Directory UMM :Data Elmu:jurnal:M:Mathematical Social Sciences:Vol37.Issue3.May1999:

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Growth rates in multidimensional spatial voting

* Maria Tataru

Department of Mathematics, Northwestern University, Evanston, IL, 60208, USA Received 18 November 1997; accepted 20 May 1998

Abstract

In the model of multidimensional spatial voting, define the sets F (a) to be the set of pointsq n

reachable from a[R through a sequence of m or fewer q-majority vote elections. Let the core be the set of points not reachable from any other alternative. The main idea in this paper is the

Keywords: Multidimensional spatial voting; Sets; Elections

1. Introduction

The basic constituents of spatial voting models are: we are given a group of N voters whose preferences are represented as follows. Each voter has an ideal point (called also a

bliss point) in the n-dimensional real space R which corresponds to the outcome that he

would regard as best. Each axis of R represents an issue and the numerical value on that

axis illustrates the position on that issue. Moreover, every point in R can be proposed as

an outcome and be voted upon. The preferences of voters over the points in R are expressed in terms of the distances of the points from their ideal points. Among the many functions one could use in measuring distances, we will select the Euclidean distance since it is both widely used and convenient.

Let the social choice correspondence be the q-majority rule, (N ) /(2),q,N, where

any coalition of size q, from the society of size N, is decisive. For instance if

q5[N / 2]11, where [x] is the greatest integer function, then we obtain the simple majority rule and if q5N we have the unanimity rule. No strategic behavior of the

voters is assumed so sincere voting is incorporated in our analysis. *Tel.: 1847-491-8544; Fax: 1847-491-8906.

E-mail address: [email protected] (M. Tataru)

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The model of spatial voting is often used in political science to describe the competition of politicians for voters. Thus,

(i) each candidate is identified by a point in R representing his position on the n issues constituting the political ‘‘horizon’’;

(ii) each voter has an ideal point in R representing his opinion on these issues, and he compares various candidates by the distance from his ideal point to their position.

In this model, a solution concept defined as the core has been extensively studied. (See Schofield, 1985 for a detailed discussion).

Definition 1. A point x is a core point for a q-rule if for all other alternatives y, it is impossible to find q voters who prefer y to x.

The core is the set of all core points or in other words, the core is the set of points not beaten by any other alternative.

The problem of determining the dimension of the issue space for which the core is generically empty and the dimension for which core points exist for an open set of preferences, has received considerable research attention (see McKelvey, 1979; McK-elvey and Schofield, 1986; Plott, 1967; Saari, Preprint; Schofield, 1985). All these results indicate that we must expect the core to be empty for many practical purposes. If the core is empty, then for any x there is a x that will be chosen over x by a1 2 1 winning coalition. Similarly, there is a x that beats x and so on . . . . The absence of the3 2 core has quite interesting consequences as has been shown by R. McKelvey, N. Schofield and others. These results are referred to as the ‘‘chaos theorems’’ because of the negative interpretation that all order is lost Richards, 1994. It has been proved that a

majority trajectory can be build from any alternative x to any alternative y[R , that is,

any alternative can be rendered the majority winner if one can freely introduce new alternatives into the voting agenda and can stop the process at any time. Moreover, sequences of votes do not reach an equilibrium and the final outcome may have little relationship to voter preferences.

To start, consider N voters, each with preferences representable by circular indiffer-ence curves in the n dimensional Euclidean space. Alternative x is preferred to y by voter i if and only if:

xs _y↔i_x₂_ai_#i_y₂_ai.

i i i

wherei i is the Euclidean norm in R .

