has the so-called uniqueness property (compare Exercise 9 from Chapter 9). In particular, for n - 2, we obtain that there exists a complete measure /z2 on the euclidean plane R 2, such that: (1) #2 is an extension of the Lebesgue measure A2; (2) #2 is invariant under the group of all isometric transformations of R2 9 (3) #2 is concentrated on some subset X of R 2 having the property that all linear sections of X are small; more precisely, the cardinality of each linear section of X is strictly less than c.
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Note t h a t the question concerning the existence of a measure #2 on the plane, satisfying conditions (1) - (3), was formulated by R.D.Mabry (personal communication). Moreover, we see that the above-mentioned support X of #2 has a much stronger property: for every analytic curve I in R 2, the cardinality of l N X is strictly less than c. Let us point out that in [96] an analogous question was considered for extensions of A2 which are invariant under the group of all translations of R 2. Namely, it was demonstrated in [96] t h a t there exists a translation-invariant extension of A2 concentrated on a subset of R 2 whose all linear sections are at most countable. R e m a r k 1. If the Continuum Hypothesis holds, then we can also conclude that the above-mentioned subset X of R 2 has the property t h a t all its linear sections are at most countable. Obviously, the same is true for the sections of X by analytic curves lying in R 2. Thus, in this case, the measure #2 is concentrated on a set with countable linear sections. In this context, the following question seems to be natural: does there exist a measure v2 on the plane, extending A2, invariant under the group of all isometric transformations of R 2 and concentrated on a set with finite linear sections? It turns out that the answer to this question is negative. The proof of the corresponding result (and a more general statement) can be found in [94] (see also Exercise 10 of this chapter). R e m a r k 2. By assuming some additional set-theoretical axioms, it is possible to obtain a much stronger result than Theorem 2. Namely, let us suppose that, for any cardinal ~ < c, the space R n (n _ 1) cannot be covered by a ~-sequence of An-measure zero sets. In fact, it suffices to suppose the validity of this assumption only for n = 1, which means that the real line R cannot be covered by a family of Lebesgue measure zero sets, whose cardinality is strictly less than c. For instance, this hypothesis follows directly from Martin's Axiom (see [64] or Appendix 1). We say that a group G of transformations of R n is admissible if each element g from G preserves the a-ideal of all An-measure zero subsets of R n. There are many natural examples of admissible groups. For instance, if G coincides with the group of all affine transformations of R n, then G is admissible. If G coincides with the group of all diffeomorphisms of R n, then G is admissible, too. Evidently, the cardinality of these two groups is equal to c. It is not hard to check that there are also admissible groups G with card(G) = 2 c. Let us fix an admissible group G with card(G) = c. Assuming the above-mentioned hypothesis and applying an argument similar to the proof
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of Theorem 2, we easily obtain the following statement" if n _ 1, then there exists a subset X of R ~ with card(X) = c, such that" 1) X is almost G-invariant; 2) X is ~n-thick in Rn; 3) for any An-measure zero subset Z of R n, we have
card(Z N X) < c. Condition 3) shows, in particular, that the above-mentioned set X is a generalized Sierpifiski subset of R n. (For the definition and various properties of Sierpifiski sets, see [155] or [165] where the dual objects to the Sierpifiski sets - the so-called Luzin sets - are discussed as well; the corresponding material about these sets is also presented in Chapter 5 of this book.) Now, we wish to return to a problem of Luzin posed by him many years ago (see [195]). We have already been slightly in touch with this problem in Chapter 4 where its close connection with a Sierpifiski partition of the plane R 2 was briefly discussed. Here we are going to consider this question in more details. First of all, let us recall the formulation of this beautiful problem. Namely, Luzin asked whether there exists a function r
R--+ R
such that the whole plane R 2 can be covered by countably many isometric copies of the graph of r We have already mentioned that, under C H , this problem admits a positive solution even in a much stronger form (compare statement (iii) and Exercise 4 of Chapter 4). Actually, a positive answer to the Luzin question follows from the existence of a Sierpifiski type partition for the euclidean plane. Here we are going to present the full solution of this problem, within the theory Z F C . Recall that the final result was obtained by Davies. (See his two works [31] and [32] devoted to the Luzin problem.) In the sequel, we need some auxiliary notions and facts about special subsets of R 2. Let X be a subset of R 2 and let p be a straight line in R 2. We shall say that X is uniform with respect to p if, for each line p~ parallel to p, we have
card(p'n X) < 1.
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Obviously, for a set X C R 2 and for any two parallel straight lines p and q in R 2, the following assertions are equivalent: 1) X is uniform with respect to p; 2) X is uniform with respect to q. Further, we shall say that a set X C R 2 is uniform if there exists at least one straight line p in R 2 such that X is uniform with respect to p. It is not hard to observe that the next two assertions are also equivalent: (1) there exists a function r : R -+ R such that the plane R 2 can be covered by countably many isometric copies of the graph Fr of r (2) the plane R 2 can be covered by countably many uniform subsets of R 2" Therefore, the Luzin problem is reduced to the problem of the existence of a countable family of uniform sets, whose union is identical with R 2. In our further considerations, we will deal with the latter problem. Let p be an arbitrary straight line in R 2 and let z be a point of R 2. W e denote by p(z) the unique line in R 2 which is parallel to p and contains z. A countable family P = (Pk)k
card([Z]) < card(Z) + w. In particular, i/ Z is infinite, then card([Z]) - card(Z). An easy proof of this statement is left to the reader. L e m m a 3. Let P - (Pk)k
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set can be well-ordered by some relation ~_ in such a way that, for any point z E Z, the set o.f natural numbers {k < w
9the line pk(z) c o n t a i n s at least one point z ~ -~ z}
is always finite.
