G. a. Hedlund -- Endomorphisms and Automorphisms of the Shift Dynamical System

Endomorphisms and Automorphisms of the Shift Dynamical System

by

G. A. HEDLUND*

Yale University New Haven, Connecticut

Introduction

Let X(Se) be the set of all bisequences over a symbol set 6 a, where 1 < card S# < 0% and let cr be the shift transformation. If the product topology induced by the discrete topology of 6: is assigned to X(6a), X(6 a) is homeomorphic to the Cantor discontinuum and ~ is a homeomorphism of X(6Q onto X(6Q. The discrete flow (X(SQ, a) is the symbolic flow over 5: or the shift dynamical system over S a.

The shift dynamical system (X(S:), a) has been analyzed rather thoroughly, both in its topological and in its measure-theoretic aspects [1, 7, 9, 10, 13, 14]. Recent work of Smale [27] shows that the shift dynamical system is ubiquitous.

Any closed subset of X(SQ which is invariant under a defines a subdynamical system. There is an endless variety of these and they have served as useful models to indicate possible structures of dynamical systems, particularly minimal sets [7, 8, 9, 10, 11, 12, 15, 20, 22, 23, 24, 25]. These systems are characterized by the fact that the phase space is totally disconnected and the transformation is expansive (Section 2). However, there are dynamical systems, notably geodesic flows on compact manifolds of negative curvature, for which the phase space is a manifold, yet the orbits can be characterized by symbolic bisequences [2, 3, 4, 5, 20, 21, 28]. Properties of such dynamical systems and their subdynamical systems can be determined from knowledge of the properties of symbolic flows.

A natural question in connection with any dynamical system is that of the existence and properties of continuous transformations which commute with the group action. In the case of the system (X(S:), ~), an obvious example of such a transformation is obtained by simply permuting the symbols. A generalization of this is to define a mapping of blocks (words) of symbols of a specified length into single symbols and to extend this mapping in a natural manner to infinite sequences. It has been shown by Curtis, Hedlund and Lyndon that these mappings, composed with powers of the shift, constitute the entire class of continuous transformations which commute with the shift (Section 3). This fundamental

* This research was partially supported by the Army Research Office (Durham).

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MA'rHEMATICAL Sy~fFA~k~ THEORY, Vo1. 3, NO. 4. Published by Springer-Vcrlag New Y o r k Inc.

Endomorphisms and Automorphisms of the Shift Dynamical System 321

structure theorem makes possible the derivation of numerous results concerning the shift commuting transformations.

Let *(d/') be the set of all continuous transformations of X(~9 ~) into X(SQ which commute with the shift. These 'are the endomorphisms of the dynamical system (X(~q'), g). The subset of ~(S#) consisting of those members which are onto maps will be denoted by E(S#). The subset of (I)(6:) consisting of those members of ~(~9 ~) which are one-to-one maps will be denoted by A(Sa). It is shown in Section 5 that A(6Q c E(6 a) and hence the members of A(6:) are homeomorphisms of X(6 a) onto X(6a). These are the automorphisms of (X(SQ, cr).

The set qb(S:) is countable and the set ~(6:) - E(6Q has infinitely many members, as does the set E(SQ, excluding powers of the shift and the set A(S:) (Section 6). It appears that the set qb(~9 a) - E(dQ is in some sense more numerous than the set E(S:) and that the set A(6 a) is relatively sparse, but nothing has been proved in this direction. Since the members of A(S#) are onto homeomorphisms, the set A(S:) is a group and is of some complexity. Any finite group is isomorphic to some subgroup of A(6Q (Section 6) and A(6 a) contains two elements ~ and fi, each of order two, such that ~/3 is of infinite order (Section 20).

There are sharp differences in behavior with respect to multiplicities between the members of E(6:) and the members of (I)(6:)- E(SQ. If 9 e (I)(6:)- E(SQ, there exists a periodic point x such that the set 9-1(x) is uncountable (Section 5). On the other hand, if 9 ~ E(S:), though card 9-1(x) may vary from point to point, it is uniformly bounded over X(6 a) (Section 5). But for such 9, even though card 9 - t(x) is not everywhere the same, A. M. Gleason and L. R. Welch have shown that card 9-1(x) is almost everywhere constant in the sense that there exists an integer K(9) such that card 9-1(x) = K(9) if x is bilaterally transitive (Section 11).

If 9 E E(S#) and 9(Y) = 9(z) with y # z, then the orbits of y and z must be separated in at least one sense (Section 8). In particular, the inverses of a recurrent point must be separated in both senses. On the other hand, if 9 ~ *(~9~) -E(~9~), then there may be distinct bilaterally asymptotic points with the same image

• under 9. For all 9 ~ qb(~9~), orbit properties such as periodicity, almost periodicity,

recurrence and transitivity are preserved under 9. If 9 ~ E(6Q, then the (usually multivalued) correspondence 9-1 likewise preserves these properties (Section 12), but this is not at all the case if 9 ~ ¢(6:) - E(S~).

The set of mappings qb(S:) is a semigroup under composition. If 9~b ~ E(6a), then 9 e E(6Q and ~b e E(6Q (Section 13). With those block mapping which define members of E(S:), L. R. Welch has associated certain integer-valued functions which are multiplicative under composition (Section 14). With the aid of these it is possible to answer questions as to the existence of roots of members of E(S,0. For example, the shift transformation cr has no continuous roots if card 6 a is a prime (Section 18).

Finally (Section 16) O. S. Rothaus has succeeded in characterizing those members of E(6Q which have a cross-section.

This paper gives the detailed proofs promised in [17]. As stated in [17], there have been a number of contributors to developments

in this area, results obtained over a period of some ten years, and which have

322 G.A. HEDLUND

not appeared in print. The writer has tried to make accurate citations of those who were the originators of fundamental results, but, under the circumstances, it is difficult to be completely accurate. In any case, it should be clear that this paper is the work of many.

1. Bisequence Space and the Shift Dynamical System

Let I denote the set of all integers. For i c I , let I i = { j [ j ~ L j >- i}. The cardinal of a set E will be denoted by card E.

Let S e 11 and let Sa be a set with card 6 a = S. The set Se is the symbol set and any element o f r a is a symbol. A convenient choice of Sa is the set {0, 1,. • . , S - l } .

A bisequence over SP is a function on I to 6 a. Let X(6 a) denote the set of all bisequences over oq', that, is, X(S a) = S ~x. The set X(6 a) is the bisequence set over S a. If x ~ X(5 p) and i ~/, then x(i) will often be denoted by x~.

We define a metric d on X(Sa). Let x, y ~ X(S~'). If x = y, then d(x, y) = O. If x ~ y, let k be the least non-negative integer such that either x k ~ Yk or X-k ~ Y-k, and define d(x, y) = (1 +k)-1 . It is easily verified that d is a metric of X(S a) and that the metric topology induced by d coincides with the product topology induced by the discrete topology of SP. If S = 1, X(SP) contains just one point. If S > 1, then X(6 a) is a compact, totally disconnected, perfect, metric space and hence is homeomorphic to the Cantor discontinuum. The metric space X(6 a) is the bisequence space over S a.

The shift or shift transformation is the mapping cr: X(ra )~X(S , ~) defined by [cr(x)]i = xi+l, x e X(S#), i ~ L The discrete flow (X(ra), ~) is also called the symbolic flow over So or the shift dynamical system over S a.

2. Subdynamical Systems of (X(~), o) and their Characterization

A transformation group (X, T), where X is a metric space with metric d, is expansive provided that there exists a positive real number 8 such that corresponding to x, y c X, with x ~ y, there exists t E T such that d(xt, yt) > 3. The discrete flow (X(ra), o) is expansive. For if x, y ~ X(6~), with x ~ y, then there exists i ~ I such that xi ~ Yi and hence d(cri(x), oi(y)) = 1.

Let Y be a closed, nonvacuous subset of X(Se) which is invariant under o. Then the restriction of g to Y is a homeomorphism of Y onto Y and defines a discrete flow (Y, g) which is variously called a subflow of (X(ra), o), a subdynamical system of (X(6~), ~), or a subspace restriction of(X(rP), o). Clearly, Yis a compact, totally disconnected, metric space and (Y, ~) is expansive. We show that any such discrete flow is isomorphic to a discrete subflow of some symbolic flow.

2.1 THEOREM. Let Z be a compact, totally disconnected, metric space and let ~o be a homeomorphism of Z onto Z. Then the following statements are equivalent.

(1) (Z, ~) is expansive. (2) There exists a symbol set S a and a subdynamical system ( Y, or) o f (X(S~), o)

such that (Z, ~o) is isomorphic to (Y, or).

Proof. Since expansiveness is preserved under isomorphism in the case of a compact space, (2) implies (i).

Assume (1). Let p be a compat ib le metric o f Z. By hypothesis there exists 8 > 0 such that if zl , z2 e Z with z t # z2, then there exists n e I such that [9"(zi), 9n(z2)] _> 8. Since Z is a compact , totally disconnected, metric space, there exists a decompos i t ion ~ = {Ui[ i = 0, 1,. • . , S - 1 } o f Z into disjoint non- vacuous sets such tha t each Ui is open and closed and each Ui is o f d iameter less than & Corresponding to z ~ Z, define $(z) = x ~ X(S~) by x n = i provided 9"(z) ~ U~. Let $(Z) = r c X(S~).

We show that $ is cont inuous. Since the members of q / a r e disjoint and closed, there exists ~ > 0 such that if u ~ Ui and v E Uj with i # j , then p(u, v) > ~.

Let z 1 ~ Z and let ~ > 0. Choose k e 11 such that (1 + k ) - 1 < ,. Since 9 is cont inuous, there exists 3 > 0 such that if z ~ Z and p(z , z l ) < fl,

then p[gn(z), 9"(zl)] < ~, {n = 0, + 1 , " ", + k }. Let m s / , Ira] _< k. Then 9re(z1) e Ui

for some i and 9"(Z) ~ U j for s o m e j . I f i # j we would have p[gm(z), 9r"(zl)] > ~, which is not the case, so tha t i = j , and hence [~(z)]n -- [~(zl)],(n = 0, + 1 , " ", + k), which implies tha t d[~(z), ~(zl)] < (i + k ) - 1 < e provided tha t o(z, zl) < 3. Thus ~ is cont inuous at each z e Z.

Since Z is compac t and ~ is continuous, Y = ~(Z) is compac t and hence closed in X(5~).

To show t h a t ¢ is h o m e o m o r p h i s m o f Z onto Y, it is sufficient to show that is one-to-one. Let u, v e Z with u # v. There exists n e I such that p[9"(u),

9"(v)] _> & But then 9"(u) ~ Ui, 9" (v ) e U j with i # j and thus [$(u)], = i # j = [~(v)],; hence ~(u) # ~(v). Thus $ is a h o m e o m o r p h i s m of Z onto r = ¢(Z).

We show that for all z s Z, e~(z) = $9(z). Let n e I. Then [~9(z)], is the subscript i o f the set Ui containing 9" ÷ x(z). But (e~(z)] n = [~(z)], + 1 is likewise the subscript i o f the set Ui containing 9 n + l(z). Thus ~r~(z) = ~9(z).

The set Y -- ~ ( Z ) is invar iant under ~r. For let y s Y. Then there exists z e Z such that ~(z) = y. But then ~r(y) = e$(z) = d/9(z) e Y.

Thus (Y, e) is a subdynamical system o f (X(S~), ~) and ~ defines an iso- morph i sm o f (Z , 9) onto (Y, e).

The p r o o f is completed. This result will also be found as a corol lary to a theorem o f Will iam L. Reddy

[26].