Let q,N be the number of votes needed to replace the status quo with a new

alternative. Let F denote the social choice rule given by the q-majority rule. Forq

example, the simple majority rule is obtained if q5(N11) / 2 and N is odd and for

q5(N / 2)11 and N even. Then

F (a)5hyuys _aj,

q q

is the set of points in R that are preferred to a by a minimum of q voters. In other

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a the set of alternatives that beat a by q-majority votes or equivalently, F (a) is the setq

of alternatives reachable from a in zero or one q-majority vote elections. 2

Then F (a)q 5F (F (a)) is the set of alternatives reachable from a in two or fewerq q

q-majority vote elections. Note that if the coalition formed by the voters a , a , . . . ,a1 2 q

gave rise to F (a), then it is possible that a different coalition of q voters (maybe a ,q 3 2

a , . . . ,a4 q12) is used to define F (a).q m

Generalizing, let F (a) be the set of points reachable from a through a sequence of mq m

or fewer q-majority vote elections. The most dramatic results that concern the sets F (a)q

are about points that are reachable ‘‘eventually’’:

^ _(a)₅

<

_{F (a).}

q q

m51

In Kelly, 1987, J.S. Kelly calls attention to the fact that these voting sets can be used to think about many common voting theory ideas. He also points out that little is known about the structure of these sets. McKelvey and Schofield, 1987 found sufficient conditions for^ _{(a) to be the whole alternative space for all a, but they do not prove}

their results by looking at the constituent sets. Since we know from them that these sets grow to fill the space, it is natural to ask about growth rates. After all, this result loses

ux2yu

much of its appeal should it require, say e steps to have an agenda starting at x and ending at y.

In this paper I continue the work started by McKelvey and Schofield and generalize the classical ‘‘either / or’’ result to all q-rules. In addition, we provide an estimate for the growth rate of the voting sets described above. This last part generalizes the results in McKelvey, 1976 in which McKelvey proves that for simple majority rule the reachable sets must grow at least at a linear rate.

The introduction presents the model and the problem to be solved. In the second section, I prove the following dichotomy: either the core is nonempty or the voting sets

F (a) grow linearly in m. More specifically, I find that the number of steps to get fromq

one point of the alternative space to another is proportional to the distance between

them. It is proved that for the case of the issue space R with an arbitrary q rule, if the

core is empty, then the sets F (a) grow linearly in m to fill the whole alternative space.q

The constant of proportionality then, gives a measure on the growth rate of the sets

F (a). This growth rate depends on the geometric positioning of the voters ideal pointsq

Tataru, 1996. The correlation between them is studied in a forthcoming paper. This paper only considers Euclidean preferences, so an interesting question, not yet answered, is whether similar results hold for a more general class of preferences.

2. Core points versus linear growth in R

Definition 2. Let B(x , r) be the ball of center x1 1 and radius r and ≠B(x ,r) its1 boundary:

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≠B(x , r)1 ; hx;ix2x1i5rj.

2 For any set A,R define

B(A, r)5

<

B( y, r).

y[A

The next assertion shows how sets of alternatives are related with the F corre-q

spondence and provides a valuable technical tool.

Theorem 3. Let x be any alternative in R . Then1 , for any r.0,

F (B(x , r))q 1 5B(F (x ), r),q 1

where B(F (x ), r)q 1 5hx[R u there exists y[F (x ) withq 1 ux2yu#rj.

Proof. The claim of the theorem is illustrated in Fig. 1 for the particular case: three

voters, R the issue space and majority rule. Here the small trefoil shaped set, centered at

x is F (x ) and the larger set is F (B(x , r)).1 q 1 q 1

To prove the theorem in the general case, note that the set of points that are preferred to x by any coalition of q of the a ’s will be in F (x ). Denote by # _{the family of q}

1 i q 1

coalitions from the set of a ’s. Then, F (x )i q 1 5<j[# F (x ) so F (x ) is formed as theq 1 q 1 union of the sets obtained from each possible coalition. Therefore, it is sufficient to

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consider a fixed coalition, say a , a ,...,a and prove the theorem for the restriction of F1 2 q q

obtained as the set of points that are preferred to x by a , a , . . . ,a . The proof will1 1 2 q

hold for any other coalition.

For the first implication we need to prove that if we let y0[F (x ) with xq 0 0[B(x , r),1 then y0[B(F (x ), r). That is, there exists a yq 1 1[F (x ) such thatq 1 uy12y0u#r. Without

any restriction of generality, take x0[≠B(x , r).1 Define the curve:

x(t)5(12t)x01tx ; t1 [[0, 1].