P r o o f . We use the method of transfinite induction (on c a r d ( Z ) ) . The case when c a r d ( Z ) - w is trivial: here it suffices to equip Z with any ordering isomorphic to the canonical well-ordering of w. Suppose now that the assertion of the lemma has already been established for all infinite Pclosed sets Z with c a r d ( Z ) < a, where a is some uncountable cardinal. Take any P-closed set Z such that c a r d ( Z ) - a. Of course, we may identify g with the least ordinal of the same cardinality. Let
z-
{z~-~ <
~}
be an injective enumeration of all elements of Z and, for each ordinal ~ < g, let
z~-
{zi2 6 r <
~}.
Further, for any point z E Z, we put ~(z)-inf{~
9~ > w ,
zE[Z~]}.
We shall say that ~(z) is the index of z (in Z). Let us mention that this notion is well defined because the equality Z=U{[Z~]
9w < ~ < ~ }
holds true in view of the P-closedness of Z. By L e m m a 2, if w _< ~ < a, then card([Z~]) - card(~) < and, taking into account the inductive assumption, we can equip the set [Z~] with a well-ordering _~ in such a way that for every point z E [Z~], the set {k < w
9pk(z) contains at least one point z ~ - ~ z}
is necessarily finite. Now, we are able to introduce a suitable well-ordering for the set Z. Namely, if z E Z and z ~ E Z are two distinct points, then we put z ~ -~ z if and only if
(~(z') -~ ~(z)) v ( ~ < ~)(~ - ~(z) - ~(z') ~ z' -~ z).
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Let us check t h a t our relation -~ possesses the required properties. First of all, by the definition of -~, it is not hard to see t h a t the corresponding relation _ is a well-ordering of Z. Let now z E Z and ~ - ~(z). We must verify that the set of natural numbers {k < w
9pk(z) contains at least one p o i n t z ~ -~ z}
is finite. Suppose to the contrary that the above-mentioned set is infinite. T h e n there exist pairwise distinct lines ql E
P, q2 E P , ..., qm E P , ...
such that, for some points Zl E Z, z 2 C Z,
..., Zm E Z ,
...,
the relations Zl E ql(z), z2 E q2(z), ..., Zm E a m ( z ) ,
...,
Zl "~ Z, Z2 ~ Z, ...~ Zm "~ Z~ ...
are fulfilled. Observe now that, according to the inductive assumption for the well-ordering ~ , the equalities
can be satisfied only for finitely many natural numbers m. This immediately implies that there are at least two distinct points from {Zl,Z2,...,Zm,...} whose indices are strictly less than ~. W i t h o u t loss of generality, we may assume that ~(Zl) < ~, ~(Z2) "~ ~, Zl r Z2Let us denote = m a x ( ~ ( z l ) , ~(z2)).
Then, obviously, we get
{z ,z2}
c
[z,]
Furthermore, it is evident that the straight lines q l ( z l ) and q2(z2) meet each other at the point z. Therefore, we must have z E [Z~], from which it follows ~(z) <_ rI - m a x ( ~ ( z l ) , ~ ( z 2 ) ) < ~ - ~(z),
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and the last relation yields a contradiction. The contradiction obtained completes the proof of L e m m a 3. 4. Let P = ( P k ) k K w be an admissible family of straight lines in R 2. Then there exists a family (Zk)k<~ of subsets o ] R 2, satisfying the following relations: 1) each set Zk (k < w) is uniform with respect to the line Pk; Lemma
2) u { z k
. k < ~} - R 2
P r o o f . Clearly, the plane R 2 is a P-closed set. Therefore, by virtue of the preceding lemma, R 2 can be endowed with a well-ordering -4 such that, for each point z E R 2, the set {k < w
9pk(z) contains at least one point z ~ -~ z}
is finite. Keeping in mind this circumstance, we may denote by k(z) the smallest natural number k for which the straight line pk(z) contains no point Z I --~Z.
Now, for any natural number k, we put zk -
{z e R2
9k ( z ) - k } .
Let us check that the sets Zk (k < w) are the required ones. First, it is evident that
u { z k - k < ~ } - R 2.
Therefore, it remains to demonstrate that every set Zk is uniform with respect to the line Pk. To see this, suppose otherwise and pick any two distinct points z E Zk and z' E Zk for which pk(z) = pk(z'). Since the disjunction z -,~ z ~ V z t -~ z trivially holds, we may assume without loss of generality that z ~ -~ z. Under this assumption, we get z' e p k ( z ' ) - p k ( z ) - pk(z)(Z),
z' -~ z,
which contradicts the definition of k(z). The contradiction obtained ends the proof of the lemma. As mentioned earlier, Lemma 4 yields a positive solution of the Luzin problem. So, we can formulate the following statement first obtained by Davies [31].
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Theorem
4. There exist a function
r
R-~R
and a countable family (gn)n<~ of motions of the euclidean plane R 2 , such that U{gn(F~) " n < w} - R 2, where F, denotes the graph of r Thus, we conclude that the problem of Luzin is solvable within the theory Z F C . Some interesting results closely related to this problem are indicated in exercises below (see also Chapter 14). A generalization of Theorem 4 for euclidean spaces of higher dimension can be found in [32].