3. A Class of Mappings which Commute with the Shift

Let n E 11. An n - b l o c k over 5 f is an ordered set x t x 2 • • • x n, where x i ~ ~9 ~

(i = 1, 2," • ", n). Let A = a l a 2. • "an be an n-block o v e r ~ and let B = b i b 2 , • .bm

be an m-block over d f . Then A appear s in B if and only if there exists an integer i,

0 < i < m - n , such that b i + l b i + 2. • "bi+n = a la2" • "an.

Let x E X(d f ) and let A be an n-block over dr. Then A appears in x if and only if there exists i~ I such that x i+ t x i + 2. • . x i + n = A .

Let n e I1 and let ~ , ( 5 f ) denote the set o f all n-blocks over ~ . Let f be a mapp ing of ~,(~9 ~) into ~ 1 ( S f ) = 5 p. The set o f all such mappings for a given n ~ 11 and a given symbol set ~ will be denoted by F ( ~ , n). We note that card ~n(Sf ) = S n and card F ( ~ , n) = S s".

L e t f E F(o O~, n) and let m e 11. Corresponding t o f a n d m, we define a mapp ing f , , : ~ m + , - 1(~9~)--~=(5f) as follows. L e t B = b l " " " bin+n- 1 e ~ m ÷ n - 1(~) and let

324 G . A . HEDLUND

a~ = f ( b ~ b g + l ' " b i + , _ 0 (i = 1, 2 , . . . , m ) . Then A = ala2. . .am~m(~C/7) and we definef, , (B) = A.

L e t f e F(50, n). Corresponding to f , we define a m a p p i n g f ~ :X(SP)-+X(50) as follows. Let x ~ X(50) and let y e X(50) be defined by Yi = f (x ix i+ 1"" "xi+,_ 1) (i e I) . By definition, fo~(x) = y.

3.1 T H E O R E M . Let f e F(50, n). Then f~o: X(5'O--~X(5/~) is continuous and commutes with the shift o.

Proof. L e t f e F(50, n), let x ~ 2 ( 5 p) and l e t f~ (x ) -- y. Let ~ > 0 and let k ~ 1 1 such that (1 + k ) -x <e. Choose 3 > 0 such that 3<(1 + k + n ) -1 and let u ~ X(50) with d(x, u ) < & Then x i = u i (i = O, +_ 1 , . . . , +_(k+n)). Let f~o(u ) = v. Then y~ -- vi (i = 0, _+ 1 , . . . , _+ k); hence d(y, v) = d(fo~(x), f~(u)) < E. I t follows that f~o is continuous.

Let x e X(50), y = f~ (x ) . Then for i e I, [~f~(x)] i -- [¢(y)]i = Yi+ x = f ( x i+ 1 "" " xi+,), [fooa(x)]* = f([e(x)]i" " "[a(x)]i+,- 1) = f ( x i+ 1"" " xi+,). Thus afoo(x ) =

fore(x) a n d f ~ commutes with ~. Let F~o(50, n) -- { f ~ o l f e F ( S , n)}.

3.2 Remark. I f 1 < m < n then F~o(50, m) c F~o(50, n). Proof. L e t f e F(50, m). Define g: ~ ( 5 Q - - > ~ 1 ( 5 Q b y g ( x l . . , x,) = f ( x x . . , xm).

Thenf~o = g o e Foo(50, n).

3.3 L E M M A . Let n e I t and let 50 be a symbol set with card 50 > 1. Then a k e Foo(50, n) i f and only ifO < k <_ n - 1.

Proof. Let n e /1 and let k be an integer, 0 <_ k _ n - 1. Define f : N , ( 5 0 ) ~ .~1(5 p) b y f ( x l x 2 " " x , ) = Xk+I. Then ek = f~o ~ Foo(50, n).

N o w suppose tha t k ~ I and cr k s F~(50, n). Then there e x i s t s f s F(50, n) such that ek = foo, and hencefo~ ~ - k = ~ - kfo ~ is the identity map on X(SQ and [~- ~ o (x)]i

= [fo~(X)]i-k = f ( X i - k ' " "Xi -k+,- 1) = Xl for all x e X(Sg) and all i s L Choosing i = k + l, we h a v e f ( x l - . - x ~ ) = Xk+ 1 for all x. Suppose that k < 0 or k > n - 1 . Define x by choosing xi = 0 if i ¢ k + 1 and Xk+ 1 # f ( x l " " "X,). I t follows that a k ¢ F~(~9 °, n),

The p roo f is completed. Let F~(SQ denote the set ~ 2 = 1F~o(50, n). We observe that eke F~(5") pro-

vided k > 0, but when card 50> 1, ek ¢F~(50) if k < 0 . Let F*(50) denote the set {d"q~ I m s /, ~0 s F~(50)} and let q5(50) be the

set o f all cont inuous mappings of X(50) into X(50) which commute with e. Clearly, F~(50) c F * ( S Q c ~(SQ, and if card 5 ° = 1, these sets are identical;

otherwise F®(50) is a p roper subset o f F*(50). The basic result that the last two sets are identical is due to M. L. Curtis, G. A. Hedlund and R. C. Lyndon.

3.4 T H E O R E M . F*(50) = 0(50). Proof. Let 50 = { 0 , . , . . . , S - 1 }. Let ~o ~ ~ ( 5 0 . For each i e 50, let U~ =

{x] x ~ X ( 5 0 ) , x o = i). Then U~ is open and closed, U f ~ U j = ~ if i # j , and S - X X(50) = ~ i = o U v Let Vi = q~-~(U~) (i e50). Then each Vi is open and closed,

S - 1 V i ~ Vs = ~ if i # j , and X(5~) = ~ ~ = o Vv F r o m the propert ies of the sets V~, there exists k e I x such that if x ~ V~ and

y e Vj with i # j , then d(x, y) >_ (1 + k ) -~ Let x e X(50) and let m be a non-negat ive integer. The central (2m + O.block

of x is the block x-m" • "X_lXoXx" " "xm.

For each i e 6 a, let ~2~ be the collection o f all (2k + 1)-blocks B over 6 a with the property that there exists x E V~ such that B is the central ( 2 k + 1)-block of x. I f i ~ j , then ~ j = o. For if B E ~ i ~ M s , there exists x e V~, x with central (2k + 1)-block B, and y ~ Vj, y with central (2k + 1)-block B. But then d(x, y ) < (1 + k) = 1, which is not the case. Since X(S/~) s -1 / 1 S - 1 ~ = Ui=0Vi , it follows that w i = o i is the set ~2k+ 1(5a) o f all (2k + 1)-blocks over SP.

Let f : M2k+ l(Se) ---~9° be defined by f (A2k ÷ l) = i provided that A2k + 1 e ~ i . Let x E X(Sce), ~o(x) = y andfo~(x) = z. We show that Yo = Z-k" There exists

a unique i e 6 a such that x e V~ and hence, if B is the central ( 2 k + 1)-block of x, then B e M i , f ( B ) = i and z_~ = i. But since ~o(x) = y and x ~ Vi = ~- i (u~) ,

we have y = ~o(x) e Ui and Yo = i. Thus Yo = i = Z_k. N o w let u ~ X(6~), c~(u) = v, f~ (u) = w and n ~ L Let x = d'(u), y = e~(v),

z = ~r~(w). Then ~o(x) = ~0d'(u) = e"p(u) = y and fo~(x) = foyerS(u) = ¢r~fo~(u) = e"(w) = z. We have shown that Y0 = Z-k. Since y = e'(v), Yo = [e~(v)]0 = v,. Since z = d'(w), z_~ = [~"(W)]-k = W-k+,. We conclude that v, = w,_ k for all n e l a n d hence ~o(u) = v = ~r-k(w) = e-kfo~(U ). Since u was arbitrary, it follows that ~o = tr-~/o~, which is a member o f F*(Sa). Thus O(oq o) ~ F*(6e). Since F*(Ne) ~ O(6a), the p roo f is completed.

It follows f rom Theorem 3.4 that card O(6 a) = ~o. For card F(N~,n) = S s" and hence F~(S e) and F * ( 5 e) are countable sets. Since ~r ~ s F*(S¢'), n e / , and e" ¢ e'~ if n ¢ m, it follows that card F*(Se) = ~o.

Standing Hypothesis. Throughout the remainder of this paper we assume that card o c~ > 1.

4. Properties of Foo(,~', 1)

W h e n f e F (5 a, 1) , f i s a mapping o f S p into b ° and the behavior off~o is easily determined.

4.1 T H E O R E M . Let f e F( 6 a, 1). Then the following statements are equivalent.

(1) f is a permutation o f S& (2) f t , : Mm(SP)--~,,(6¢~) is onto f o r each m e I 1. (3) card fro- I(B) = 1 f o r each m e I 1 and each B e ~m(6e). (4) fo~ is onto, i.e., f~o(X(SP)) = X(Na). (5) foo is a homeomorphism o f X ( Sa).

Proof. Assume (1). Let m e l l , let B = Y i Y 2 " " Y m ~m( : ,q ' ) and let xi = f - l ( y ~ ) (i = 1, 2 , . . - , m). Then A = X l X z ' " x , , e M , , ( S ~ and fro(A) = B. This proves (2).

Assume (2). This implies that fm is a permutat ion o f ~m(oq a) for each m e 11 and this implies (3).

Assume (3). Let y e X(Sa). Let x e X ( 6 a) be defined by x i = f Z l(yi) = f - l(yi) (i e 1). Then foo(x) = y. This proves (4).

Assume (4). To prove (5) it is sufficient to prove that foo is one-to-one. Suppose there exist x, y e X(6a) , with x ¢ y, such that f~o(x) = foo(Y). Since x # y, there exists i e / s u c h that x~ ~ y~ and we havef(x~) = f(y~), But then there exists s E 6 ¢" such that f - l(s) = ~. Let z e X ( b °) be defined by zi = s, i e L Then z Cfoo(X(Na)), contrary to (4). This proves (5).

326 G.A. HEDLUND

Assume (5). I f f i s not a permutation, hence not onto, then, as in the preceding proof, f® is not onto. This proves (1).

The proof is completed. Let n ~ 11 and l e t f ~ F(SP, n). Since fo~ is continuous and X ( ~ ) is compact,

f ~ ( X ( ~ ) ) is compact and hence closed. Let Y =f~(X(9~)) . Since f~ commutes with ~, Yis invariant under ~ and thus (Y, ~) is a subdynamical system of (X(S°), o). If n = 1 a n d f i s not a permutation of~9 °, then Y is a proper subset of X ( ~ ) . In this case it is easy to determine the structure of (Y, ~) and the multiplicities off~o.

4.2 Remark. Let f ~ F(S/' , 1) and let f ~ be not onto. Let J " = f ( S '~) and thus card J - < card 5 °. Let Y = f ~ (X( 5#)). Then the subdynamical system ( Y, o) is the shift dynamical system (X(~-"), or).

4.3 Remark. Let f ~ F( 5 a, 1) and let foo be not onto. Then there exists x ~ X ( Se) such that x is f i x ed under ~ and the set f~o l(x) is uncountable.

Proo f Sincefo~ is not o n t o , f is not a permutation of Sa and there exist i , j ~ 5 a with i # j such tha t f ( i ) = k = f ( j ) .Le t x be the bisequence defined by x, = k, n ~ L Then the se t f~ l(x) includes all bisequences y defined by y, = i or j (n e I). This set is uncountable.

L e t f e F(5 a, 1) and let f® be not onto. Then ~-- = f ( S ¢~) is a proper subset of 5 p. Let ~--= { t i [ i = 1, 2 , . - - , T}, T = c a r d J ' < S . Let I V / = c a r d f - l ( q ) (i = 1, 2,- . . , T. We can assume that Nx < N2 < • "" < Nr. Then Nr > 1. Let r = f~(X(~9~)).