The idea is to find y15y(1) as the endpoint of a curve y(t) that satisfies: y(0)5y0

iy9(t)i#r

5

y(t) is preferred to x(t) for any t[[0, 1], The second condition guarantees that uy12y0u#r.

Because y0[F (x ),q 0 iy02a1i#ix02a1i iy02a2i#ix02a2i iy02aqi#ix02aqi.

Consider the above inequalities. Without loss of generality suppose that up to j, 0#j#q we have equalities, and for j,m,q the inequalities are strict. In order to

determine the curve y(t) we distinguish several cases.

Case 1. If j5q, thenix02ami5iy02amifor m51, 2, . . . ,q. Let Y be the linear space

n n

generated by a , a , . . . ,a in R . Then there exists an isometry6 _{in R that preserves}

1 2 q 0

the a ’s (therefore preserves Y as well) and maps x into y . In this case define y(t) to bei 0 0 the image of x(t) through 6 _{and y} ₅_{y(1). Clearly,} u_x(t)₂_au₅u_y(t)₂_au _{for all t.}

0 1 j j

Case 2. If j,q, henceix02ami5iy02ami for m51, 2, . . . , j, andiy02aii,ix02aii

for i5j11, . . . ,q, then let U be the linear subspace generated by a , a , . . . ,a .1 2 j

Sinceix 2a i5iy 2a i for m51, 2, . . . , j it follows that there exists an isometry 6

0 m 0 m 1

that preserves U and maps x into y . Define the curve y(t) to be:0 0

y(t)56 _(x(t)). 1

Note that 6 _{preserves the distance:}_i_x(t)₂_a _i₅_i_y(t)₂_a _i _{for m}₅_{1, 2, . . . , q, m}±_i.

1 m m

Subcase 2.1. If the relation

ix(t)2aii,iy(t)2aii, i5j11, . . . ,q

is preserved for any t[[0, 1], then we can define y15y(1) and it satisfies the desired

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Subcase 2.2. Otherwise, there exists some value 0#t1#1 such that (2) holds up to t1 and at t :1 ix(t )1 2ai₀i5iy(t )1 2ai₀i, for some i0[j11, . . . ,q. We stop at t and go back1 to the beginning with the appropriate increased value for j.

Note that in any case, y(t) satisfiesiy9(t)i#r. This is so because for x(t) the property

holds by hypothesis and y(t) is obtained from x(t) by an isometry, so the distance is preserved. Also this process will only take a finite number of steps because the value of j cannot increase to more than q.

Conversely, we need to prove that given r, x1 and y1[F (x ), if we considerq 1

y0[B( y , r) that is1 iy02y1i#r, then y0[F (B(x , r)). Thus we need to find an x suchq 1 0 that y is preferred to x and0 0 ix02x1i#r.

The idea of the proof is similar to the previous argument. Define the curve:

y(t)5(12t)y01ty , t1 [[0, 1],

and look for x on a curve x(t), t0 [[0, 1] that satisfies:

x(1)5x1 ux9(t)u#r

5

y(t) is preferred to x(t) for any t[[0, 1].

From here on, the proof proceeds as in the direct implication. 2

Corollary 4. For any set A,R , F (B(A, r))q 5B(F (A), r).q

Proof. The proof follows immediately using Theorem 3 and Definition 2:

F (B(A, r))q 5

<

F (B(x, r))q 5

<

B(F (x), r)q 5B(F (A), r).q x[A x[A

Important for our analysis is the following iterated version of Theorem 3. 2

Theorem 5. Let x be any alternative in R . For any r1 .0, m[N:

m m

F (B(x , r))q 1 5B(F (x ), r).q 1

Proof. The proof uses mathematical induction. Indeed, for m51 the result is true by Theorem 3. Assume that the identity above holds for m21 and show that it is true for m as well. Using the induction hypothesis and Corollary 4:

m m m m

F (B(x , r))q 1 5F (Fq q 21(B(x , r)))1 5F (B(Fq q21(x ), r))1 5B(F (x ), r),q 1 which concludes the proof.

Next, we can use the result above to get information about the rate of increase for the

sets F (a). In Theorem 7, the main result of this section, we show that the number ofq

steps needed to reach an arbitrary point x starting from a is proportional with the distance between the two points.

