THREE RESULTS IN RECURRENCE
Randall McCutcheon
Proofs of some recent results of Kriz and Forrest are presented which distinguish between four dierent kinds of \sets of recurrence" arising naturally
in connection with topological and measure-preserving dynamical systems.
0. Suppose that (X; ; ; T ) is an invertible measure preserving system,
that is, (X; ; ) is a measure space with (X ) = 1 and T : X ! X is a
bijection mod 0, with (A) = (TA) = (T A) for all measurable sets A.
(By bijection mod 0 we mean that there exists X 0 2 , (X 0 ) = 1, such
1
that the restriction of T to X 0 is a bijection.) Then
Theorem 0.1. (Poincare) If A 2 with (A) > 0, then for some
natural number n, (A \ T nA) > 0.
This is the classical Poincare recurrence theorem. A renement is
Theorem 0.2. If E is an innite subset of the natural numbers and
(A) > 0, then for some k in the set
E E = fn m : n > m; n; m 2 E g
we have (A \ T k A) > 0.
Proof. The sets T nA, n 2 E , being of equal positive measure, cannot
be pairwise disjoint in the nite measure space X , hence for some n; m 2 E ,
n > m, (T nA \ T m A) > 0. Since T is measure preserving we have
(A \ T n m A) > 0.
Theorem 0.2 motivates the following denition.
Denition 0.3. Suppose E N = the set of natural numbers. E will
be called a set of measure theoretic recurrence if for any invertible measure
preserving (X; ; ; T ) and A, (A) > 0, there exists n 2 E such that
(A \ T nA) > 0.
Hence the set E E of Theorem 0.2 is a set of measure theoretic
recurrence. Our concern here is not really invertible measure preserving
systems but subsets of N.
Denition 0.4. Suppose that B N. The upper density of B is the
number
# f1; 2; ; N g \ B
d(B) = lim sup
N
N !1
A subset E N
is called density intersective if whenever d(B ) > 0 we have
d B \ (B + n) > 0 for some n 2 E .
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Proposition 0.5. E is a set of measure theoretic recurrence if and
only if E is density intersective.
The proof of Proposition 0.5 is implicit in [Furstenberg, 1981] and explicit in [Bergelson, 1987]. Suppose now that instead of an invertible measure preserving system, we have a minimal topological dynamical system
(X; T ). Here X is a compact metric space and T : X ! X is a homeomorphism such that for every non-empty proper closed subset C X ,
C 6= T (C ).
Denition 0.6. A subset E of N is said to be a set of topological
recurrence if whenever (X; T ) is a minimal topological dynamical system,
and U is an open subset of X , there exists n 2 E such that U \ T nU 6= .
Proposition 0.7. E is a set of topological recurrence if and only if for
any minimal system (X; T ), the set of points x 2 X with the property that
whenever U is open and contains x, there exists r 2 E such that T r x 2 U ,
is residual in X .
Proof. Set
A = fx 2 X : there exists r 2 E such that d(x; T r x) < 1 g
n
n
Then each An is open and dense so that n=1 An is the required residual
set. The converse is trivial since residual sets are necessarily dense.
Denition 0.8. By an r-coloring of N we mean a partition
N=C1 [ C2 [ Cr . The indices i, 1 i r, are the colors. A subset E N is said to be r-intersective if for every r-coloring of N there
exists a color i such that (Ci Ci) \ E is non-empty. E is said to be
chromatically intersective provided it is r-intersective for all r 2 N.
Denition 0.9. E N is said to be syndetic if there exists k 2 N
such that for all n 2 N,
E \ fn; n + 1; ; n + kg =
6 If T is a homeomorphism of a compact metric space X , a point x 2 X is
called a uniformly recurrent point if for any open neighborhood U of x, the
set fn : T nx 2 U g is syndetic.
Proposition 0.10. (Birkho) If T is a homeomorphism of a compact
metric space X , then there exists a uniformly recurrent point x 2 X .
Proposition 0.11. Suppose X is a compact metric space and T :
X ! X a homeomorphism such that x 2 X is uniformly recurrent. Then
the orbit closure fT nx : n 2 Zg is a minimal T -invariant closed subset of
X.
T1
2
For proofs see, for example, [Furstenberg, 1981], p.29. We now have
the following characterization of sets of topological recurrence which is the
topological analog of Proposition 0.5.
Proposition 0.12. Suppose E N. Then E is a set of topological
recurrence if and only if E is chromatically intersective.