4.4 Remark. I f N 1 > 1, then for each y ~ Y, the set f ~ 1 is uncountable. We assume that N1 = 1. Then there exists an integer k, 1 < k < T, such that

1 = N t = " " = N k , Nk+l> l .

4.5 Remark {cardfd l (y) l y e Y, c a r d f £ ~ ( y ) < ~ } =

{Nk~.]'N~.+2~" " N ~ l p, ~ I o, i = k + 1 , . . . , T}.

Let W = {tili = 1 , . . . , k}.

4.6 Remark. {Yl Y z Y, c a r d f ~ X(y) = 1 } = X(W).

5. Multiplicities of the Mappings fm and foo

From Theorem 4.1 it follows that i f f E F(50, 1), then foo is onto if and only if each of the mappings fro is onto. The same result holds for arbitrary n.

5.1 THEOREM. Let n ~ 11 and let f ~ F( ~ ¢, n). Then foo is onto i f and only

i f each f~: ~m+, - 1(~9°)--~m(~9 ¢) is onto for all m ~ 11. Proo f We assume that f~ is onto. Let m E/1 and let B = bxb2 . . , bm ~ ~m(~P).

Let x ~ X(5 ¢) be defined by xi = 0 if i > m or i _< 0 and x l x 2 • • • xm = bib 2 • • • bm.

There exists y ~ X(Sg) such that f ~ ( y ) = x. Let A --YlY2"''Yra+n-l" Then f~(A) = B; hencef~ is onto.

Assume that fro : 5~m+,- 1(~'~')--~5~m(5 p) is onto for each m ~ I1. Let x e X(dg) and let k ~/1. Let m = 2 k + l . Then B = X _ k ' ' ' X k E~m(~P~ ) and there exists A = Y - k ' ' ' Y k + , - 1 E ~ m + . - l ( ~ ) such that fro(A) = B. Define y E X ( 5 ~ ) by Y - k ' ' ' Y k + n - 1 = . 4 , Yi = 0 if i < - k or i > k + n - 1 , and let f~o(y)= u. Then u_ k" " "Uk = X_ k" " "Xk and d(x, u) < (1 + k ) - 1. Since k is arbitrary in 11 it follows that the set foo(X(S¢))is dense in X(d,¢). But sincefoois continuous and X(dg) is


compact, f~ (X(de) is compact and hence closed, and f~ (X(de) = foo(X(de)) = x(de).

The proof is completed. W h e n f e F(de, n) with n = 1, it follows from Theorem 4.1 t h a t f ~ is onto if

and only if cardf~- l(B) = 1 for each m e I i and each B ~ Mm(de). That this is not the case when n > 1 is easily verified by consideration of particular examples. For arbitrary n, it has been shown by W. A. Blankenship and O. S. Rothaus that

f ~ F(de, n) and foo onto implies that card f~- l (B) = S "-1 for each m e I 1 and each B e ~m(de), thus implying the stated result for n = 1. This result will be derived as Theorem 5.4.

Let A = a l ' " "am e ~m(de) and let B = b l . . . b, EM,(de). Then AB will denote the (m + n)-block a 1 ' • • ambl" " "b,. Let M and d be collections of blocks over de. Then d M will denote the collection {AB[ A ~ d , B ~ M}. Generalizations of these notations are obvious.

5.2 LEMMA. Let n ~ I l, let f e F( ~ a, n) and let there exist an m-block B over de such that card f~- l (B) = k > 0 and card f~S+li(Bb) > k for each b ~de. Then card f,S+ll(Bb) = k for each b ~ de.

PrOof. Let B be an m-block over 6 e such that card f,,- I(B) = k > 0 and let f,,- I(B) = {Ci, C2," • ' , Ck} = cg. Then fm+~x(B5 e) = c~de. For f,,+ ~(Cc) e Bde if and only if C is an (m + n - 1 ) - b l o c k such that f , ,(C) = B, or equivalently, if and only if C s ~ .

Thus card f~+ll(Bde) = kS = ~b ~ z card fm) l (Bb). Now suppose that card f~+ll(Bb ) > k for every b ede. If, for any b Ede, cardfm+ll(Bb)>k, we would have card f ~ ? l ( B d e ) > kS. We conclude that card fm+ll (Bb) = k for each b e de and the lemma is proved.

5.3 LEMMA. Let n e 12 and let f e F( de, n). Let there exist an m-block B over de such that card f m ~(B) = k > 0 and card fZml+,_ i(BDB) = k for every D E ~ , _ l(de). Then k = S" - 1.

Proof. Let f~ - l (B) = { C 1 , ' " , Ck} = cg. Let ~ = M,- l (de) . We show that

(1) f2-~+._ I ( B ~ B ) = ~¢£.

For let A s c~c~; thus A = C~C~, where C~, C s E c~. Since each C~ is of length m + n - 1, A is of length 2m + 2 n - 2 andf2m+,_ l(A) = E is of length 2m +n - 1. Since .fro(C3 = B = f,,(Cs), E is of the form BDB, where D is of length n - 1, so thatf2m+,-I(A) e BNB.

Now suppose that f2m+,- l (A) = BDB, where D e ~ . Then A is of length 2 m + 2 n - 2 : A = A1A ~, where A 1 and A2 are of length r e + n - 1, fro(A1) = B = fro(A2), so that Al, A2 e ~ and AIA2 ec~c~. This completes the proof of (1).

Clearly card c£~ = k 2. I f card f~-m~+,_I(BDB) = k for each D ~ , then since card ~ = S" - i, we have cardf2~+,_ I (B~B) = k S " - 1. From (1) it follows that k ~ = k S " - ~ ; thus k = S" - ~.

The proof is completed.

5.4 T H E O R E M . Let n ~ I~ and let f e F(de, n). Then the following statements are equivalent.

(1) fo~ is onto. (2) card fm~(B) = S "-~ for all B e ~ , , ( d e ) and for all m e I 1.

328 G.A. HEDLUND

Proof. It follows from Theorem 5.1 that (2) implies (1). We assume (1). I fn = 1, then (2) follows from Theorem 4.1. We assume that n _ 2. Let

k = inf {cardf~-1(Bin) IBm • ~(oo~), m E/1 }.

It follows from Theorem 5.1 that k > 1. There exist an integer m and an m-block B over ~ such that card f~- I(B) = k. Then cardf,7+ll(Bb) > k for each b•~W, and it follows from Lemma 5.2 that card f~+ll(Bb ) = k for each b e ~ . By induction we infer that card f~+ i(BA) = k for every/-block A and hence, in particular, card f2~t+n_1(BDB)= k for every D •~n_1(~9~). It follows from Lemma 5.3 that k = S n-1 and thus cardfp-l(Bp) > S ~-1 foreachBp e ~ p ( ~ ) and each p • 11.

Suppose there exist p • I 1 and B •~p(d~) such that card f 7 I(B) > S n- 1. Then

(2) ~ c a r d f v- l(Bp) > S ~ +~ - 1, ape ~p (~')

since card f p - l ( B ) > S ~ - i and card fp-'(B~)>_ S ~-1 for each Bp•~p(~9"). But fp- t (~p (~90 ) = ~ p +~ +1 (~9°), so that

card fp- l(Bp) = card fp- '( ~p(~90) = card~p+~_t(~9' ) = S p+~- l, Bp e ~p(~)

contrary to (2). We conclude that card fp- I (B) = S ~- 1 for all B • ~'~(~9 a) and all m • 11.

The proof is completed. The fact that for f® onto, each of the mappings fm is an exactly s~-l-to-1

map does not imply t h a t f , has the same property. Examples to be constructed subsequently show that for any n • 13, there exists f • F(~9 ~, n) such that f~o is onto but not an exactly k-to-1 map of X(6 a) onto X(Sf). But from Theorem 5.4 we can show that S ' -1 is an upper bound for the multiplicity offoo whenfoo is onto.

5.5 THEOREM. Let n •11, let f e F ( 6 a, n) and let f ~ be onto. Then card f ~ l(x) < S ~- 1 for all x • X(S f ) .

Proof. We suppose that foo is onto and that card f ~ l(x)> S ~-1 for some x • X(6a). Then there exist r e 11, r > S ~- 1, and a set {y(')l a = 1, 2,. •., r ) of r members of X(6 a) such that foo(y (~)) = x (a = 1, 2 , . . . , r). For k ~ 11, let B2k(a)+ 1 = Y~)k'" "Y(k ~)" Since y(~) # y(a) if a # /3, there exists k > n - 1 so large that B(~) # u(P) if a # ft. Let A be the (2k-n+2)-b lock X_k' ' 'Xk_n+l. Since 2 k + l ~t"2k+ 1

f~(y~) = x for each ~, we have f2k_n+ 2~,tU(~)2k + Va = A for each ~. But then card f2~l_~+2(A) = r > S ~-1, contrary to Theorem 5.4.

The proof is completed. In view of Theorem 3,4 we can draw conclusions concerning those members of

¢ ( ~ ) which define onto maps. Let E(S a) denote the set of all continuous mappings of X(S~) onto X(SP)

which commute with or. These are the onto endomorphisms of (X(S~), or). Clearly, E(~9 ~) c O(~9~).

5.6 THEOREM. Let 9 • E ( d P ) . Then there exists M e l i such that card 9-1(x) < M for all x • X(df) .

Proof Let q~ ~ E(Se). According to Theorem 3.4, there exist p ~/, n ~/1, f ~ F(5:, n) such that ~0 = ~r'fo ~ =fo~o e. Since o~ is a one-to-one map of X(Sa) onto X(oq') and ~0 is onto, it follows thatfoo is onto. Now ~(y) = x if and only if foo (Y) = ~r- ~(x), so that, by Theorem 5.5 card ~0-1 (x) = cardfo~ l(~r-,(x)) < S "- If we let M = S "-~, the proof is completed.

Remark 4.3 becomes valid for n > 1 provided the condition that x be fixed under o is replaced by the condition that x be periodic. The proof of this depends on several lemmas which "are useful in other contexts.

5.7 LEMMA. Let m ~ I 1 and let B be an m-block over S#. For q ~ Ii , q > m, let ~ (B , q) be the set of all q-blocks over 5: in which B appears and let N(B, q) = card ~(B , q). Then

lim N(B, q) = 1. q--* o~ S q

Proof. Let ~o(q) = S-qN(B, q). Let C s ~ ( B , q). Then C c ~ ( B , q + l ) for each c ~5 a, so that N(B, q+ 1) > S .N(B , q), and hence qo(q+ 1) > ~0(q). Since card~g(5: ) = S ~, it follows that N(B, q) < S q and ~o(q) < 1. Thus lim~_,~d0(q) = ~, w h e r e 0 < ~ < 1.

Let k e 11 and let A be a block of length km. Then A = BxB2" • • Bk, where each B1 is an m-block. Let ~¢~ be the collection of all such blocks A for which Bi = B, Bj # B if j i. If i # j, then ~¢i ~ ¢ ' y = ~. Let d ( k ) = ~ = 1 ~¢i. Then B appears in every member of~C(k).

Let a = S " - 1, b = Sm. Then

k k

card~¢(k) = ~ card~¢i = ~, a i- lbk-i = (a k_bk) / (a_b) = skm__(S m - - 1)k. i = 1 i = 1

Since N(B, km) > card d ( k ) , we have

9,(kin) = S-k '~N(~,km) >_ s-km[skm-(Sm-1)~l = l - l - ~ ,

and hence limk_,~(km) = 1 = ~ = limq_, ~o(q). The proof is completed.

5.8 LEMMA. Let n ~ II, let f ~ F( S, n) and suppose that foo is not onto. Let k, t ~ Ix. Then there exists an m-block A with m > t such that card f,,- I(A) > k.