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Definition 6. Given a[R , let

r(m);r(F (a))q ; sup ix2ai.

x[F_q(a)

The quotient (r(m) /m) gives a measure on the rate of increase for the sets F (a). In otherq m

words, r(m) is the smallest radius of a ball so that F (a)q ,B(a, r(m)). For our purposes,

r(m) measures the distance to the farthest points in F (a).q n

For any a[R there exists a constant C$0, which does not depend on a, such that:

r(F (a))q

]]]

lim 5C.

m→` m

Proof. First, we prove that r(m) is a subadditive function of m: r(m1n)#r(m)1r(n) for any m, n[N.

By definition, r(n) and r(m) with m, n[N satisfy:

n m

F (a)q #B(a, r(n)) and F (a)q #B(a, r(m)).

By using Theorem 5 we obtain:

m1n m n m m

Fq (a)5F (F (a))q q #F (B(a, r(n)))q #B(F (a), r(n))q #B(B(a, r(m)), r(n))

5B(a, r(m)1r(n)).

Hence we showed that:

m1n

Fq (a)#B(a, r(m)1r(n)).

Finally, using the definition of r(m1n), we conclude that: r(m1n)#r(m)1r(n) for any m, n[N.

Next, the idea is to use the lemma below, a standard result in analysis, to obtain the claim in the theorem.

Lemma 8. If r is a nonnegative, subadditive function then the limit

r(m)

]]

lim

m→` m

exists and is nonnegative.

Proof of Lemma. See Appendix A.

Let us prove next that the constant C does not depend on the initial point a. Thus the

growth of the sets F (a) does not depend on the starting point.q n

The idea is to show that given any two points a, b[R :

m m

r(F (a))q r(F (b))q

]]] ]]]

lim 5lim .

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m m

Let r 5ub2au. Then a[B(b, r) and F (a)[B(F (b), r) Hence, using Theorem 5 and Definition 6 we conclude that:

m m m

r(F (a))q #dist(B(F (b),r), a)#r 1 r 1r(F (b)).q

Similarly, we start with b[B(a, r) and obtain that

m m

r(F (b))q #2r 1r(F (a)).q

Therefore the following identity holds:

m m m

r(F (b))q #2r 1r(F (a))q #4r 1r(F (b)).q

Divide by m and let m→`to obtain the desired result:

m m

r(F (a))q r(F (b))q

]]] ]]]

lim 5lim .

m→` m m→` m

n m

So far we have showed that for any a[R , the sets F (a) are increasing at each step at aq

rate slower or equal to m. If C±_{0, then these sets grow to fill the entire alternative space} and their growth is of order exactly m. If C50, we know that these sets are growing slower than m. This however, does not preclude slower growth rates such as log m or

` m

]

Œ_{m. What we will prove next is that if C}₅_{0 then in fact}_< _{F (a) is a bounded set.}

m51 q n

Theorem 9. For any point a[R , there are two possibilities:

` m

• either <m51 F (a) is a bounded set and the set of core points is nonempty andq

convex,

• or C.0 and the sets F (a) grow to fill the entire alternative space.q

n m

Proof. For any a[R , note that a[F (a) for all mq [N. We distinguish two cases.

Case 1. Suppose a[≠F (a) for all mq [N.

Define G (a) to be the set of points that do not lose strictly to a:q

]]]₂₁ _c

G (a)q ;(Fq (a)) .

Note that F (a)q #G (a). Also, G (a) is a bounded set since it is obtained as finiteq q

union of intersections of disks. Indeed, if d5maxiai2aithen G (a)q ,B(a, 2d ). The idea

is to show that if a[≠F (a), for all m, then:q m

F (a)q #F (a)q #G (a).q m

Let x[F (a) and assume that xq [⁄ G (a). Then x loses strictly to a and therefore to aq

neighborhood of a as well. This implies that a neighborhood of a is contained in

m11 m11

Fq (a) (since the points that beat a are in Fq ) which contradicts the fact that

m11 m

a[≠Fq (a). This implies that F (a) stays in the bounded set G (a).q q m 0

Case 2. If there exists an m0[N such that a[Int(Fq (a)).

m 0

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2m₀ m₀ m₀ m₀ m₀

Fq (a)5Fq (Fq (a))$Fq (B(a, r))5B(Fq (a), r)$B(B(a, r), r)5B(a, 2r).