Proof. Suppose E is a set of topological recurrence. Let Xk =
f1; ; kgZ with the product topology, a compact metrizable space. Dene T : Xk ! Xk by (T)n = n+1. Suppose we are given a k coloring
N = C1 [ [ Ck . Let x 2 Xk , be the point with xn = j precisely when
n 2 Cj . Let Xx = fT nx : n 2 Zg. Xx is closed and T -invariant so there
exists a uniformly recurrent point 2 Xx. By the foregoing proposition
T acts minimally on X . If U = f 2 X : 0 = 0g, then U is an
open neighborhood of and there exists r 2 E such that U \ T r U 6= .
Therefore 0 = r for some 2 Xx . Since 2 Xx, this in turn implies
that xn = xn+r for some n, so n; n + r 2 Cxn and E is a set of chromatic
recurrence.
The converse is from [Forrest, 1990]. Suppose E is chromatically intersective. IfS1now (X; T ) is minimal with U X open, we have (by
i
minimality)
i=1 T U = X . By compactness, there exists k such that
Sk
i
x 2 X and consider a k-coloring of N such that n 2 Ci
i=1 T U n= X . Let
i
implies T x 2 T U . For some r 2 E , and some i, and some n 2 Ci,
n + r 2 Ci, therefore T n x and T n+r x lie in the same shift of U . It follows
that U \ T r U 6= and E is a set of topological recurrence.
The rst of three results we will prove is Theorem 1.2 below, which was
rst proved by Kriz ([Kriz, 1987].) Because of Propositions 0.5 and 0.12
it implies that sets of topological recurrence need not be sets of measure
theoretic recurrence. His treatment of the problem was in graph theoretic
terms, and resolved a question asked by V. Bergelson in 1985 (see [Bergelson, 1987] for the details of Bergelson's conjecture.) Forrest gives another
proof ([Forrest, 1990]) where he also observes that the result also resolves
negatively the following question of Furstenberg ([Furstenberg, 1981], p.76.)
Question 0.12. If S is a subset of the natural numbers N with positive
density, does there necessarily exist a syndetic set W such that W W S S?
Answer. No. By Theorem 1.2 we have a set R which is a set of
chromatic intersectivity and a set S N of positive density such that
R \ (S S ) = . For any syndetic W , R \ (W W ) 6= since nitely many
shifts of W color N and each of these shifts has the same dierence set as
W . It follows that W W cannot be contained in S S .
1. All of the proofs presented here depend heavily on an elegant idea of
Rusza's (indeed the proof of Theorem 1.2 follows [Rusza, 1987] more or less
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exactly.) All proofs of this theorem seem to make use of the following result
of Lovasz, which had been conjectured by Kneser.
Proposition 1.1. ([Lovasz, 1978]) Let E be the family of r-element
subsets of f1; 2; 2r +kg. Given any k-coloring of E , there exists a disjoint
pair of elements from E which are of the same color.
Theorem1 1.2. ([Kriz, 1987]) For every > 0 there exists A N such
that d(A) > 2 ; and C N, which is chromatically intersective, that
satisfy (A A) \ C = . (Therefore C is not density intersective.)
Proof. We claim that for every > 0; k 2 N there exist natural
numbers M; N , A0 f0; 1; 2; :::; M (N 1)g, B0 f0; 1; 2; :::; M (N 1)g,
and C 0 f0; 1; 2; ; M 1g, such that C 0 is k-intersective, (A0 + C 0) \ A0 =
, (B0 +C 0)\B0 = , (A0 ) > 21 , and (B0 ) > 12 . (here is normalized
counting measure on f0; ; MN 1g:)
We will prove the claim. Let r and N 2N be large and let p1 ; ; p2r+k
be large odd primes (how large will be specied shortly.) Putting M =
p1 p2r+k and taking 0 to be neither odd nor even let
A0 = fa : M a < (N 1)M and a(modpi ) is even for
< r indices and odd for all other i:g
B0 = fb : M b < (N 1)M and b(modpi ) is odd for
< r indices and even for all other i:g
We require N; r, and the pi 's to be so large that (A0 ) > 12 and
(B0 ) > 21 . Note that if a 2 (A0 [B0 ) then a 6= 0 (mod pi ), 1 i 2r +k.
We now let
C 0 = fc : 0 c < M and c (mod pi ) is 1 or pi 1 for 2r indices i:g
If a 2 A0 and c 2 C 0, then (a+c) (mod pi ) is even or zero for at least r indices
i, 1 i 2r + k, hence (A0 + C 0) \ A0 = . Similarly (B0 + C 0) \ B0 = .