Proof We assume the hypotheses of the lemma and that the conclusion is false, that is, cardf~- I(A) _< k for all m-blocks A, m _> t. Sincef~ is not onto, it follows from Theorem 5.1 that there exists a q-block B such tha t fq- l (B) = ~, and hence card fq-I(B) = 0. This implies that card fro-I(D) = 0, where D is any m-block in which B appears.

Let ~(B, m) be the set of all m-blocks over 5 : in which B appears and let '(B, m) be the set of all m-blocks over ~9 ° in which B does not appear. Let N(B, m)

= card ~(B, m). Since ~ . ( 5 : ) = ~(B, m ) u ~ ' ( B , m) and card ~m(5:) = S", card ~' (B , m) = S" - N(B, m). Now ~ , +,_ 1(5:) = f~- 1(:~,,(S)) = f~- i (~(B, m)) wf,~a(~ ' (B, m)) and ~(B, m ) c ~ ' ( B , m) = ~, so that

Sin+.- 1 = card ~m+, - 1(5a) = card fro l(~(B, m)) +card fro I (~ ' (B, m)).

330 G . A . HEDLUND

If m > q, c a r d f m l ( ~ ( B , m)) = 0, so that S m÷"-I = cardfr71(~ ' (B , m)). By assumption, cardfm-l(A) < k for all m-blocks A, m > t. Thus i fm _> t+q,

S m ÷, - 1 < k card ~ ' (B , m) = k[S m - N(B, m)];

consequently sra+n-1

N(B, m) < S " k '

o r

N(B, m) S" - 1 - - < _ I - - -

S " k This implies that

lim sup - - m'-* oo

contrary to Lemma 5.7. The proof is completed.

N(B, m) S "-1 < 1 - - -

S m k ,

5.9 THEOREM. Let n ~ 11 and let f ~ F(Sg, n). Then the following statements are equivalent.

(1) foo is onto. (2) There exists M ~ I 1 such that card f m 1(B) < M for each m ~ 11 and each

Proof. It follows from Theorem 5.4 that (1) implies (2). It follows from Lemma 5.8 that (2) implies (1).

5.10 LEMMA. Let n ~ I 2, m e l 1 and let ~ be a collection o f (m + 2 n - 2)- blocks over 5 '~ such that card ~ > S 2"- 2. Then there exist distinct blocks E and E* in ~ such that E = ADB and E* = AD*B, where A and B are ( n - 1)-blocks.

Proof. Since card ~ , _ 1(5 ~°) = S n- 1, we can write ~ , _ 1(5 ¢) = {A~I i = 1, 2, • • . , S"- 1 }. For 1 < i < S"- 1, let d~i be the collection of all members of d ~ with initial ( n - 1)-block A~. Suppose that for all i, card ~ _< S "-1. Then

S n - 1

c a r d g = ~ cardS~ < S " - I S "-1 = $2"-2, ' i = 1

contrary to hypothesis. Thus there exists an integer j such that card @j > S n- 1

But then at least two distinct members of@j must have the same terminal ( n - 1)- block B. Let these be AjDB and A~D*B. Let A = A j, E = ADB, E* = AD*B. The statement of the lemma is proved.

5.11 LEMMA. Let n ~ 12, let f ~ F( 5 a, n) and let m ~ 11. Let there exist an ( re+n-1 ) -b lock C over 5 a such that card fm+X._l(C)> S 2n-2, Then there exist distinct ( re+n-1) -b locks P and P* and an (n-1)-block A such that

Jr,,+ 2(n- 1)(APA) = fm+ 2(n- 1)(AP*A) •

Proof. Let ~ = fm+~,_ 1(C). Then d ~ is a collection of (m + 2n - 2)-blocks and, by hypothesis, card N > S 2"-2. It follows from Lemma 5.10 that there exist distinct blocks E and E* in d ~ such that E = ADB and E* = AD*B, where A and B are (n-1)-blocks. Then D and D* are m-blocks and D ¢ D*. Since E, E* e f - l _ 1(C), fro+,- I(E) = C = f , ,+,_ I(E*).

Let P = DB, P * = D*B. Then P and P* are distinct ( m + n - 1)-blocks. Let f . _ I(BA) = Q. Then

fm + 2(.- 1)(APA) = fro+ 2(.- I)(ADBA) = CQ,

fm+2(,-1)(AP*A) = f,,+2(,-1)(AD*BA) = CQ.

The p r o o f is completed. Let x e X(S#). Then x is periodic provided there exists oJ E 11 such tha t

xi = x~+o, for all i e L Then oJis a period o f x and the least such integer is the period of x.

Let m e I1, and for each i e I, let B~ be a block of length m over ~9 °. Then • . . B_ 1/}oBI. • • is that member x o f X(A ~) defined by

X k m X k m + l " " " X k m + m _ I = Bk, ]eel.

I f {Bil i E I} is any collection of blocks over ~9 a, the meaning of . - . B_ 1BoBt. • • is clear.

5.12 T H E O R E M . Let n e 11, let f ~ F ( 5 p, n) and let fo~ be not onto. Then there exists x E X( Sg) with x periodic such that the set f ~o l(x) is uncountable.

Proof. I f n = 1, the conclusion follows f rom R e m a r k 4.3. We assume tha t n > 2. F r o m L e m m a 5.8 there exists an ( re+n-1) -b lock C

such that card fm-+~,_ 1(C) > S 2"- 2. It follows f rom L e m m a 5.11 that there exist distinct ( r e + n - 1 ) blocks P and P * and an ( n - 1 ) - b l o c k A such that

fro+ 2(n- 1)(APA) = D = fm+2( , - 1)(AP*A).

Let Y be the subset o f X ( 6 p) defined by

"" "AQ-1AQoAQ1AQ2" " "

where, for i E L Q i is either P or P* . Let x ~ X ( ~ ) be defined by x . . . . D f ) D D . . . . Then x is periodic, and i f y e Y, thenfo~(y) = x. The set Yis uncountable, and since Y c fo~ *(x), the set f~-*(x) is uncountable.

By virtue of Theorem 3.4, we can draw conclusions concerning non-onto members o f O(~90. I f x ~ X(S~) then x is periodic, in the sense in which we have defined it, if and only if x is a periodic point o f the dynamical system (X(A~), ~); that is, there exists a positive integer ,o such that ~°(x) = x.

5.13 T H E O R E M . Let ~ ~ 0 ( 5 ¢') and suppose that 9 is not onto. Then there exists a periodic point x o f X( SP) and an uncountable subset Y o f X( SP) such that 9(Y) = x.

Proof. Let go ~ O(Af). According to Theorem 3.4, there exist m E I and n ~ I1 such that ~o = ~mf~, w h e r e f E F ( ~ , n), If~0 is not onto, thenfoo is not onto and according to Theorem 5.12, there exist an uncountable subset Y of X(o O~) and z e X ( ~ ) , z periodic, such that fo~(Y) = z But then 9(Y) = ~mfo~(Y) = ~m(z). Let x = ore(z). Then x is periodic and the theorem is proved.

Let H ( A f) be the set o f all members of O(~ O~) which are one-to-one. An automorphism of (X(Af), ~) is a h o m e o m o r p h i s m of X(A f) onto X ( o of)

which commutes with ~. Let A(A f) be the set o f all au tomorph i sms of (X(o<f), ~). Clearly A(A f) c E(~9~).

332 G.A. HEDLUND

5.14 THEOREM. A(50 = H ( 5 0 . Proo f Clearly A(50 ~ H ( 5 0 . Let 9 • H ( 5 0 . It follows from Theorem 5.13 that 9 is onto, and hence 9 is a

homeomorphism of X ( 5 0 onto X(50 .

6. Existence Theorems for the Classes A(Aa), E(5 °) and ap(5#)

For a given symbol set 5#, let X(5 a) = {d'l n E I}. Then X(dO c A(5#) c E(SP) c ~(5#). We show that each of the classes A(5°) - Z(5~), E(5¢) -A(S#) , tl)(5#)- E(~9 °) is infinite and we develop some ineresting properties of the group A(~).

Let n • I 2 and let f• F(~, n). Then f depends on x. if and only if there exist an (n-1)-block A over 5: and a, b • ~9 ~ such that f(Aa) # f(Ab);f depends on x x if and only if there exist an (n - l)-block A over ~9 ~ and a, b • 5# such thatf(aA) f (bA) .

6.1 Remark. Let n • 12 and let f • F(5#, n). Then the following statements are equivalent.

(1) f does not depend on x, . (2) There exists k, 1 <_ k <n, such that foo • Foo(5 °, k).

Proo f Assume (1). T h e n f ( A a ) = f ( A b ) for all A • ~ , _ 1(50 and all a, b • 50. Define g: ~ , _ 1 ( 5 0 ~ 1 ( 5 : ) = 5 : by g ( A ) = f ( A 0 ) (A • ~ , - 1 ( 5 0 ) . Then g ~ F(5 : , n - 1). Let x E X ( 5 0 and let g®(x) = u. Then ui = g ( x : . "x i+, - 2) (i • 1). Let fo~(x) = v. Then vi = f ( x i ' " x i + , - 1 ) = g ( x i ' " x i + , - 2 ) = ui ( i • I) . Thus u = v, fo~ = g~o and f~ • F®(5 °, n - 1). This proves (2).

Assume (2). Then there exist k, 1 < k < n, and g • F (5 °, k) such thatfo~ = go~. Let A • ~ , _ 1(50 and let a, b • 5 p. Choose x, y • X (5 0 such that x l " ' " x,_ ix, = Aa and Yx'" "Y , - IY , = Ab. Let f®(x) = u = g~(x) , foo(Y) = v = goo(Y). Then u 1 = f ( A a ) = g ( x l . . . x k ) and v 1 = f ( A b ) -- g ( Y l " " Y k ) = g ( x l " " x k ) . Thus f ( A a ) = f ( A b ) and f does not depend on x,. This proves (1),

6.2 THEOREM. The set ~b( 5 0 - E ( 5 0 is o f infinite cardinality. Proo f For each n • /2 , we define f ("+1) • F(5 : , n + 1) as follows. Let ~9 ° =

{0, 1 , " ", S - 1 }. Let A be an n-block overS:, and f ( " + ~)(A0) = 0, f ( " + ~)(A 1) = 1, f ( " + 1)(B) = 0 for every B • ~ , + 1(50 except A0 and A1. T h e n f (" + 1) depends on x,+ ~ and it follows from Theorem 5.4 thatf(~ " + ~) is not onto; thusf~ "+ ~) ¢ E ( 5 0 . It follows from Remark 6.1 that if m, n • /2 and m # n, thenf(~ re+l) # f( ,+x).

The proof is completed. We construct a large class of members of E ( 5 0 - A ( 5 0 .

6.3 Definition. Let n •I~ and l e t f E F ( 5 : , n). T h e n f i s a mapping O f ~ n ( 5 0 into 5# and for each n-block XlX2" • . x , over 5 #, f ( x l x 2 " " .x,) is defined and is a member of 5:. If n > 1 and x2," " ", x, are fixed elements of S#, then g(x 0 = f ( x l ~ 2 " • "Y~,), x~ •50 , defines a mapping g of S: into 5".

If, for every choice of if2,. •. , ft,, g defines a permutation of 5 ~, then f is permutive in x 1. I fn = 1 thenfdefines a mapping o f£ # into 5# a n d f i s p e r m u t i v e in x~, provided f is a permutation of S:.

I f n > 1 and 21,22," • ", 2,_ x are fixed elements of 5O, then g(x,) = f ( 2 ~ . . . 2n- xXn), xn • 5O, defines a mapping of 5o into 5O. If, for every choice Of2xr~2, . . . , 2n-x, g defines a permutation of5o, t h e n f i s permutive in x n.