Inductively, we obtain that:

km₀

Fq (a)$B(a, kr),

which implies that

km₀

r(km )0 5r(Fq (a))$kr.

Divide by km , let k0 →` and use Theorem 7 to conclude that

km₀

r(Fq (a)) r

]]] ]

C5lim $ .0.

km m

k→` 0 0

At this point we answered the question raised by J. Kelly in Kelly, 1987 about how these voting sets vary with the initial point and in time.

` m

Next we will try to understand what is the structure of these sets when<m51F (a) isq

a bounded set. In this case some sets don’t grow as they are preserved under iterations of

F .q

Definition 10. A stable set is a set A with the property that F (A)q 5A.

` m

Note that a core point is a stable set with a single element. The set A5<m51F (a) isq

stable by Theorem 9 and the full space is another stable set. The following results distinguishes the two.

Proposition 11. For any bounded stable set A there exists a stable subset A1#A so that int(A )1 5[.

Proof. Let A be a stable set. If int(A)5[ then the subset A1 is exactly A. Suppose

int(A)±_[ _{and let}

r5suphsuthere exists x[A such that B(x, s),Aj, A15hxuB(x, r),Aj.

By hypothesis A is bounded so the above supremum exists. This implies that A ±_[_{. We} 1

prove that A is stable with empty interior.1

If x[A , then by construction B(x, r)1 ,A. Using Theorem 3 and the fact that A is

stable:

B(F (x), r)q 5F (B(x, r))q ,F (A)q 5A.

According to the definition of A , we have that F (x)1 q [A ; this means that A is stable.1 1 Suppose, by contradiction, that int(A )±_[_{. Therefore there exists an} _e _{so that B(x,}

e),A . By definition of A :1 1

B(x,e 1r)5B(B(x,e), r),A,

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with ideal points at 0, 1, 2 and 3. The core is the interval Kelly, 1987; McKelvey, 1979 and any point in it satisfies the conditions of Proposition 11.

Theorem 12. If C50 then the set of core points is nonempty, closed and convex.

` m

Proof. If C50, then A5<m51 F (a) is stable and bounded by Theorem 9. Byq

Proposition 11 there exists A , a stable set with int(A )1 1 5[. We prove that every point

a[A1 is a core point. For any a[A , F (a)1 q [A1 since A1 is a stable set. F (a) isq

constructed as a union of intersections of q discs. This intersection is either a point, and in this case F (a)q 5a, hence a is a core point, or this intersection is a set with interior.

However, the second possibility contradicts the hypothesis that int(A )1 5[, so this case cannot occur.

To show that the set of core points is convex, let a, b be two core points. We will prove that x5ta1(12t)b is also a core point. Denote by r5iaxiand s5ibxi. Then B(a, r) is a stable set because a is core point so F (a)q 5a and by Theorem 5, F (B(a,q

r))5B(F (a), r)q 5B(a, r). Similarly B(b, s) is a stable set. Therefore their intersection is a

stable set. But this intersection consists of the point x. This shows that x is core point, hence the set of core points is convex.

Acknowledgements

This work was partially supported by a fellowship from AAUW Educational Foundation.

Appendix A

Auxiliary material

Lemma 13. If r: R→R is a nonnegative, subadditive function then the limit

r(n)

]]

lim

n→` n

exists and is positive.

Proof. By hypothesis, the function r5r(n) is subadditive which means that r(m1n)#r(m)1r(n) for any m, n[R.

This implies that, for example:

r(n)5r((n21)11)#r(n21)1r(1)#r(n22)12r(1)#r(n23)13r(1)

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Similarly:

r(n)5r((n22)12)#r(n22)1r(2)#r(n24)12r(2)#r(n26)13r(2)

] # ? ? ? #r(n22k)1kr(2)# ? ? ? # r(2).