EXERCISES 1. Show that there is a nowhere dense set Z C satisfying the relation card(Z N l) - c
R 2
of A2-measure zero,
for all straight lines 1 C R 2. Starting with this fact and applying the method of transfinite induction, demonstrate that there exists a nowhere dense Mazurkiewicz set of $2measure zero. 2. For any straight line l in the plane R 2, let ~(1) be a cardinal number such that 2 <_ n(1) < w. Prove that there is a subset X of R 2 satisfying the relation card(l N X ) - ~(1) for each straight line l in R 2. This beautiful result is due to Sierpifiski and it generalizes some theorems of Mazurkiewicz and Bagemihl. Prove, in the theory ZF, that there exists a set Y C R 2 satisfying the relations: (a) c a r d ( Y ) = c;
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(in E) if, for each point z E Z, we have
zr
\ (z}).
For example, the set of all vertices of a convex polygon in R 2 is convexly independent. By using the Ramsey combinatorial theorem (see [179]), prove that for each natural number n >_ 2, there exists a natural number N(n) having the following property: Any set X C R 2 of points in general position, with card(X) > N(n), contains a convexly independent subset whose eardinality is equal to n. This beautiful result is due to ErdSs and Szekeres [44]. They also conjectured that the smallest possible value of N(n) is 2 n-2 + 1. So far, there is no significant progress in proving (or disproving) their conjecture. By using an infinite version of the Ramsey theorem [179] (more precisely, by using its countable version), demonstrate that every infinite set X C R 2 of points in general position contains an infinite convexly independent subset. 12. By applying the method of transfinite induction, show that there exists a set X C R 2 having the following properties: (~) ca,'d(X) - c; (b) X is a set of points in general position; (c) no subset of X of cardinality e is convexly independent. For this purpose, denoting by a the least ordinal of cardinality e, take an injective family {L~ 9~ < a} consisting of all convex curves in R 2 and construct by transfinite recursion a subset of R 2 which almost avoids the members L~ of this family (compare the classical constructions of Sierpifiski and Luzin sets presented in Chapter 5). Finally, conclude that if the Continuum Hypothesis holds, then there exists a set in R 2 of cardinality e, whose points are in general position and which contains no uncountable convexly independent subset. 13. Suppose that the cardinal c is real-valued measurable. Starting with the result of Kunen (see Chapter 4), demonstrate that there exists a family {Li " i C I} of Jordan curves in R 2, satisfying the following relations: < r
(b) U { L i ' i e I } - R
2.
14. Let E be a set and let {Xi " i C I} be an injective family of nonempty subsets of E. We shall say that this family is a homogeneous
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119
covering of E if there exists a cardinal number a such that card({i E I " x E X i } ) =
for all elements x E E. In this case, we also say that a is the index of homogeneity of { X i " i E I } and that { X i 9 i E I } is a a-homogeneous covering of E. For example, if ~ - 1, then the a-homogeneity of {Xi 9i EI ) simply means that {X~ 9i EI} is a partition of E. Show t h a t there exists no partition of the plane R 2 consisting of Jordan curves. On the other hand, show within the theory Z F that, for every natural number k _> 1, there exists a 2k-homogeneous covering of R 2 all whose elements are circumferences congruent to the unit circumference Sl C R 2. Moreover, for every natural number k >_ 2, construct a k-homogeneous covering of R 2 consisting of pairwise congruent circumferences. (Use the m e t h o d of transfinite recursion.) Generalize these results to the case of the euclidean space R n where n _> 2. Namely, prove that: (a) there exists no partition of R n whose all elements are homeomorphic images of the unit sphere Sn-1 C Rn; (b) for any natural number k >_ n, there exists a k-homogeneous covering of R n whose all elements are spheres congruent to Sn-1. Also, prove that for n >_ 1 and for any natural number k, there is no k-homogeneous covering of R n whose all elements are balls in R n. 15. Let P be a family of straight lines in the plane R 2, such that E p}) - c
card({p E 7 ) ' z
for every point z E R 2. Show that, for each natural number k _> 2, there exists a subfamily of :P which is a k-homogeneous covering of R 2. Formulate and prove a suitable analogue of this result for the euclidean space R n where n > 3. 16. Let us identify the plane R 2 with the field C of all complex numbers and consider two mappings r defined as follows:
r
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Check that r is a translation of R 2 and r is a rotation of R 2 about its origin 0. Denote by G the semigroup of motions of R 2, generated by {r r and put X = G(0). Taking into account the fact that e i is a transcendental number, demonstrate that r
n r
= o,
r
u r
= x.
In other words, the countable unbounded set X C R.2 admits a decomposition into two sets which are G-congruent with X. (Such a decomposition is impossible for a nonempty bounded subset X of R2.) This result is due to Mazurkiewicz and Sierpifiski. It does not rely on the Axiom of Choice (compare Chapter 14).