The extension of this definition to permutive in x~, 1 <_ i <_ n, is obvious.

6.4 Remark. Let n ~ 12 and let f • F( SO, n). I f f is permutive in x I then f depends on x 1. I f f is permutive in x n, then f depends on x n.

6.5 Remark. Let n • 12. Then there exists f e F(5O, n) such that f is permutive in both x 1 and x n.

Proof. Let rr by the cyclic permutation of 5O -- {0, 1 , . . . , S - i } defined by ~-(s) = s + 1 (s = 0, 1 , . . . , S - 2 ) , 7r(S- 1) = 0. For n = 2, def inef~ F(5O, n) by f ( t s ) = rrt(s) (s, t e 5O). Then f is permutive in both x 1 and x,.

Assume that n ~/2. Define g e F(SO, n) by g(xxx2"" "Xn) = f(xtxn) . Then g is permutive in both x~ and xn.


6.6 THEOR EM. Let n e I1, let f e F(SO, n) and let f be permutive in Xx[X,]. Then foo is onto.

Proof. We assume tha t f i s permutive in x 1. I fn = 1, it follows f rom Theorem 4.1 that foo is onto.

We assume that n > 1. According to Theorem 5.1 it is sufficient to prove that f, . : ~m+n_x(5o)->~m(so) is onto for each m > 1. Let x x . . . x , , ~ m ( 5 o ) , Y,, + 1" " "Y,, + n- X ~ ~ , - X(5O)" Since f is permutive in xx, there exists y,, e 5O such that f ( Y m Y m + l ' " y , , + n - 1 ) = x,,. For the same reason, there exists Ym-1 such thatf(Y,,-x"" "Ym+n-2) = x, ,_t . By continuation ofthis process, we conclude that there exists y~ e 5O (i = 1,. •. , m) such that

f ( Y i Y i + x ' " Y i + n - t ) = xi (i -- 1, 2 , . . . , m) and hence

fro(Y1"' " Y,,+n- x) = xx " " x,,"

It follows that f~o is onto. The proof when f is permutive in xn is similar.

6.7 THEOR EM. Let n e I1, let f • F(5O, n) and let f be permutive in both x x and x, . Then f ~ is an exactly S n- a-to-one mapping o f X(5O) onto X(5O).

Proof. We assume that f is permutive in both Xl and xn. I f n = 1, then f defines a permutation of 5o and it follows from Theorem 4.1 tha t f~ is an exactly one-to-one map of X(5O) onto X(5O). So Theorem 6.7 is proved if n = 1.

We assume n • I2. Let x s X(5O) and let Yx'" "Yn- Z • ~ . - 1(5°) • Since f is permutive in x,, there exists y. e 5O such that f (Yx"" "Yn) = xx. Proceeding inductively, for each i s I n, there exists Yi e s o such t h a t f ( y j ' " y j + n _ x ) = xj ,

j ~ I 1 . Since f is permutive in x 1, there exists Yo s 5O such that f ( y o . . . y . _ ~) = Xo.

Proceeding inductively, for each i • I, i < 0, there exists y~ e 5" such that

f ( Y ~ ' " Y j + , - 1 ) = xj , j <_ O. Thus there exists y • X(SO)such that f ( y ) = x and Y x ' " Y n - ~ is any given

member of ~ n - 1(5O). Since card ~ . _ 1(5") = S " - x, there exists a subset Y of X(SO) with card Y = S n-1 and f ~ ( Y ) = x. Since, from Theorem 5.4,

334 G.A. HEDLUND

card f ~ l ( x ) <_ S " - t for all x sX(Aa), we conclude that card f g t ( x ) = S n-1 for all x ~ X(Se) and the theorem is proved.

6.8 T H E O R E M . The set E( SP)- A( SP) is of infinite eardinality. Proof. Let n e/2. By remark 6.5, there exists f (") ~ F ( 9 o, n) such t h a t f (") is

permutive in both xt and x,. By Theorem 6.7, f~(") is an exactly S"-1-to-one mapping of X(Sz) onto X(SZ). Thus f~" )e E(Aa)-A(Aa).

Suppose that m, n e I z with m <n. Then, s incef (m) is permutive in x,, a n d f ~") is permutive in x,, it follows from Remark 6.4 t h a t f (m) depends on Xm a n d f ~") depends on x,. By Remark 6.1,f~ ") CF®(SP, m), and thusf~ m) ¢ f~"). This proves the theorem.

We seek to find members of A(Se) which are not powers of cr, that is, homeomorphisms of X(Sa) onto X(5 a) which commute with o and are not powers of e. As the following theorem shows, in case card 5 p = 2, there are no candidates from among those members of Fo(5 p, n) which are permutive in xx except for those two in Fo(SP, 1) which exist by virtue of Theorem 4.1.

6.9 T H E O R E M . Let card 5/' = 2 and let n ~ I x. Let f ~ F( Sg, n), let f be permutive in x x and let f o be a one-to-one map of X ( SP) onto X ( Sa). Then there exists g ~ F(5 a, 1) such that f.o = g®.

Proof. We assume that the theorem is false, and thus there does not exist g s F ( 5 a, 1) such tha t f~ = go. Let k be the least integer such thatf~o ~ F o ( 9 °, k). Then k > 1 and there exists g E F(5 a, k) such that f o = go.. It follows from Remark 6.1 that g depends on Xk. Sincefo is one-to-one, so is g~.

Since card 5 P = 2, we can let 5,° = {0, 1 }. Let a ~ 5#. Then d is defined to be 1 if a = 0 and d = 0 if a = 1. Since g depends on Xk, there exists a k-block al"" "ak such that g(ax'" "ak) ~ g(al"" "ak-Xdk). Sincef is permutive in xl , so is g, and thus g(0a2" "'ak) = g(la2"' 'ak-tdk). Let A = a2""ak, B = a2"''ak-Xdk. Thus g(OA) = g(1B).

Let P be the set of all integers p > k with the property that there exist p- blocks A v = Oa2. . .a v and Bp = lb2" "bp such that gv_k+ x(Av) = gv_k+ x(Bv). Then k c P. We show that P = {ml re>k}. Clearly, i f p > q > k a n d p e P, then q e P. Thus it is sufficient to show that P is not bounded.

Suppose P is bounded. Let m be the least upper bound of P and thus there exist m-blocks A,, = 0C m_ x, B,, = 1D m_ x such that gin-k+ x(Am) = grn-k+ x(Bm). Since m is maximal,

a = gm-k+2(OAm) ~ g,.-k+2(1Bm) = b,

e = g,,_k+2(1A,,) ~ gm_k+2(OBm) = d.

Thus a = b and e = d. Since g is permutive in xx, e = d and d = b, so that a = d and ¢ = b, and we have shown that

gm-k+2(eAm) : gm-k+2(eBm), e = 0 or 1.

This process can be continued to obtain

gin(DAm) = g,,(DBm), all D SMk_X(SP).

But then g(DO) = g(D1) for all D c ~k-x(Sa) , and this contradicts the fact that g depends on x k. We conclude that P = {m[m > k}.

Endomorphisms and Automorphisms of the Shift Dynamical System 3 35

Let p ~ P. There exist ( p - 1)-blocks y(1 p) . . "Y(p~l and z(~ p) . . • z(p) over 50 such - - p - 1

. . . . . (P) a Def iney (p) e X(50) by setting that g~-k+ I(0Y(~ p)" "Y(p~ 1) = gp-k+ I(lz~ p) ~p- l J" y~P) = 0 if i < 0 or i > p. To define z ~p), set Z(o p) = 1 and z~ p) = 0 (i _> p). Since g is permutive in x~ for each i < - 1, there exists z~ p) ~ 5" such that

g ( z ~ ) . . . z ~,) ~ g(y~t,) . . . . . (~) ~ (i < 1). i + k - l J = , , V i + k - l ] - - - -

Let g~(y(P)) = u (p), g~ ( z (p)) = V (p). Then u} p) = v} p) (i < p - k ) . This implies that d(u (p), v (p)) < (p - k + 2)- i. Thus limp_. ~d(u (p), v (p)) = O.

Since X(S a) is a compact metric space, any infinite sequence o f points contains a convergent subsequence, so that we can assume that limp_.®y(P)= y, limp~ooz (p) = z, limp=,~u (p) = u, limp._,o~v (p) = v. Since limp_,~od(u ~p), v (p)) = O, it follows that u = v. Since y(o p) = 0, Yo = 0, and since z(o p) = 1, Zo = 1. Thus y ¢ z. N o w go~(Y) = go~ (limp~o~Y (p)) = limp-,o~g(y (p)) = l i m r , o~u(P) = u and go~(z) = goo(limp-,ooz (~)) = l imp_,~g~(z (p)) = limp_,~ov (p) = v = u, so that go~(Y) = go~(z). This contradicts the hypothesis that f~o, and hence go~, is one-to-one.

The p r o o f o f the theorem is completed. This theorem shows that if f e F(5" , n) and if f is permutive in x~, then

f ~ e A(50) only i f f ~ is either the identity mapping or the mapping q~: X(Sg)-+ X(50) defined by q~(x) = y, where Yi = xi (i ~ I) .

However, the preceding theorem is no longer true if card 5 ~ > 2. We construct an example, with card 5 ° = 3, of a mapping f : ~'z(50)-+0~'~(50) = 50 such that : ( 1 ) f i s permutive in Xl, (2) f d e p e n d s on Xz and (3 ) f~ is one-to-one. It follows f rom Remark 6.1 that there does not exist g e F ( 5 e, 1) such t h a t f ~ = g~o.

Let 5 ~ = {0, 1, 2 } and define f as follows:

f ( 0 0) = 0, f ( 0 1) = 2,

f (1 0) = 1, f (1 1) = 1,

f ( 2 0 ) = 2, f ( 2 1) = 0,

f ( o 2) = o,

f ( 1 2 ) = 1,

f ( 2 2 ) = 2.

Clear lyf is permutive in xl a nd fde pe nds on x2. We show tha t f~ is one-to-one. Suppose that there exist y, z e X ( 5 a) such that y ¢ z andfo~(y) = x = f®(z) .

Since y ¢ z, there exists an integer i such that Yi ¢ zi. Since f ( ab ) = 1 if and only i f a = 1, we cannot have xi = 1, for this would imp lyy l = 1 = z~. Thus x~ is either 0 or 2 and Yi and z~ can assume only the values 0 or 2. We can assume that y~ = 0 and zi = 2.

Suppose that xi = 0. Since f ( z i z i + O = f ( 2 z ~ + 0 = 0, we have zi+l = 1. This implies that xi+ 1 = 1 and hence Yi+l = 1. But then x~ = f ( Y ~ Y i + l ) = f ( 0 1) = 2 ~ x i. Thus this ease is impossible.

N o w suppose that xi = 2. Since f (y iY i+ 1) = f(Oyi+ 1) = 2, we haveyi+ 1 = 1. This implies that x~+ 1 = 1 and hence zi+ 1 = 1. But then xi = f ( z i z i+ 1) = f (2 1) = 0 ~ x~, so this case is also impossible.

We conclude that f~o must be one-to-one and has the desired properties. Clearly f ~ is not a power o f or, and thus we have an example o f a member o f A ( 5 0 ) - T ( 5 0 ) , card 50 = 3.

We show that even in the case card 50 = 2, the group A(50) has a complex structure. In particular, we show that every finite group is isomorphic to some subgroup of A(50).

336 G.A. HEDLUND

6.10 Definition. Let n ~ 11 and let B = b:b 2 • • .b, be an n-block over S:. Let m a Ix, m _< n. The initial m-block o f B is the block bl"" "bin. The terminal m-block o f B is the block b,_m+ x'" .b,.