2 By induction we obtain that:

]

r(n)# r(k), for any k51, 2, 3 . . .

which implies

r(n) r(k)

]] ]

lim sup # , for any k51, 2, 3, . . .

n→` n k

Hence,

r(k) r(n)

] ]]

lim inf #lim sup .

n→` k n→` n

But, by definition of lim sup and lim inf

r(k) r(k)

] ]

lim inf #lim sup .

n→` k k→` k

Thus, we conclude that

r(k) r(k)

] ]

lim inf 5lim sup ,

k k

k→` k→`

which proves that

r(n)

]]

lim

n→` n

exists. The fact that this limit is positive is obvious since r is nonnegative by hypothesis.

References

Kelly, J.S., 1987. Voting Sets. Social Choice and Welfare 4, 235–239.

McKelvey, R., 1979. General conditions for global intransitivities in formal voting models. Econometrica 47, 1085–1112.

McKelvey, R., 1976. Intransitivities in multidimensional voting models and some implications for agenda control. Journal of Economic Theory 12, 472–482.

McKelvey, R., Schofield, N.J., 1986. Structural instability of the core. Journal of Mathematical Economics 15, 179–198.

McKelvey, R., Schofield, N.J., 1987. Generalized symmetry conditions at a core point. Econometrica 55, 923–933.

Plott, C.R., 1967. A notion of equilibrium and its possibility under majority rule. American Economic Review 57, 787–806.

D. Richards, Intransitivities in multidimensional spatial voting: period three implies chaos, Social Choice and Welfare (1994), 11, 109–119.

D.G. Saari, The generic existence of a core for q rules, NU Preprint. N.J. Schofield, Social Choice and Democracy, Springer-Verlag, 1985.

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Subcase 2.2. Otherwise, there exists some value 0#t1#1 such that (2) holds up to t1

and at t :1 ix(t )2a1 i₀i5iy(t )1 2ai₀i, for some i0[j11, . . . ,q. We stop at t and go back1

to the beginning with the appropriate increased value for j.

Note that in any case, y(t) satisfiesiy9(t)i#r. This is so because for x(t) the property

Conversely, we need to prove that given r, x1 and y1[F (x ), if we considerq 1

y0[B( y , r) that is1 iy02y1i#r, then y0[F (B(x , r)). Thus we need to find an x suchq 1 0

that y is preferred to x and₀ ₀ ix₀2x₁i#r.

The idea of the proof is similar to the previous argument. Define the curve:

y(t)5(12t)y01ty , t1 [[0, 1],

and look for x on a curve x(t), t0 [[0, 1] that satisfies:

x(1)5x1

ux9(t)u#r

5y(t) is preferred to x(t) for any t

[[0, 1].

From here on, the proof proceeds as in the direct implication.

2 Corollary 4. For any set A,R ,

F (B(A, r))q 5B(F (A), r).q

Proof. The proof follows immediately using Theorem 3 and Definition 2:

F (B(A, r))q 5

<

F (B(x, r))q 5

<

B(F (x), r)q 5B(F (A), r).q

x[A x[A

Important for our analysis is the following iterated version of Theorem 3.

Theorem 5. Let x be any alternative in R . For any r1 .0, m[N:

m m

F (B(x , r))q 1 5B(F (x ), r).q 1

m m m m

F (B(x , r))q 1 5F (Fq q 21(B(x , r)))1 5F (B(Fq q21(x ), r))1 5B(F (x ), r),q 1

which concludes the proof.

Next, we can use the result above to get information about the rate of increase for the

sets F (a). In Theorem 7, the main result of this section, we show that the number ofq

steps needed to reach an arbitrary point x starting from a is proportional with the distance between the two points.

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Definition 6. Given a[R , let m

r(m);r(F (a))q ; sup ix2ai. m

x[F_q(a)

The quotient (r(m) /m) gives a measure on the rate of increase for the sets F (a). In otherq m

words, r(m) is the smallest radius of a ball so that F (a)q ,B(a, r(m)). For our purposes, m

r(m) measures the distance to the farthest points in F (a).q n

For any a[R there exists a constant C$0, which does not depend on a, such that:

m r(F (a))q ]]]

lim 5C.

m→` m

Proof. First, we prove that r(m) is a subadditive function of m:

r(m1n)#r(m)1r(n) for any m, n[N.