We begin this chapter with consideration of non Lebesgue-measurable subsets of the euclidean plane R 2, which have small linear sections. We will investigate these sets from the point of view of measurability with respect to appropriate invariant extensions of the Lebesgue measure on R 2. Our interest in such subsets of the plane is inspired by a classical Sierpifiski partition of R 2 whose properties were thoroughly discussed in Chapter 4. In the literature there are many examples of paradoxical subsets of a finite-dimensional euclidean space, having small sections by certain affine hyperplanes of this space. One of the earliest examples is due again to Sierpifiski who constructed a function
f :R~R such that its graph is a A2-thick (or A2-massive) subset of the plane R 2. This construction was presented in Chapter 2. Here A2 denotes, as usual, the standard two-dimensional Lebesgue measure on R 2, and we recall that a subset X of R 2 is A2-thick (or A2-massive) in R 2 if the inner A2-measure of the set R 2 \ X is equal to zero (see the corresponding definition in [62]). In particular, the A2-thickness of the graph F / o f f implies that F / i s nonmeasurable with respect to A2 and: therefore, f is not measurable in the Lebesgue sense (see Theorem 4 from Chapter 2). At the same time, any straight line in R 2 parallel to the line {0} • R meets F / at exactly one point. Another interesting example was given by Mazurkiewicz who constructed a subset Y of R 2 having the property that for each straight line I in R 2, the set 1 N Y consists of exactly two points. In the sequel, any such subset of the plane will be called a Mazurkiewicz set. 102
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103
The descriptive structure of a Mazurkiewicz set turned out to be rather complicated. In general, one cannot assert that a Mazurkiewicz set is necessarily nonmeasurable with respect to &2. Indeed, there are Mazurkiewicz subsets of the plane which have &2-measure zero (see Exercise 1 of this chapter). Moreover, by using the classical Fubini theorem, one can easily conclude that if a Mazurkiewicz set is )~2-measurable, then it must be of ~2-measure zero. Slightly changing the argument of Mazurkiewicz, we shall demonstrate below the existence of a Mazurkiewicz set which is )~2-thick (consequently, it is nonmeasurable with respect to &2). Actually, the same argument works not only for the plane R 2 and the measure &2 on it, but also for the euclidean n-dimensional space R n, where n __ 2, and for the standard n-dimensional Lebesgue measure &n on this space. In fact, various analogous constructions of sets in R n which have small sections but are large in some sense can be carried out by utilizing the method of Mazurkiewicz (compare Exercise 3 of this chapter). T h e o r e m 1. Let n > 2. There exists a subset X of R n such that: 1) each affine hyperplane in R n intersects X in exactly n points; consequently, X is a set of points in general position; 2) X is thick with respect to )~n. P r o o f . Let (~ be the least ordinal number of cardinality continuum (denoted by c), let {L~ 9~ < c~} be the family of all affine hyperplanes in R n and let {B~ 9~ < a} be the family of all those Borel subsets of R n which have strictly positive An-measure. We are going to construct, using the transfinite recursion method, a family {X~ "~ < a} of subsets of R n satisfying the following conditions" (1) if r < ~ < c~, then Xr C X~ (in other words, this family is increasing by inclusion); (2) card(X~) <_ card(~) + w for any ordinal ~ < c~; (3) for each ~ < c~, the set X~ is a set of points in general position in the space R n ; (4) if ~ < a, then card(X~ N L~) - n; (5) if ~ < c~, then X~ N B~ % 0. Suppose that, for an ordinal ~ < c~, the partial family of sets <
has already been constructed. Consider the set Bf. Let B~ be a subset of B~ of cardinality continuum, whose all points are in general position.
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Note that the existence of such a subset of B~ can easily be established by utilizing Lemma 1 of this chapter, which is presented below. Further, let us denote
Yr = u { x r
r < ,1}.
Evidently, Y~ is a set of points in general position and
card(Y~) < card(~) + w. Consequently, we may write
card(Y~
n
L~) < n.
Now, by using ordinary recursion, it is not hard to define a finite subset Z~ of L~ such that: (i) card((Y~ U Z~) n L~) = n; (ii) Y~ U Z~ is a set of points in general position. Note that in order to define Z~, it suffices to apply a simple geometric fact stating that no affine linear manifold L in R n can be covered by a family of affine linear manifolds whose cardinality is strictly less than c and whose members all have dimension strictly less than dim(L) (compare Lemma 1 below in which a more general result is formulated). Further, since
card(Y~ U Z~) < c,
card(B~) - c
and B~ is a set of points in general position, there exists a point x~ E B~ for which the union {x~ } U Y~ U Z~ is also a set of points in general position. So, we may put
x~ = {x~} u Y~ u z~. Proceeding in this way, we are able to construct the required family of sets
{ x ~ : ,1 < ~}. Finally, we define
x = u { x ~ : ~ < o~}. It can easily be verified, in view of our construction, that the set X satisfies relations 1) and 2) of the theorem. This finishes the proof. In particular, for n = 2, the theorem proved above yields a Mazurkiewicz subset of the euclidean plane R 2, which is thick with respect to the Lebesgue
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measure A2. Hence, this subset turns out to be nonmeasurable with respect to ~2. Now, according to the standard construction of extending the Lebesgue measure (see, for instance, Exercise 10 of Chapter 2), any Mazurkiewicz set can be included in the domain of a translation-invariant extension of ~2, and we can assert that any Mazurkiewicz set necessarily becomes a set of measure zero with respect to such an extension (compare Exercise 4 of this chapter). The same result remains valid for those extensions of ~2 which are invariant under the group of all isometric transformations (motions) of R 2. Actually, we are interested in subsets of R 2 having small linear sections in connection with the following natural question: how small are these subsets from the point of view of invariant extensions of the Lebesgue measure We are going to demonstrate in our further considerations that there are invariant extensions of ~2, concentrated on sets with small linear sections in all directions. In fact, a more general result will be obtained stating t h a t the corresponding sets can have small sections by all analytic curves lying in R 2. Note t h a t our further argument is applicable to the space R n, where n >_ 2, and to the s t a n d a r d Lebesgue measure An on this space. In the sequel, we need the following simple auxiliary statement which is useful in many analogous situations. L e m m a 1. Let Z be a An-measurable subset of the Euclidean space R n with An(Z) > 0 and let {Mi : i E I} be a family of analytic manifolds in R n, such that: 1) card(I) is strictly less than c; 2) for each index i E I, the dimension of Mi is strictly less than n. Then the relation
Z\u{M
: i E I } #O
is satisfied. In particular, Lemma 1 states that the space R ~ cannot be covered by a small (in the sense of cardinality) family of analytic submanifolds of R n whose dimensions are strictly less than n. The proof of this lemma is not difficult and can be carried out by induction on n. Here the classical Fubini theorem plays an essential role. We leave the corresponding purely technical details to the reader.