6.11 Definition. Let m, n E I : , let B be an m-block over 5# and let C be an n-block over 50. Then B and C overlap provided there exists k ~ Ix such that either the terminal k-block of B coincides with the initial k-block of C or the initial k-block of B coincides with the terminal k-block of C.

Let n ~ I 1 and let A be an n-block over 5#. Let m ~ I I. Then A m is the (ran)- block AxA2"" "Am, where Ai = A (i = 1, 2 , . . . , m).

6.12 LEMMA. L e t nEI2, let A = 1 "+i and let B = ( 1 0) m i f n = 2 m , B = (1 0) m / f n = 2 m + l . Let

= AON,(~e)OB.

Then no two members o f ~ overlap. Proof Let C, D e 5 , C # D; thus

C = AOcl" • • c,OB,

D = A O d x ' " d , OB.

Suppose there exists k e I x such that the terminal k-block of C coincides with the initial k-block of D. Since C # D and C, D ~ '2 ,+2 , ,+3 , it follows that 1 < k < 2 n + 2 m + 3 .

Suppose k > n + 1. Then the initial k-block of D contains 1" + x as a sub-block. But no terminal k-block of C contains 1" +: as a sub-block provided k < L(C). T h u s l < k _ < n .

We cannot have k = 1, for the initial 1-block of D is 1 and the terminal 1-block of C is 0. Thus 2 < k _< n.

I f 2 _< k _< n, the initial k-block of D contains 1 1 as a sub-block. Since the length 0B is _> n, the terminal k-block of C is a sub-block of 0B. But 1 1 is not a sub-block of 0B.

We conclude that there cannot exist k ~ 11 such that the terminal k-block of C coincides with the initial k-block of D.

Interchanging the roles of C and D, we conclude that there cannot exist k ~ Ix such that the initial k-block of C coincides with the terminal k-block of D.


6.13 THEO R EM. (Curtis, Hedlund, Lyndon) Every finite group is isomorphic to some subgroup o f the group A( 5 °) o f automorphisms o f (X(oq'), ~).

Proof Let G be an arbitrary finite group. For n e 11, let H(5: ' , n) be the symmetric group on ~ , ( 5 : ) . Since card ~ ' , ( 5 a) = S" and S _> 2, there exists n ~ 12 such that G is isomorphic to a subgroup of H(5 : , n). To prove the theorem, it suffices to show that A(5 ¢') contains a subgroup isomorphic to H(5 : , n), n ~/2.

Let n e/2, let ~r ~ H ( 5 : , n) and let x ~ X(~q~). We define q~,(x) = y e X(5: ) as follows. Let ~ be defined as in Lemma 6.12. Each block in cg is of l eng th j = 2 n + 2 m + 3 . Let i ~ L I f there exist C~(Y and k_< i < k + j - 1 such that Xk" " "Xk+j-X = C = AOe x • " .e, OB, then it follows from L e m m a 6.13 that k is unique. Let *r(cx'" "c,) = dx '" -d , . We define Y k ' ' ' Y k + j - i = AOdx'" "d, OB. I f no such k exists, we define Yi = x~.

We remark that y~ is uniquely determined by the block x ~ _ j " ' x i + j .

The mapp ing ~.: X(5"~)-+X(5 e) is continuous. For let E > 0 and choose m ~ 11 such that (1 +rn) -x <e. Let 3 = (1 + m + j ) -1. Suppose, x, z ~ X(Sa) with d(x, z)

< 3. Then xi = z~ if [i1 _< m + j . Let ~o,(x) = y, ~ ( z ) = u. Then y, = u, if li[ _< m, which implies that d[~0,(x), ~o,(z)] = d(u, v) < ( l + m ) -1 <E.

We show that ~o, commutes with a. Let x ~ X(Sf) , ~%(x) = y, a(x) = u. Then [a~(x)] i = Yi+I and u i = x i+l , i ~ L Let ~,(u) = v. We wish to show that i E I

implies y~+ ~ = v,, or, equivalently, yi = v,_ 1. Suppose there exists an integer k such that k < i < k + j - 1 and X k ' ' ' X k + j _ l ~ , and thus X k ' ' ' X k + j _ ~ = C

= A0cl"" "cnOB = Uk-l"" "Uk+j-2. But then V k - ~ ' ' ' V k + j - 2 = AOdl"" "d.OB =

Y k ' ' ' Y k + i - 1 , where d l " " d , = rr(cl""cn) . Since k - 1 _< i - 1 < k + j - 2 , it follows that v i_ ~ = y~.

Suppose there does not exist an integer k such that k < i < k + j - 1 and

Xk ' ' 'Xk+j_ 1 ~(~. Then y~ = x~. But Xk ' ' 'Xk+j_ 1 ~- Uk-l"" "Uk+j-2, SO that there does not exist an in tegerp such t h a t p < i - 1 <_ p + j - 1 and Up. • • Up+j_ ~

~ . This implies that vi-~ = u~_~ = x~ = y~. This completes the p r o o f that 9~ commutes with ~.

We show that 9~ e A(6a). Let x ~ X(6a), cp~(x)= y. We observe that Y i ' " Y i + j - ~ ~cg if and only if x i . . . x ~ + j _ 1 ~ . For if x ~ . . . x i + j _ ~ ~ , then x i ' " x i + ~ - i = AOc~ . . . c ,OB and Y i ' " Y i + j - ~ = AOd~" .d~OB, where d~. . "dn = ~r(c~- • .c,), so that Yi" " "Yi+j- ~ e ~ . Conversely, suppose that y~" • "y~+~_ 1 ~ ~ and x ~ ' " x i + ~ _ l Cry. I f any of the blocks Xk' ' 'Xk+~_ ~, where i < k <_ i + j - 1

or i < k + j - 1 < i + j - 1, is in c~, then Yk" " "Yk + j - 1 ~ c~ and we have two members of C which overlap, contrary to L e m m a 6.12. But then for each m, i _< m __ i + j - 1, Ym = Xm and x i" • "xi+j_ 1 = Yi" " " Y i + j - ~ ~c~. Thus y~" • "y~+~_~ ~cg if and only if x f ' . x i + j _ 1 ~ c~. It follows f rom this that ~0~-,~o~ is the identity mapp ing and hence ~o~ is a h o m e o m o r p h i s m of X(~9 a) onto X(Sa), so that cp~ ~ A(Se).

I t is easily verified that if ~,/3 ~ H(S~, n), then ~ 0 ~ = ~0a~. For ~ t ransforms any block AOa~" • .a~OB of C~ appear ing in x ~ X ( 6 ~) into the block AOb~. • • b,OB,

w h e r e b l"" .b~ = ~(a~.. "an), in the same posit ion in y = ~ ( x ) . Such blocks appear in y only in the same positions in which they appear in x and ~oa trans- forms the block AOb~.. .b~OB in y into the block AOc~.. .c~OB, c l . . . c , = /3(b~"" b~) = f l~(c~" • cn), in the same posit ion in z = ~o~(y). Now ~o~ t ransforms AOal" • • anOB appear ing in x into the block AOci. • • cnOB appear ing in the same posi t ion in u = 9a~(x). I f i ~ I and there does not exist k ~ I such that k _< i _< k + j - 1 and Xk' ' 'Xk+~_ ~ ~ , then Yi = x~, z~ = Yi, ui = x~, so that z i = u i. Thus ~o~ = ~ o ~ .

The mapp ing A: H(5~ , n ) -+A(5 a) is defined by A(z r )= ~ and the set A ( H ( 5 a, n)) is a subgroup of A(5 a) i somorphic to H ( 5 a, n).

This completes the p r o o f of Theorem 6.13. I t follows f rom Theorem 6.13 that given any k ~ 11, there exists an element of

A(S ~) of order k. This implies the following corollary.

6.14 C O R O L L A R Y . The set A ( S ~ ) - ~ ( 5 P ) is infinite. The elements of A(5 a) commute with ~r, so that E(S ~) is a normal subgroup of

A(6 v) and the quotient group A(S~)/E(S~) is well defined. Let G be a subgroup o f A(5 a) and let G ' = {gZ(Sa) /g ~ G }. Then G ' is a subset o f A(da)/Z(~9 e) and G '

338 G.A. HEDLUND

is a subgroup of A(S~/Z(Sz) , provided that we define the group operation by (glE) o (g2Y,) = (glg2)Z. Let o¢ denote the identity mapping of X(S:). Then G is isomorphic to G' provided that Gc~Z(Se) = ( J} .

We have shown that H(6:, n) is isomorphic to the subgroup G = A(H(S:, n)) of A(S,Q. It is evident that Y.(S0c~G = (0¢ } and thus G' is isomorphic to H(6 a, n). This proves the following statement.

6.15 COROLLARY. Every finite group is isomorphic to some subgroup o f A ( ~ / Z ( ~ .

We will develop further properties of A(5 #) subsequently.

7. Classification of Points of ( X ( ~ ) , ~)

Since (X(Sa), ~) is a particular transformation group or dynamical system, the various concepts with respect to a point x e X(5 a) or its orbit, such as periodic, almost periodic, etc., are applicable. For the sake of completeness, we state characterizations of these properties for the particular system under consideration, characterizations which could be taken as definitions in the present case (cf. [10].)

7.1 Remark. Let x ~ X(5"). Then the following statements are pairwise equivalent:

(1) x is a periodic point of (X( Sa), or). (2) There exists co e l i such that x(i+o~) = x(i) ( i~ I ) . (3) There exists ~ ~ I 1 such that cr°'(x) = x. The set of periodic points of (X(S:), a) will be denoted by P(Sa).

7.2 Remark. Let x ~ X(5:) . Then the following statements are pairwise equivalent:

(1) x is a regularly almost periodic point of (X(Sa), a). (2) l f n ~1 i, then there exists p ~11 such that x( i+pj) = x(i) (1il -< n , j El) . (3) I f n ~ I, then there exists p ~ I1 such that x(n +pj) = x(n) ( j ~ I).

The set of all regularly almost periodic points of (X(Sf), a) will be denoted by RAP(S:).

7.3 Remark. Let x ~ X(Sa). Then the following statements are equivalent.

(1) x is an isochronous point o f (X( S#), ~). (2) I f n ~ 11, then there exist p ~ I l and q e I such that x ( i+p j+q) = x(i)

([il -< n, j e I).

The set of all isochronous points of (X(Sa), g) will be denoted by 1(50.

7.4 Remark. Let x ~ X(Sa). Then the following statements are pairwise equivalent.

(1) x is an almost periodic point o f (X(5 : ) , or). (2) I f n ~ Is, then there exists a syndetic (relatively dense) subset E of I such

t ha t x ( i+ j ) = x(i)([il <_ n, j E E ) . (3) l f A is a sub-block of x, then there exists k ~ I 1 such that A appears in every

k-sub-block of x.

(4) I f n • 11, then there exists k> 0 such that every n-sub-block of x appears in every k-sub-block of x.

The set of all almost periodic points of (X(5¢), a) will be denoted by AP(Sg). The following remarks are easily verified.

7.5 Remark. P ( 5 a) c RAP(S~) c I(5/') c AP(50 .

7.6 Remark. P( 5 a) is an invariant countable dense subset o f X( Sa). The existence of points in RAP(5~) - P(Sa), I ( 5 a) - RAP(S~), AP(5 a) - 1 (50

is less obvious. However, for the case card 5 a = 2, and hence for any finite 5 a, it is known that each of these sets is non-vacuous. The existence of regularly almost periodic points of X ( 5 0 which are not isochronous follows from Theorem 12.55 of [10]. The existence ofisochronous points which are not regularly almost periodic follows from Remark 12.54 and Theorem 12.55 of [10]. The existence of points which are almost periodic but not isochronous follows from Theorem 12.39 of [10].