We need to show that C 0 is k-intersective. Let
D = fd : 0 d < M 1; d = 2(mod pi) for exactly
r indices i and 1 for all other ig
D is in obvious 1 1 correspondence with the family E of r element subsets
of f1; 2; ; 2r + kg. Therefore, by Proposition 1.1, given any k coloring of
D, there exists a monochromatic pair d1 > d2 such that d1(modpi) = 2 =
d2(modpi ) never occurs, 1 i 2r + k. It follows that (d1 d2)(modpi ) is
1 or pi 1 for 2r indices, that is, d1 d2 2 C 0 and C 0 is k intersective. This
establishes our claim. We note that A0 + C 0; B0 + C 0 f0; 1; ; MN 1g.
Let > 0. Let (k )1
1 converge to zero very quickly, (how quickly will
be specied shortly.) For every k and we have numbers Mk , Nk , and sets
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Ak , Bk , and Ck having the properties stated in the previous paragraph
(with k instead of .) Notice each n 2 N is uniquely expressable as a sum
n = 1 + M1 N12 + M1N1 M2N2 3 + + M1N1 Ml Nll+1
where i 2 f0; 1; 2; ; Mi Ni 1g, 1 i l + 1.
The (k ) are chosen so that one of the sets
A1 = f1 + M1 N12 + M1N1 M2N2 3 + + M1 N1 MlNl l+1 : i 2 Ai
for an even number of i and i 2 Bi or i = 0 for all other ig
A2 = f1 + M1 N12 + M1N1 M2N2 3 + + M1 N1 MlNl l+1 : i 2 Ai
for an odd number of i and i 2 Bi or i = 0 for all other ig
has upper density > 21 . Let A be that set. Let
C = C1 [ M1N1 C2 [ M1 N1M2 N2C3 [ C is k intersective for all k. (It is routine to show that if J is k-intersective
then zJ is for any integer z.) Also (A + C ) \ A = . To see this, suppose
that
n = 1 + M1 N12 + + M1 N1 MlNl l+1 2 A
and
c = M1 N1 Mj Nj cj+1 2 C
now if j+1 = 0 then j+1 + cj+1 is non-zero and in neither Aj+1 nor
Bj+1, so n + c 2= A. If on the other hand j+1 2 Aj+1 (Bj+1), then
j+1 + cj+1 2= Aj+1 (Bj+1 ) and again n + c 2= A. This completes the proof
of Theorem 1.2.
2. We prove in this section two results of Forrest, which again provided
answers to questions of Bergelson.
Denition 2.1. A subset C N is called a set of strong recurrence
if, whenever B N with d(B) > 0,
lim sup d (c + B) \ B > 0
c2C;c!1
If in fact for all such B
lim sup d (c + B) \ B d(B)
2
c2C;c!1
then C is called set of nice recurrence.
By Theorem 2.5 below these two properties are not equivalent. Bergelson proved ([Bergelson, 1985]) that if D is a set of strong recurrence and
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A N2 has positive density then there exists an innite B D such
that B B A A. He then asked whether there were sets of (measuretheoretic) recurrence which are not sets of strong recurrence. If not, then
the hypotheses of the aforementioned result could be weakened. Forrest
showed however that there are such sets.
Our proof of Theorem 2.4 diers from Forrest's in that it uses products
of cyclic prime-order groups similar to that used in the proof of Theorem
1.2. We do follow Forrest in the use of the following lemmas.
Lemma 2.2. ([Kleitman, 1966]) If C f0; 1g2M , and for each a; b 2
C , ai = bi for at least 2J indices i, then
jC j X 2M iM J
i
Lemma 2.3. (Central Limit Theorem) If (X; ) is a probability space,
Ai X , i 2 N are independent subsets of measure , 0 < < 1, then for
every real number a
lim x :
k!1
k
X
i=1
p
1Ai (x) k a k(1 ) = F (a)
where F (a) = p12 a1 e 2t dt is the distribution function determined by
the standard normal distribution. F is a continuous and increasing, with
F (0) = 21 , F (a) ! 0 as a ! 1, and F (a) ! 1 as a ! 1. Another form
for F (a) is
R
2
lim k (x1 ; x2 ; ; xk ) 2 X k :
k!1
k
X
i=1
p
1Ai (xi ) k a k(1 )
Theorem 2.4. ([Forrest, 1991]) There exists a set of recurrence C N which is not a set of strong recurrence.
Proof. Let T be the unit circle in the complex plane. Let be the
usual Lebesgue probability measure on T. Then k is the usual measure on
the k torus Tk . We claim that
(1) Forpall M > 0 and > 0, there exist arbitrarily large k; N , and J , with
J M Nk, and W Tk , such that k (W ) = and for all r 2 R, where
R = f(x ; x ; ; xk ) 2 Tk : jarg xi j N
(2.1)
for at least k 2J indices ig
1
2
1
2
2
6
T
we have k (Wr) W < 3T. R has the property that for any measurable
subset B Tk with BB 1 R = , we have
k ( B ) 2Nk
2
Nk
2 iNk J i
1
X
(2:2)
By Lemma 2.3 (with = 21 ) the right-hand side is asymptotically
J , which will go to zero as pJ ! 1, or as M ! 1.