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Some applications of Lemma 1 to certain questions of the geometry of euclidean spaces may be found in [87]. Note that the lemma does not hold true for topological manifolds in R n. For example, there are models of set theory in which the plane R 2 can be covered by a family of Jordan curves, whose cardinality is strictly less than c. (Here a Jordan curve is any homeomorphic image of the unit circumference Sl C R2.) Starting with the above-mentioned lemma, we are able to establish the following result. T h e o r e m 2. Let G be the group of all analytic diffeomorphisms of the space R n (n > 1) onto itself. Then there exists a subset X of R n such that: 1) X is almost G-invariant; in other words, we have (Vg E G ) ( c a r d ( g ( X ) A X ) < c),
where the symbol A denotes the operation of symmetric difference of sets; 2) card(X) = c and X is )~n-thick in Rn; 3) for any analytic manifold M in R n with dim(M) < n, we have card(M 0 X ) < c.
P r o o f . We use the method developed in [75], [82] and [100]. Let a denote the least ordinal number of cardinality continuum. Since the equality card(G) = c holds, we may write a=u{a
:
where {G~ 9 ~
card(G~) <_ card(~) + w; (b) the family {G~ : ~ < a} is increasing by inclusion. Also, we introduce the family {Y~ : ~ < a} consisting of all those Borel subsets of R n which are of strictly positive An-measure. Finally, we denote by {M~ : ~ < a} the family of all analytic manifolds in R n whose dimensions are strictly less than n.
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Let us now define, by the method of transfinite recursion, an injective c~-sequence {x( : ( < c~} of points in R n, satisfying the following relations: (i) for any ( < c~, the point x( belongs to Y(; (ii) for any ~ < c~, we have
9~ r G~((u{Mr
: ~ < ( } ) u {~r : ( < (}).
Note that Lemma 1 guarantees, at each ~-step of our recursion, the existence of a point x~. So the recursion can be continued up to c~. Proceeding in this way, we are able to construct the required c~-sequence of points {x~ : ~ < c~}. Now, putting
x = {ar
: ~ < ~},
we can easily check that the set X is the desired one. (All details are left to the reader.) This finishes the proof of Theorem 2. The next statement is a trivial consequence of the theorem just proved. T h e o r e m 3. Let G denote the group of all isometric transformations (motions) of the space R n. Then there exists a subset X of R n such that: 1) X is almost G-invariant; 2) card(X) = c and X is An-thick in Rn; 3) for each analytic manifold M in R n with dim(M) < n, we have
card(X n M) < c. In particular, for any affine hyperplane L in R n, the inequality card(X n L) < c is true. It immediately follows from Theorem 3 that if the Continuum Hypothesis holds, then, for each analytic manifold M in R n with dim(M) < n, we have card(X n M) < w, where X is the set from Theorem 3. Let us point out one straightforward application of Theorem 3.
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Various statements related to the classical Fubini theorem lead to the following intriguing question: does there exist a measure #n on R n extending the Lebesgue measure An, invariant under the group of all isometric transformations of R n and concentrated on some subset X of R n with small sections by all hyperplanes? Here the smallness of sections of X means that, for any hyperplane L in R n, the cardinality of L N X is strictly less than the cardinality of the continuum. Theorem 3 immediately yields a positive answer to this question. Indeed, let G be the group of all isometric transformations of R n and consider the G-invariant a-ideal of subsets of R n, generated by the set R n \ X, where X is the set from Theorem 3. We denote this a-ideal by 3". Then, for any set Z E if, the inner An-measure of Z is equal to zero (since X is almost Ginvariant and An-thick in Rn). Taking this fact into account and applying the standard methods of extending invariant measures (see, for instance, [82], [100], [212] and [214]), we infer that there exists a measure #n on R n satisfying the relations: 1) ~Un is complete and extends )~n; 2) #n is invariant under the group G; 3) ff C dom(#n); 4) for each set Z E ,7, we have # n ( Z ) - O . Relation 4) implies at once that
u . ( R n \ X ) =0. In other words, our measure/~n is concentrated on the set X. At the same time, we know that X has small sections by all hyperplanes in R n and, moreover, by all analytic manifolds in R n whose dimensions are strictly less than n. In addition to this, the measure #n being complete and metrically transitive has the so-called uniqueness property (compare Exercise 9 from Chapter 9). In particular, for n - 2, we obtain that there exists a complete measure /z2 on the euclidean plane R 2, such that: (1) #2 is an extension of the Lebesgue measure A2; (2) #2 is invariant under the group of all isometric transformations of R2 9 (3) #2 is concentrated on some subset X of R 2 having the property that all linear sections of X are small; more precisely, the cardinality of each linear section of X is strictly less than c.