Remark 12.41 of [10] states that if card SP = 2, then the set of all non- isochronous almost periodic points of X(5 a) is dense in X(5~'). The methods used there enable us to prove that for any finite 5 a, each of the sets R A P ( S t ' ) - P ( 5 0 , I ( 5 ~) -RAP(~9~), A P ( S a ) - I(S~) is dense in X(Sa).

Let N e I1. An N-block substitution over 5 z' is a mapping ~: 5 t '~MN(50. Let x • X ( ~ ) and let e be an N-block substitution over 5a. We define

y = a*(x) • X ( 5 0 by y ( m N ) . . . y ( m N + N - 1) = e(x(m)) (m • I).

7.7 Remark. Let x • X(5¢'), let N • 11 and let ~ be an N-block substitution overs a. I f x is periodic, then ~*(x) is periodic, l f a is one-to-one and e*(x) is periodic, then x is periodic.

Proof. Assume that x is periodic; thus there exists ~o • I i such that x(i+ oJ) = x(i) ( i • I ) . But then e(x(i+~o))= ~(x(i)), and hence if y = e*(x), then y ( i N + o ~ N ) . . . y ( i N + o J N + N - 1) = y ( i N ) . . . y ( i N + n - 1) (i s I), which implies that ~oN is a period of ~*(x).

Suppose ~ is one-to-one and that y = ~*(x) has oJ as a period. Then o~N is a period of y, y( iN+ oJN) . . . y ( iN+ ~oN+ N - 1) = y( iN) . . " y ( i N + n - 1) (i • / ) , and hence o~(x(i + co)) = e(x(i)) (i • 1). Since ~ is one-to-one, it follows that x(i + o~) = x(i) (i ~ I), so that x is periodic with oJ as a period.

7.8 Remark. Let x • X ( 5 0 , let N • I i and let ~ be an N-block substitution over 5 ~. I f x is regularly almost periodic, then ~*(x) is regularly almost periodic. I f ~ is one-to-one and e*(x) is regularly almost periodic, then x is regularly almost periodic.

Proof. Assume that x is regularly almost periodic. Let n • L There exist K, i • / , 0 < i < N - 1, such that n = K N + 1. From Remark 7.2 there exists p • I 1 such that x(K + pj) = x(K) ( j ~ I). But then y (KN + p j N ) . . . y ( K N +pjN + N - 1) = y ( K N ) . . . y ( K N + N - 1 ) ( j • I ) , and hence y ( n + p j N ) = y ( n ) ( j • I ) . By Remark 7.2, y = ~*(x) is regularly almost periodic.

Now suppose ~ is one-to-one and y = ~*(x) is regularly almost periodic. Let n = I i. Choose P • I1, P > (n+ 1)N. By Remark 7.2 there exists p • I1 such that y ( i + p j ) = y ( i ) ( l i [ <-P, j • l ) , so that in particular, y ( - n N + p j ) ' " y ( n N + N - l + p j ) = y ( - n N ) . . ' y ( n N + n - 1 ) ( j • l ) . But this implies that

340 G.A. HEDLUND

~(x(i+pj)) = ~(x(i))([il -< n, j ~ I) and hence, since ~ is one-to-one, x( i+pj ) = x(i) (lil <- n, j ~ I), so that x is regularly almost periodic.

7.9 Remark. Let x ~ X ( St) , let N ~ 11 and let ~ be an N-block substitution over St . I f x is isochronous, then a*(x) is isochronous. I f ~ is one-to-one and ~*(x) is isochronous, then x is isochronous.

Proof The proof is similar to that of Remark 7.8.

7.10 Remark. Let x ~ X ( S t ) , let N ~ 11 and let ~ be an N-block substitution over St . I f x is almost periodic, then ~*(x) is almost periodic. I f ~ is one-to-one and ~*(x) is almost periodic, then x is almost periodic.

Proof Let x ~ X, let x be almost periodic and let y = ~(x). Let n s 11. Choose p ~ 11 such that p N > n. By Remark 7.4, there exists a syndetic subset E of I such that x ( i + j ) = x(i) (1il -< p, j ~ E), hence, y ( i N + j N ) . . . y ( i N + N - 1 + i N ) = y( iN) . . . y ( iN+ N - 1) (lil -< p , j ~ E). Let E* = { j N I j ~ E}. Then E* is syndetic and we have y ( i + k ) = y(i) (li] -< n, k ~ E*). Thus y = ~(x) is almost periodic.

Suppose that ~ is one-to-one and that y = g*(x) is almost periodic. By the Inheritance Theorem 4.04 of [10] it follows that y is almost periodic under cr N. Consequently, if n ~ 11, there exists a syndetic subset E o f / s u c h that y ( - n N ) . . . y ( n N + N - 1) = y ( i N - n N ) . . . y ( i N + n N + N - 1) (i ~ E). Hence c¢(x( i -n ) ) . . . ~(x(i+n)) = ~ ( x ( - n ) ) . . . ~ ( x ( n ) ) (i ~ E). Since ~ is one-to-one, it follows that x ( i - n ) " " x(i + n) = x ( - n). . . x(n) (i ~ E). This implies that x is almost periodic and the proof is completed.

7.11 THEOREM. Each o f the sets RAP(SO-P(S t ) , I ( S t ) - R A P ( S t ) and A P ( S t ) - I ( S t ) is everywhere dense in X ( S O.

Proof. As stated (cf. [10]), there exists x ~ X(St*), where card St* = 2, such that x is regularly almost periodic but not periodic. Let St be an arbitrary finite symbol set with card5 a > 2. Let N e 11 and let B e ~N(St)" Define ~: St~MN(St) such that ~ is one-to-one and ~(Xo) = B. This is clearly possible. By Remark 7.8, ~*(x) is regularly almost periodic and by Remark 7.7, ~*(x) is not periodic. Given E>0 and z e X(SO, we can choose N and B such that the orbit of ~*(x) under rr has at least one of its points within distance E of z. But each point in the orbit of ~*(x) is regularly almost periodic but not periodic. We conclude that the set R A P ( S t ) - P ( S t ) is everywhere dense in X ( S ) .

The proofs of density of the sets I (S t ) -RAP(S~) and AP( St) - I( St) are similar.

7.12 Remark. Let x E X ( St). Then the following paired statements are pairwise equivalent.

(1) x is a positively recurrent point o f (X(S t ) , ~). (2) Let n ~ 1 o. Then given any i ~ 1, there exists j ~ L j > i, such that x ( j + k )

= x([k I < n). (1 ') x is a negatively recurrent point o f (X(s t ) , (r). (2') Let n ~ I o. Then given any i E L there exists j ~ I, j < i, such that x ( j + k)

= x ( k ) ([kl -< n). (1") x is a recurrent point o f (X (SO, ~). (2") x is both positively recurrent and negatively recurrent.

The set of all positively recurrent [negatively recurrent] [recurrent] points of (X(~9°), a) will be denoted by PR(5 a) [NR(Sg)] [R(5¢)].

7.13 Remark. AP(2T) c R(5 a) = pR(Sf )nNR(Sf ) . The set R ( S P ) - A P ( 5 e) is everywhere dense in X( 5°). The set R( 5 °) is an invariant residual G6 subset of X ( ~ ) .

7.14 Remark. Let x ~ X( Sf). Then the following paired statements are pairwise equivalent.

(1) x is a positively transitive point of (X(~9~), ~). (2) Let n E 11 and let B ~ ~n(~9°). Then there exists i E I o such that x ( i ) . . .

x ( i + n - 1) = B. (1') x is a negatively transitive point of (X( Sa), ~). (2') Let n ~ 11 and let B ~ M,( Sf). Then there exists i c I o such that ( x - i - n + 1)

• . . x ( - i ) = B .

(1") x is a bilaterally transitive point of (X( Sa), ~). (2") x is both positively transitive and negatively transitive. (1") x is a transitive point of (X(Sa), or). (2") Let n E 11 and let B c~ , (S#) . Then there exists i c I such that x(i+ 1). . .

x ( i + n) = B.

The set of positively transitive [negatively transitive] [bilaterally transitive] [transitive] points of (X(Sg), ~) will be denoted by PT(S~ [NT(SP)] [BT(Sg)] [T(,~)].

7.15 Remark. PT(5 p) c pR(Sf). NT(5 °) c NR(Sa). BT(Se) c R(SP). BT(5~) = PT(Sa)c~NT(Se). The set BT(5 ~) is an invariant residual G~ subset of X(Sa).

7.16 THEOREM. T(5 f) = PT(Sf)wNT(S,a). Proof. It is clear that PT(5 ~) c T(Sa) and NT(5 f) c T(SP). Let x ~ T(5 p) and suppose that x is neither positively transitive nor negatively

transitive. Then there exist m ~ 11 and B c Mm(5 a) such that x(i). • • x(i + m - 1 # B for all i E I0, and there exist n c/1 and A ~ , ( 5 a) such that x ( - i - n + 1). . . x ( - 1 ) # A for all i c lo . Let A = a l " " a , , B = bl"" .b,,. Since x c T(S~), the block AB = al"" "a, bl"" "b,, appears in x and there exists j E I such that x ( j ) " . x ( j + n - 1)x(j+ n). . . x ( j + m + n - 1) = al" " "a, b l " - bm; hence x ( j ) . . . x ( j + n - 1) = A and x ( j + n ) " . x ( j + m + n - 1) = B. It follows that j + n - 1 > 0 a n d j + n < 0, which is impossible. We infer that either x E pT(Se) or x c NT(Sf).

The proof is completed. The following result is easily verified.

7.17 Remark. The dynamical system (X(5°), ~) is (topologically) mixing.

8. Invariance of Properties of (X(~9°), ~) under ,~ e ~p(~o) and under 9-1

8.1 THEOREM. Let 9eqb(Sa); that is, q~ is a continuous mapping of X(S f ) into X (S f ) which commutes with ~. Let Y be any of the invariant sets P(Sa), RAP(5:) , 1(5~), Ap(Sf), pR(Sa), NR(Se), R(Sf). Then 9(Y) is invariant and ~o(Y) c y.

Proof. It is obvious that for any invariant subset Y of X(Sf), the set 9(Y) is invariant.

342 G.A. HEDLUND

The proofs that ~ Y ) c y in the various cases cited are similar to each other and we present only one.

Let y • AP(Se) and let z = ~o(y). Let U be a neighborhood of z. Since ~o is continuous, there exists a neighborhood V o f y such that ~ V ) c U. Since y is almost periodic, there exists a syndetic subset E of I such that o~(y) • V (i • E) But then o~(z) = o~(~(y)) = ~(cri(y)) • U (i • E). It follows that z is almost periodic, and hence ~0(AP(Aa)) c AP(Se).

The simplest .example of ~0 ~ ¢p(6~) is that which maps X(Se) into a single point of X(6~), which is necessarily a fixed point under or. If this is not the case, the structure of ~(X(S~) is fairly complex.

8.2 THEOREM. Let ~o • ~( A °) and suppose that card ~X(AP)) > 1. Let Y = ~o(X(A°)). Then the following statements are valid.

(1) Y is a closed invariant subset of X(A~), and hence (Y, o) is a subdynamical system of (X(AP), ~).

(2) The set of periodic points o f (Y , ~) is dense in Y. (3) There exist bilaterally transitive points of ( Y, o). (4) (Y, o) is topologically mixing. (5) Y is a compact, totally disconnected, perfect, metric space and hence is

homeomorphic to the Cantor discontinuum.

Proof. (1) The set X(oq °) is compact; hence Y = ~0(X(6e)) is compact and thus closed. We have already remarked that Y is invariant under o.