F p Nk
Nk
2
We now make some observations before proving (1). For all N there
exists a measure-preserving map gN : T! f0; 1g2N such that if gN (x) i 6=
gN (y) i for every i; 1 i 2N , then j arg xy 1 j 2N . It follows that
if j arg xy 1 j > 2N then gN (x) i = gN (y) for at least one i. (The
required maps are pie partitions with opposite slices going to reversed bit
patterns. For example, if N = 1 send the rst quadrant to (1; 1), the third
to (0; 0), the second to (1; 0), and the fourth to (0; 1).) Let
hN;k : Tk ! f0; 1g2Nk = f0; 1g2N
k
be the map induced by gN . Now, for x; y 2 Tk , if hN;k(x) i 6= hN;k (y) i
for at least 2Nk 2J indices i, then jarg xj yj 1 j 2N for at least k 2J
indices j , hence xy 1 2 R.
T
We verify (2.2). Let B Tk . If BB 1 R = then for all x = (xi ),
y = (yi ) 2 B, j arg xi yi 1 j > 2N for at least 2J indices i, 1 i k,
hence hN;k (x) i = hN;k (y) i for at least 2J indices i, 1 i 2Nk, and
by Kleitman's lemma
2Nk
jhN;k(B)j i
iNk J
X
But hN;k is measure-preserving, so (2.2) is satised.
We now complete the proof of (1). By Lemma 2.3 we may pick t > 0
small enough that if
p
Wk = fx 2 Tk : Re xi 0 for at least k2 + t k indices ig
(2:3)
. Fix a large N to be determined and let k,
then limk!1 k (Wk ) > 21 p
J ! 1 according to J M Nk. In particular we assume that J < k2.
For each choice of k; J , consider the set R of (2:1). supr2R k Wk \ (Wk r)
is attained for such an r with ri = 1 for 2J indices i, and
jarg ri j = 2N
for the other k 2J indices. (Actually k Wk \ (Wk x) is increasing in Re
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xi , 1 i k.) For such an r, and x 2 Wk \ (Wk r), we have x 2 Wk and
xr 1 2 Wk . A look at (2.3) shows us that one of the following must occur:
p
indices i, 1 i 2J , or
(a) Re xi > 0 for at least J + 21 t k p
1
1
(b) Re xi ri > 0 for at least J + 2 t pk indices i, 1 i 2J , or
(c) Re xi > 0 for at least k2 J + 21 t k indices i, 2j + 1 i k, and the
same is true of Re xi ri 1 .
p
Since J increases as k and t > 0, the probability of (a), which is
to say the measure of the set of points x = (xi )ki=1 2 Tk satisfying (a),
goes to zero by Lemma 2.3. The same is true for (b). As for (c), when
2J + 1 i k, the probability that Re xi > 0 and Re xi ri 1 > 0 is
2 (N +1), as is the probability that these
p are both negative. (c) implies
that the former happens for at least t k more indices i than the latter.
By Lemma 2.3 the probability of that happening for x 2 Tk can be made
arbitrarily small by choosing N large enough. p
In other words, N can be
chosen so that for large enough k and J M Nk, k Wk \ (Wk r) is
small for all r 2 R. The proof of claim (1) is completed by picking k odd
and letting
W = fx 2 Tk : Re x1 > 0 for at least k2 indices ig
Thus far it has been convenient to work in the torus Tk because of
the naturalness of applying Lemma 2.3. We will denote by Cq the cyclic
group f0; ; q 1g under addition mod q. Let q be normalized counting
measure on Cq . What we really will need is the following:
(2) For every > 0, there exists M = p1 p2 pk , where pi are distinct
primes, 1 i k, and R CM , such that if A CM , M (A) > ;
then there exists r 2 R such that M A \ (A + r) > 0; and W CM ;,
M (W ) 21 , such that for every r 2 R, M W \ (W + r) < and
M W c \ (W c + r) < (2) is true for the same reasons as (1). Instead of working in Tk , however,
one works in the natural subgroup Cp1 Cpk Tk , where p1 ; ; pk
are large enough. This subgroup is of course isomorphic to CM .