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Note t h a t the question concerning the existence of a measure #2 on the plane, satisfying conditions (1) - (3), was formulated by R.D.Mabry (personal communication). Moreover, we see that the above-mentioned support X of #2 has a much stronger property: for every analytic curve I in R 2, the cardinality of l N X is strictly less than c. Let us point out that in [96] an analogous question was considered for extensions of A2 which are invariant under the group of all translations of R 2. Namely, it was demonstrated in [96] t h a t there exists a translation-invariant extension of A2 concentrated on a subset of R 2 whose all linear sections are at most countable. R e m a r k 1. If the Continuum Hypothesis holds, then we can also conclude that the above-mentioned subset X of R 2 has the property t h a t all its linear sections are at most countable. Obviously, the same is true for the sections of X by analytic curves lying in R 2. Thus, in this case, the measure #2 is concentrated on a set with countable linear sections. In this context, the following question seems to be natural: does there exist a measure v2 on the plane, extending A2, invariant under the group of all isometric transformations of R 2 and concentrated on a set with finite linear sections? It turns out that the answer to this question is negative. The proof of the corresponding result (and a more general statement) can be found in [94] (see also Exercise 10 of this chapter). R e m a r k 2. By assuming some additional set-theoretical axioms, it is possible to obtain a much stronger result than Theorem 2. Namely, let us suppose that, for any cardinal ~ < c, the space R n (n _ 1) cannot be covered by a ~-sequence of An-measure zero sets. In fact, it suffices to suppose the validity of this assumption only for n = 1, which means that the real line R cannot be covered by a family of Lebesgue measure zero sets, whose cardinality is strictly less than c. For instance, this hypothesis follows directly from Martin's Axiom (see [64] or Appendix 1). We say that a group G of transformations of R n is admissible if each element g from G preserves the a-ideal of all An-measure zero subsets of R n. There are many natural examples of admissible groups. For instance, if G coincides with the group of all affine transformations of R n, then G is admissible. If G coincides with the group of all diffeomorphisms of R n, then G is admissible, too. Evidently, the cardinality of these two groups is equal to c. It is not hard to check that there are also admissible groups G with card(G) = 2 c. Let us fix an admissible group G with card(G) = c. Assuming the above-mentioned hypothesis and applying an argument similar to the proof
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of Theorem 2, we easily obtain the following statement" if n _ 1, then there exists a subset X of R ~ with card(X) = c, such that" 1) X is almost G-invariant; 2) X is ~n-thick in Rn; 3) for any An-measure zero subset Z of R n, we have
card(Z N X) < c. Condition 3) shows, in particular, that the above-mentioned set X is a generalized Sierpifiski subset of R n. (For the definition and various properties of Sierpifiski sets, see [155] or [165] where the dual objects to the Sierpifiski sets - the so-called Luzin sets - are discussed as well; the corresponding material about these sets is also presented in Chapter 5 of this book.) Now, we wish to return to a problem of Luzin posed by him many years ago (see [195]). We have already been slightly in touch with this problem in Chapter 4 where its close connection with a Sierpifiski partition of the plane R 2 was briefly discussed. Here we are going to consider this question in more details. First of all, let us recall the formulation of this beautiful problem. Namely, Luzin asked whether there exists a function r
R--+ R
such that the whole plane R 2 can be covered by countably many isometric copies of the graph of r We have already mentioned that, under C H , this problem admits a positive solution even in a much stronger form (compare statement (iii) and Exercise 4 of Chapter 4). Actually, a positive answer to the Luzin question follows from the existence of a Sierpifiski type partition for the euclidean plane. Here we are going to present the full solution of this problem, within the theory Z F C . Recall that the final result was obtained by Davies. (See his two works [31] and [32] devoted to the Luzin problem.) In the sequel, we need some auxiliary notions and facts about special subsets of R 2. Let X be a subset of R 2 and let p be a straight line in R 2. We shall say that X is uniform with respect to p if, for each line p~ parallel to p, we have
card(p'n X) < 1.
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Obviously, for a set X C R 2 and for any two parallel straight lines p and q in R 2, the following assertions are equivalent: 1) X is uniform with respect to p; 2) X is uniform with respect to q. Further, we shall say that a set X C R 2 is uniform if there exists at least one straight line p in R 2 such that X is uniform with respect to p. It is not hard to observe that the next two assertions are also equivalent: (1) there exists a function r : R -+ R such that the plane R 2 can be covered by countably many isometric copies of the graph Fr of r (2) the plane R 2 can be covered by countably many uniform subsets of R 2" Therefore, the Luzin problem is reduced to the problem of the existence of a countable family of uniform sets, whose union is identical with R 2. In our further considerations, we will deal with the latter problem. Let p be an arbitrary straight line in R 2 and let z be a point of R 2. W e denote by p(z) the unique line in R 2 which is parallel to p and contains z. A countable family P = (Pk)k
card([Z]) < card(Z) + w. In particular, i/ Z is infinite, then card([Z]) - card(Z). An easy proof of this statement is left to the reader. L e m m a 3. Let P - (Pk)k
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set can be well-ordered by some relation ~_ in such a way that, for any point z E Z, the set o.f natural numbers {k < w
9the line pk(z) c o n t a i n s at least one point z ~ -~ z}
is always finite.