By definition, r(n) and r(m) with m, n[N satisfy:

n m

F (a)q #B(a, r(n)) and F (a)q #B(a, r(m)).

By using Theorem 5 we obtain:

m1n m n m m

Fq (a)5F (F (a))q q #F (B(a, r(n)))q #B(F (a), r(n))q #B(B(a, r(m)), r(n)) 5B(a, r(m)1r(n)).

Hence we showed that:

m1n

Fq (a)#B(a, r(m)1r(n)).

Finally, using the definition of r(m1n), we conclude that: r(m1n)#r(m)1r(n) for any m, n[N.

Next, the idea is to use the lemma below, a standard result in analysis, to obtain the claim in the theorem.

Lemma 8. If r is a nonnegative, subadditive function then the limit

r(m) ]]

lim

m→` m

exists and is nonnegative.

Proof of Lemma. See Appendix A.

Let us prove next that the constant C does not depend on the initial point a. Thus the

growth of the sets F (a) does not depend on the starting point.q n

The idea is to show that given any two points a, b[R :

m m

r(F (a))q r(F (b))q

]]] ]]]

lim 5lim .

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m m

Let r 5ub2au. Then a[B(b, r) and F (a)[B(F (b), r) Hence, using Theorem 5 and Definition 6 we conclude that:

m m m

r(F (a))q #dist(B(F (b),r), a)#r 1 r 1r(F (b)).q

Similarly, we start with b[B(a, r) and obtain that

m m

r(F (b))q #2r 1r(F (a)).q

Therefore the following identity holds:

m m m

r(F (b))q #2r 1r(F (a))q #4r 1r(F (b)).q

Divide by m and let m→`to obtain the desired result:

m m

r(F (a))q r(F (b))q

]]] ]]]

lim 5lim .

m→` m m→` m

n m

So far we have showed that for any a[R , the sets F (a) are increasing at each step at aq

rate slower or equal to m. If C±_{0, then these sets grow to fill the entire alternative space}

and their growth is of order exactly m. If C50, we know that these sets are growing slower than m. This however, does not preclude slower growth rates such as log m or

` m ]

Œ_{m. What we will prove next is that if C}₅_{0 then in fact}_< _{F (a) is a bounded set.}

m51 q n

Theorem 9. For any point a[R , there are two possibilities:

` m

• either <m51 F (a) is a bounded set and the set of core points is nonempty andq convex,

• or C.0 and the sets F (a) grow to fill the entire alternative space.q

n m

Proof. For any a[R , note that a[F (a) for all mq [N. We distinguish two cases. m

Case 1. Suppose a[≠F (a) for all mq [N.

Define G (a) to be the set of points that do not lose strictly to a:q ]]]₂₁ _c

G (a)q ;(Fq (a)) .

Note that F (a)q #G (a). Also, G (a) is a bounded set since it is obtained as finiteq q

union of intersections of disks. Indeed, if d5maxiai2aithen G (a)q ,B(a, 2d ). The idea m

is to show that if a[≠F (a), for all m, then:_q m

F (a)q #F (a)q #G (a).q m

Let x[F (a) and assume that xq [⁄ G (a). Then x loses strictly to a and therefore to aq

neighborhood of a as well. This implies that a neighborhood of a is contained in

m11 m11

Fq (a) (since the points that beat a are in Fq ) which contradicts the fact that

m11 m

a[≠Fq (a). This implies that F (a) stays in the bounded set G (a).q q m 0

Case 2. If there exists an m0[N such that a[Int(Fq (a)).

m 0

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2m₀ m₀ m₀ m₀ m₀

Fq (a)5Fq (Fq (a))$Fq (B(a, r))5B(Fq (a), r)$B(B(a, r), r)5B(a, 2r).

Inductively, we obtain that:

km₀

Fq (a)$B(a, kr),

which implies that

km₀

r(km )0 5r(Fq (a))$kr.