We now complete the proof of the theorem. For a sequence fk g converging to zero (at a rate to be determined shortly,) let Mk , Wk , and
Rk be as in (2) and let Nk be so large that Mk Nk (Ak ) > 12 k and
Mk Nk (Bk ) > 12 k where
Ak = fa : Mk a < (Nk 1)Mk and a(mod Mk ) 2 Wk g
Bk = fb : Mk b < (Nk 1)Mk and b(mod Mk ) 2 Wkcg
8
Let Ck = Rk considered
c 2 Ck , we have
now as a subset of N. For each
Mk Nk Ak \ (Ak + c) < k and Mk Nk Bk \ (Bk + c) < k . The fk g are
chosen so that the following set has upper density greater than 1 :
1 + M1N1 2 + + M1 N1 Mk Nk k+1 : i 2 Ai [ Bi [ f0g
Let F be the subset of the above set consisting of those numbers that have
i 2 Ai for an even number of indices i. Then d(F ) > 21 . Now let
C = C1 [ M1N1 C2 [ M1 N1M2 N2C3 [ We claim that Ci has the property
that if D N, d(D) > i , then there
exists r 2 Ci with d (D + r) \ D > jCijiMi . Suppose then that d(D) > i .
Then there exists an sequence (at ) N satisfying at+1 at > Mi , with
d (at ) > Mii , such that for each t,
D
\ fat ; at + 1; ; at + Mi 1g > i Mi
Recall that Ci = Ri viewed as a subset of N. A glance at the denition of
Ri shows us that r 2 Ci if and only if Mi r 2 Ci. This fact, coupled with
the intersectivity property of Ri, shows there exist a; b 2 f0; 1; ; Mi 1g,
with a < b, such that at + a 2 D, at + b 2 D, and r = b a 2 Ci. It follows
from the rst observation that at + b 2 D \ (D + r). Therefore we have
[
d
(D \ D + r) > d (at ) > Mi
i
r2Ci
Hence for some r 2 Ci we have d (D + r) \ D > jCijiMi , establishing
our claim. One routinely veries that the claim is true for Ci replaced by
any constant multiple of Ci. From this fact it follows that C is a set of
recurrence.
However, by construction
lim d F \ (F + c) = 0
c2C;c!1
because, if c = M1N1 Mn 1 Nn 1cn, with cn 2 Cn, then the only elements of F \ (F + c) are those numbers
1 + M1N1 2 + + M1 N1 Mk Nk ak+1
with n 2 An and n cn 2 An (or else both these in Bn.) The set of
natural numbers with this property is less than two times the density of
9
An \ (An + cn) in f0; 1; ; MnNn 1g, that is, less than 2n. Hence C is
not a set of strong recurrence. This completes the proof of Theorem 2.4.
Finally we have another result of Forrest.
Theorem 2.5. ([Forrest, 1990]) There exists a set C~ N which is a
set of strong recurrence but not a set of nice recurrence.
Proof: Let us return to the notation of the proof of Theorem 2.4 and let
C~1 = C1; C~2 = C1; C~3 = C2; C~4 = C1; C~5 = C2; C~6 = C3; C~7 = C1; , etc.
Dene similarly N~k ; M~ k ; A~k , and B~k . Following the construction of the
previous proof with these new sets we obtain C~ and F~ . According to the
claim at the end of the proof of Theorem 2.4, C~ is a set of strong recurrence.
C~ is not, however, a set of nice recurrence, since d(F~ \ (F~ + c)) < 21 for
all c 2 C~ , just as at the end of the proof of Theorem 2.4.
References
1. V. Bergelson, Ergodic Ramsey Theory, Contemporary Mathematics, 65
(1987), 63-87.
2. V.Bergelson, Sets of recurrence of Z m -actions and properties of sets of
dierences in Z m, J. London Math. Soc. (2) 31 (1985), 295-304.
3. A. H. Forrest, Doctoral thesis, The Ohio State University, 1990.
4. A. H. Forrest, The Construction of a set of recurrence which is not a set
of strong recurrence, Israel J. Math. 76 (1991), 215-228.
5. H. Furstenberg, Recurrence in Ergodic Theory and Combinatorial Number Theory, Princeton University Press, 1981.
6. D. J. Kleitman, On a Combinatorial Conjecture of Erdos, J. Comb.
Theory 1 (1966), 209-214.
7. I. Kriz, Large independent sets in shift-invariant graphs. Solution of
Bergelson's problem, Graphs and Combinatorics 3 (1987), 145-158.
8. L. Lovasz, Kneser's conjecture, chromatic number and homotopy, J.
Comb. Theory (A) 25 (1978), 319-324.
9. Rusza, Dierence Sets and the Bohr Topology, 1987, Preprint
Department of Mathematics, The Ohio State University, Columbus, Ohio
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