P r o o f . We use the method of transfinite induction (on c a r d ( Z ) ) . The case when c a r d ( Z ) - w is trivial: here it suffices to equip Z with any ordering isomorphic to the canonical well-ordering of w. Suppose now that the assertion of the lemma has already been established for all infinite Pclosed sets Z with c a r d ( Z ) < a, where a is some uncountable cardinal. Take any P-closed set Z such that c a r d ( Z ) - a. Of course, we may identify g with the least ordinal of the same cardinality. Let
z-
{z~-~ <
~}
be an injective enumeration of all elements of Z and, for each ordinal ~ < g, let
z~-
{zi2 6 r <
~}.
Further, for any point z E Z, we put ~(z)-inf{~
9~ > w ,
zE[Z~]}.
We shall say that ~(z) is the index of z (in Z). Let us mention that this notion is well defined because the equality Z=U{[Z~]
9w < ~ < ~ }
holds true in view of the P-closedness of Z. By L e m m a 2, if w _< ~ < a, then card([Z~]) - card(~) < and, taking into account the inductive assumption, we can equip the set [Z~] with a well-ordering _~ in such a way that for every point z E [Z~], the set {k < w
9pk(z) contains at least one point z ~ - ~ z}
is necessarily finite. Now, we are able to introduce a suitable well-ordering for the set Z. Namely, if z E Z and z ~ E Z are two distinct points, then we put z ~ -~ z if and only if
(~(z') -~ ~(z)) v ( ~ < ~)(~ - ~(z) - ~(z') ~ z' -~ z).
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STRANGE SUBSETS OF THE EUCLIDEAN PLANE
Let us check t h a t our relation -~ possesses the required properties. First of all, by the definition of -~, it is not hard to see t h a t the corresponding relation _ is a well-ordering of Z. Let now z E Z and ~ - ~(z). We must verify that the set of natural numbers {k < w
9pk(z) contains at least one p o i n t z ~ -~ z}
is finite. Suppose to the contrary that the above-mentioned set is infinite. T h e n there exist pairwise distinct lines ql E
P, q2 E P , ..., qm E P , ...
such that, for some points Zl E Z, z 2 C Z,
..., Zm E Z ,
...,
the relations Zl E ql(z), z2 E q2(z), ..., Zm E a m ( z ) ,
...,
Zl "~ Z, Z2 ~ Z, ...~ Zm "~ Z~ ...
are fulfilled. Observe now that, according to the inductive assumption for the well-ordering ~ , the equalities
can be satisfied only for finitely many natural numbers m. This immediately implies that there are at least two distinct points from {Zl,Z2,...,Zm,...} whose indices are strictly less than ~. W i t h o u t loss of generality, we may assume that ~(Zl) < ~, ~(Z2) "~ ~, Zl r Z2Let us denote = m a x ( ~ ( z l ) , ~(z2)).
Then, obviously, we get
{z ,z2}
c
[z,]
Furthermore, it is evident that the straight lines q l ( z l ) and q2(z2) meet each other at the point z. Therefore, we must have z E [Z~], from which it follows ~(z) <_ rI - m a x ( ~ ( z l ) , ~ ( z 2 ) ) < ~ - ~(z),
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6
and the last relation yields a contradiction. The contradiction obtained completes the proof of L e m m a 3. 4. Let P = ( P k ) k K w be an admissible family of straight lines in R 2. Then there exists a family (Zk)k<~ of subsets o ] R 2, satisfying the following relations: 1) each set Zk (k < w) is uniform with respect to the line Pk; Lemma
2) u { z k
. k < ~} - R 2
P r o o f . Clearly, the plane R 2 is a P-closed set. Therefore, by virtue of the preceding lemma, R 2 can be endowed with a well-ordering -4 such that, for each point z E R 2, the set {k < w
9pk(z) contains at least one point z ~ -~ z}
is finite. Keeping in mind this circumstance, we may denote by k(z) the smallest natural number k for which the straight line pk(z) contains no point Z I --~Z.
Now, for any natural number k, we put zk -
{z e R2
9k ( z ) - k } .
Let us check that the sets Zk (k < w) are the required ones. First, it is evident that
u { z k - k < ~ } - R 2.
Therefore, it remains to demonstrate that every set Zk is uniform with respect to the line Pk. To see this, suppose otherwise and pick any two distinct points z E Zk and z' E Zk for which pk(z) = pk(z'). Since the disjunction z -,~ z ~ V z t -~ z trivially holds, we may assume without loss of generality that z ~ -~ z. Under this assumption, we get z' e p k ( z ' ) - p k ( z ) - pk(z)(Z),
z' -~ z,
which contradicts the definition of k(z). The contradiction obtained ends the proof of the lemma. As mentioned earlier, Lemma 4 yields a positive solution of the Luzin problem. So, we can formulate the following statement first obtained by Davies [31].