Divide by km , let k0 →` and use Theorem 7 to conclude that

km₀

r(Fq (a)) r

]]] ]

C5lim $ .0.

km m

k→` 0 0

At this point we answered the question raised by J. Kelly in Kelly, 1987 about how these voting sets vary with the initial point and in time.

` m

Next we will try to understand what is the structure of these sets when<m51F (a) isq

a bounded set. In this case some sets don’t grow as they are preserved under iterations of

F .q

Definition 10. A stable set is a set A with the property that F (A)q 5A.

` m

Note that a core point is a stable set with a single element. The set A5<_m51F (a) isq

stable by Theorem 9 and the full space is another stable set. The following results distinguishes the two.

Proposition 11. For any bounded stable set A there exists a stable subset A1#A so that int(A )5[1 .

Proof. Let A be a stable set. If int(A)5[ then the subset A1 is exactly A. Suppose

int(A)±_[ _{and let}

r5suphsuthere exists x[A such that B(x, s),Aj, A15hxuB(x, r),Aj.

By hypothesis A is bounded so the above supremum exists. This implies that A ±_[_{. We} 1

prove that A is stable with empty interior.1

If x[A , then by construction B(x, r)1 ,A. Using Theorem 3 and the fact that A is

stable:

B(F (x), r)q 5F (B(x, r))q ,F (A)q 5A.

According to the definition of A , we have that F (x)1 q [A ; this means that A is stable.1 1

Suppose, by contradiction, that int(A )±_[_{. Therefore there exists an} _e _{so that B(x,} 1

e),A . By definition of A :1 1

B(x,e 1r)5B(B(x,e), r),A,

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with ideal points at 0, 1, 2 and 3. The core is the interval Kelly, 1987; McKelvey, 1979 and any point in it satisfies the conditions of Proposition 11.

Theorem 12. If C50 then the set of core points is nonempty, closed and convex.

` m

Proof. If C50, then A5<_m51 F (a) is stable and bounded by Theorem 9. Byq

Proposition 11 there exists A , a stable set with int(A )1 1 5[. We prove that every point

a[A1 is a core point. For any a[A , F (a)1 q [A1 since A1 is a stable set. F (a) isq

constructed as a union of intersections of q discs. This intersection is either a point, and in this case F (a)_q 5a, hence a is a core point, or this intersection is a set with interior.

However, the second possibility contradicts the hypothesis that int(A )1 5[, so this case cannot occur.

To show that the set of core points is convex, let a, b be two core points. We will prove that x5ta1(12t)b is also a core point. Denote by r5iaxiand s5ibxi. Then B(a,

r) is a stable set because a is core point so F (a)5q a and by Theorem 5, F (B(a,q r))5B(F (a), r)5q B(a, r). Similarly B(b, s) is a stable set. Therefore their intersection is a

stable set. But this intersection consists of the point x. This shows that x is core point, hence the set of core points is convex.

Acknowledgements

This work was partially supported by a fellowship from AAUW Educational Foundation.

Appendix A

Auxiliary material

Lemma 13. If r: R→R is a nonnegative, subadditive function then the limit

r(n) ]]

lim

n→` n

exists and is positive.

Proof. By hypothesis, the function r5r(n) is subadditive which means that r(m1n)#r(m)1r(n) for any m, n[R.

This implies that, for example:

r(n)5r((n21)11)#r(n21)1r(1)#r(n22)12r(1)#r(n23)13r(1)

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Similarly:

r(n)5r((n22)12)#r(n22)1r(2)#r(n24)12r(2)#r(n26)13r(2)

n ] # ? ? ? #r(n22k)1kr(2)# ? ? ? # r(2).

2 By induction we obtain that:

n ]

r(n)# r(k), for any k51, 2, 3 . . .

which implies

r(n) r(k)

]] ]

lim sup # , for any k51, 2, 3, . . .

n→` n k

Hence,

r(k) r(n)

] ]]

lim inf #lim sup .

n→` k n→` n

But, by definition of lim sup and lim inf

r(k) r(k)

] ]

lim inf #lim sup .

n→` k k→` k

Thus, we conclude that

r(k) r(k)

] ]

lim inf 5lim sup ,

k k

k→` k→`

which proves that

r(n) ]]

lim

n→` n