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Theorem
4. There exist a function
r
R-~R
and a countable family (gn)n<~ of motions of the euclidean plane R 2 , such that U{gn(F~) " n < w} - R 2, where F, denotes the graph of r Thus, we conclude that the problem of Luzin is solvable within the theory Z F C . Some interesting results closely related to this problem are indicated in exercises below (see also Chapter 14). A generalization of Theorem 4 for euclidean spaces of higher dimension can be found in [32].
EXERCISES 1. Show that there is a nowhere dense set Z C satisfying the relation card(Z N l) - c
R 2
of A2-measure zero,
for all straight lines 1 C R 2. Starting with this fact and applying the method of transfinite induction, demonstrate that there exists a nowhere dense Mazurkiewicz set of $2measure zero. 2. For any straight line l in the plane R 2, let ~(1) be a cardinal number such that 2 <_ n(1) < w. Prove that there is a subset X of R 2 satisfying the relation card(l N X ) - ~(1) for each straight line l in R 2. This beautiful result is due to Sierpifiski and it generalizes some theorems of Mazurkiewicz and Bagemihl. Prove, in the theory ZF, that there exists a set Y C R 2 satisfying the relations: (a) c a r d ( Y ) = c;
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(in E) if, for each point z E Z, we have
zr
\ (z}).
For example, the set of all vertices of a convex polygon in R 2 is convexly independent. By using the Ramsey combinatorial theorem (see [179]), prove that for each natural number n >_ 2, there exists a natural number N(n) having the following property: Any set X C R 2 of points in general position, with card(X) > N(n), contains a convexly independent subset whose eardinality is equal to n. This beautiful result is due to ErdSs and Szekeres [44]. They also conjectured that the smallest possible value of N(n) is 2 n-2 + 1. So far, there is no significant progress in proving (or disproving) their conjecture. By using an infinite version of the Ramsey theorem [179] (more precisely, by using its countable version), demonstrate that every infinite set X C R 2 of points in general position contains an infinite convexly independent subset. 12. By applying the method of transfinite induction, show that there exists a set X C R 2 having the following properties: (~) ca,'d(X) - c; (b) X is a set of points in general position; (c) no subset of X of cardinality e is convexly independent. For this purpose, denoting by a the least ordinal of cardinality e, take an injective family {L~ 9~ < a} consisting of all convex curves in R 2 and construct by transfinite recursion a subset of R 2 which almost avoids the members L~ of this family (compare the classical constructions of Sierpifiski and Luzin sets presented in Chapter 5). Finally, conclude that if the Continuum Hypothesis holds, then there exists a set in R 2 of cardinality e, whose points are in general position and which contains no uncountable convexly independent subset. 13. Suppose that the cardinal c is real-valued measurable. Starting with the result of Kunen (see Chapter 4), demonstrate that there exists a family {Li " i C I} of Jordan curves in R 2, satisfying the following relations: < r
(b) U { L i ' i e I } - R
2.
14. Let E be a set and let {Xi " i C I} be an injective family of nonempty subsets of E. We shall say that this family is a homogeneous
S T R A N G E SUBSETS OF THE EUOLIDEAN PLANE
119
covering of E if there exists a cardinal number a such that card({i E I " x E X i } ) =
for all elements x E E. In this case, we also say that a is the index of homogeneity of { X i " i E I } and that { X i 9 i E I } is a a-homogeneous covering of E. For example, if ~ - 1, then the a-homogeneity of {Xi 9i EI ) simply means that {X~ 9i EI} is a partition of E. Show t h a t there exists no partition of the plane R 2 consisting of Jordan curves. On the other hand, show within the theory Z F that, for every natural number k _> 1, there exists a 2k-homogeneous covering of R 2 all whose elements are circumferences congruent to the unit circumference Sl C R 2. Moreover, for every natural number k >_ 2, construct a k-homogeneous covering of R 2 consisting of pairwise congruent circumferences. (Use the m e t h o d of transfinite recursion.) Generalize these results to the case of the euclidean space R n where n _> 2. Namely, prove that: (a) there exists no partition of R n whose all elements are homeomorphic images of the unit sphere Sn-1 C Rn; (b) for any natural number k >_ n, there exists a k-homogeneous covering of R n whose all elements are spheres congruent to Sn-1. Also, prove that for n >_ 1 and for any natural number k, there is no k-homogeneous covering of R n whose all elements are balls in R n. 15. Let P be a family of straight lines in the plane R 2, such that E p}) - c
card({p E 7 ) ' z
for every point z E R 2. Show that, for each natural number k _> 2, there exists a subfamily of :P which is a k-homogeneous covering of R 2. Formulate and prove a suitable analogue of this result for the euclidean space R n where n > 3. 16. Let us identify the plane R 2 with the field C of all complex numbers and consider two mappings r defined as follows:
r
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Check that r is a translation of R 2 and r is a rotation of R 2 about its origin 0. Denote by G the semigroup of motions of R 2, generated by {r r and put X = G(0). Taking into account the fact that e i is a transcendental number, demonstrate that r
n r
= o,
r
u r
= x.
In other words, the countable unbounded set X C R.2 admits a decomposition into two sets which are G-congruent with X. (Such a decomposition is impossible for a nonempty bounded subset X of R2.) This result is due to Mazurkiewicz and Sierpifiski. It does not rely on the Axiom of Choice (compare Chapter 14).