(2) By Remark 7.6, the set p(Aa) is dense in X(Aa). Since ~0 is continuous, it follows that the set ~o(p(Se)) is dense in Y. But by Theorem 8.1, ~0(p(SP)) c P(S~), and we conclude that the set of periodic points of (Y, or) is dense in Y.

(3) By Remark 7.15, the set BT(SP) is an invariant residual G~ subset of X(Se), and hence BT(SP) ~ ~. Let x • BT(S a) and let y = ~0(x). Then each of the sets {~i(x)l i • Io}, {o- l (x) l i • I0} is dense in X(Sa), and hence each of the sets {o~(y)[ i • Io}, {o-Z(Y)l i • I0} is dense in Y, which implies that y is a bilaterally transitive point of (Y, ~).

(4) Let Uand Vbe non-vacuous open subsets of Y. Then ~- I(U) and ~o- I(V) are non-vacuous open subsets of X(Se). Since (X(6a), o) is topologically mixing, there exists N • I such that [i] > N implies ~o-l(V)ncri(~o - I(U) ~ ~. But then Iil -> N implies that V n ~i(U) ¢ e and (Y, or) is topologically mixing.

(5) We have already seen that Y is closed and compact. Since X(Sa) is totally disconnected, so is Y.

Let x • X(Se) be defined by x(i) = 0 (i • I) and let y = ~o(x). Then x is fixed under o and hence y is fixed under cr.

Since,by hypothesis, card Y> l, there exists z • Y, z ~ y. By (3), there exists w • Y such that w is bilaterally transitive in (Y, e) and thus the orbit of w under o is dense in Y. If w = y, then the orbit-closure of w is {y }, z is not in the orbit- closure of w and w is not bilaterally transitive. Thus w ¢ y. If w were periodic, its orbit-closure would be identical with its orbit and would not include y. Thus w is not periodic. Let u • Y and let U be a neighborhood of u in Y. By (2) there exists a periodic point p • U, and since w is bilaterally transitive, there exists a point w* in the orbit of w such that w* • U. Then p ¢ w*, so that U contains a

point Y distinct f rom u. I t follows that Y is perfect and the p r o o f of (5) is completed.

We have seen that many recursive propert ies are preserved under any mapp ing 9 e * ( t a ) • We consider the converse question. I f we assume some recursive proper ty of x and if 9(Y) = x, does it necessarily follow that y has this proper ty?

The following simple example shows that , wi thout restriction on 9, the answer may be in the negative. Let x be defined by xi = 0 (i ~ I) . Let 9(Y) = x (y E X(S~)). Then 9 ~¢(6a) . S incex is fixed under ~, x is periodic, a lmost periodic and recurrent. Since y is arbi trary, y may not exhibit any o f these properties.

This is a trivial example, but according to Theorem 5.12, i fn ~ I i , f ~ F(6¢, n) andfo~ is not onto, then there exists x e X(SP), x periodic, such that the s e t f ~ l(x) is uncountable, and hence there exist m a n y non-periodic points y such that foo(Y) = x. It is easy to see how to construct y such that y is not a lmost periodic and even not recurrent.

We can put the question in a somewhat less restrictive fashion and ask if, on the assumpt ion that 9 - l ( x ) # ~, there exists some y E 9 -1 (x ) which exhibits a given proper ty of x. The following theorems give some answers.

8.3 T H E O R E M . Let n = 11 and let f s F( SP, n). Let x be a periodic point and suppose that f g l(x) # ~. Then there exists y e f £ l(x) such that y is periodic.

Proof. We can assume that n > 2. For i f f s F(Se , I), define g E F ( S a, 2) by g(blb2) = f(bl) , (bib 2 e 9~2(Sf)). Then g® = f ~ .

N o w suppose tha t x has period ~o and that z s f £ l(x). Then mo is a period of x. Let a = no.

For i e I , let B(i )= z,iz~i'"z~i+n-2, so tha t B ( ° e ~ , _ 1 ( S f ) . Since card ~ ' , _ t (6 '~) = S "-1 , there exist i, j e I , i<j, such that B (*) = B (j;. Let A = B (°, let C = z , i + , _ l " "z~j_ 1 and let y . . . . A C A C A C . . . . Then y e X ( 6 a) and y is periodic. N o w ACA = B(*)CB (j) = z , i . . . z , j + , _ 2 and f , ( j_ l ) (ACA ) =x ,~ . . . x~j_ 1. Let D = x~, . . .x~j_ 1. Then fo~(Y) . . . . DDD . . . . x. The p roo f is completed.

8.4 C O R O L L A R Y . Let 9 e ¢(~9a). Let x ~ P( 6 a) and suppose that 9-1(x) # o. Then P ( d a ) n g - l ( x ) # ~.

Proof. This follows f rom Theorems 3.4 and 8.3. Let x e X(6¢). The a-limiting set o f x under ~, denoted by ~(x), is the set of all

y ~ X for which there exists a sequence o f integers " " < n - 1 <no such that lim~.~o~"'(x) = y. The set ~(x) is closed, non-vacuous and invariant under or.

The ~o.lirniting set o f x under or, denoted by oJ(x), is the set o f all y e X(oq a) for which there exists a sequence of integers no < nt < • • • such that lim~_.o~o~,(x) = y. The set ~o(x) is closed, non-vacuous and invariant under ~r.

8.5 L E M M A . Let 9 ~ ~(S~) and let x ~ X(~a). Then 9[~(x)] = ~[9(x)] and 9[o~(x)] = ~o[9(x)].

Proof. Let z ~ 9Ion(x)]; thus there exists y ~ ~o(x) such that z = 9(Y). Since y ~ ~o(x), there exists a sequence of integers no < nl < " ' " such that limi_.o~rn'(x) = y. But 9[~r"'(x)] = ~"'[9(x)], and since 9 is continuous, limi_~9[~r"'(x)] = 9(Y).

Thus z = 9 ( Y ) = limi~oo~"'[9(x)] and z ~ o~[9(x)]. I t follows that 9Ion(x)] = o~[9(x)].

Let 9(x) = u and let z ~ o~[9(x)] = o~(u). There exists a sequence of integers

344 G . A . HEDLUND

no < nl < " '" such that z -- lim~_~"~(u). Since X(A °) is compact, we can assume, by choosing a subsequence if necessary, that limv~o~o~(x) : y ~ X(A~), which implies that y e o~(x). Now

z = lim o~'(u) = lim o~'[~o(x)] = lim ~o[o~'(x)] = ~o(y). i- '~ oO i ~ a o i--~ oo

Thus z e ~[oJ(x)] and o~[~o(x)] c ~o[,.(x)]. We conclude that ~o[oJ(x)] = ~o[~o(x)]. The p roo f that ~o[a(x)] = a[~o(x)] is similar.

8.6 T H E O R E M . Let ~o e op(Aa). Let x ~ X(Aa), let x ~ NR(AQ [PR(AQ] and let 0 < c a r d ~o-l(x)< oo. Then there exists y e ~o-~(x) such that y e NR(6P) [PR(~9~)].

Proo f Let x ~ NR(S #) and u o E ~ - l (x ) . Then x ~ ~(x), and f rom Lemma 8.5 we infer that a(x) = a[~(Uo)] .= ~o[a(Uo)]. Thus there exists ul ~ o~(uo) such that ~o(u~) = x. Since o~(Uo) is a closed invariant set, a(ul) c a(Uo).

Now ~o[a(ul)] = a[~O(Ux)] = a(x), x e a(x), and there exists u 2 ~ a(ul) such that ~o(u2) = x and a(u2) c a(u~).

This process can be continued to define a sequence of points Uo, u~. . - with the following properties:

(1) ~o(u~) = x (i e Io) ,

(2) u ,e~(u i_~) c ~(u,-2) c " '" C ~(Uo) ( i e Io ) .

Since the collection of ~o-*(x) is finite, there exist m, n e Io with m < n such that Um= U,, and hence ~(Um) = ~(un). It follows f rom (2) that ~(Um) = ~(Um+~) . . . . . ~(Un) and Um+,eo~(um)= ~(Um+,). Thus Um+t ~NR(Sa) and since ~(Um+,) = X, there exists y = Um+~ ~ ~o-X(x) such that y ~ NR(Sa).

The p roo f for the case of positive recurrence is similar.

8.7 C O R O L L A R Y . Let ~ eO(~qa), let x e R ( Se), let card ~o-X(x) = 1 and let ~o(y) = x. Then y e R( AQ.

8.8 T H E O R E M . Let rp ~ dO(A a) and let x ~ AP(b°). I f card ~o-l(x)> 0, then AP(Sa)n~o-~(x) ~ ~.

Proo f Let x s A P ( b °) and let F (x) be the orbit-closure of x under o. Then F(x) is a minimal set and thus ~(x) = F(x) = ~o(x). Let u e ~o- X(x).

Since X(S#) is compact , any non-vacuous, closed invariant set contains a minimal set, so that ~o(u) contains a minimal set M. According to Lemma 8.5, ~o[o~(u)] = ~o[~(u)] = o~(x), and it follows that ~o(M) ~ ~o(x). Now ~o(M) is a closed invariant set, and since ~o(x) is a minimal set, we have ~o(M) = ~o(x) = F(x). Thus there exists y e M such that ~o(y) = x. Since M is minimal, y is almost periodic and the theorem is proved.

9. A Fundamenta l Proper ty o f Inverses

We show that of n ~ 12, f ~ F(~9 ~, n), f ~ is onto and card f ~ l(x) > 1,then the members of f f l ( x ) display certain fundamental separation properties. These properties carry over to any ~ ~ E(~f) . These results will enable us to show that many properties carry over to inverses in case ~ is onto.

The invariancc o f periodicity is easily proved.

9.1 THEOREM. Let ~ e E(SQ, let x e p( Sz) and let ¢p(y) = x. Then y • P( S/'). Proof. Suppose ~ • E(SQ, x • P(SQ, ~(y) = x and y is not periodic. Let ~o be

the period of x. Then ak°(x) = X (k • I). Since y is not periodic, if i, k e I and i ~ k, then #~'(y) ~ ak~'(y). For if

ai°'(y) = tTk°(y), then o (i- k)~(y) = y and y is periodic. Now for all k • / , ~[gkO(y)] = ~ko~(~(y)) = gkO,(X ) = X, and hence card ~0- l(x)

is not finite, contrary to Theorem 5.6.

9.2 LEMMA. Let n • 12 and let f • F(SP, n). Then f oo is not onto provided any one of the following conditions is fulfilled.

(1) There exist A •~ ,_I (SP) , k • I 1 and distinct members P, Q of ~k (S f ) such that

A + , - x(APA) = fk+,- x(AQA).

(2) There exist A, B e ~ , _ l ( S P ), k • 11 and distinct members P, Q of ~k(5 Q such that

A+.- I(APa) = A+.- I(AQB).

(3) There exist A, B • ~ , _ x(Sf), k • 11 and k-blocks P, Q, R such that either P ~ Q o r Q # R a n d

fk+,- 1(APA) = fk+,- I(AQ B) = fk+,- I(BRB).

Proof. Assume (I). Let Co = AP and let Ci = AQ (i • / , i # 0). Let

y . . . . C_ 1CoC1 . . . . . . . A Q A P A Q . . . ,

and let D = fg+,_ I(APA) = fk+,- I(AQA) • Then fo~(Y) . . . . DDD. . ., which is periodic. It is easily verified that since P # Q, y is not periodic. It follows from Theorem 9.1 that f~o is not onto.

Assume (2). Let C = fk+,- I(APB) = fk+,- I(AQB) and let f ,_ I(BA) = D. Then fk+z(,_l)(APBA) = CD = fk+2(,_I)(AQBA). Since PB # QB, (1) is fulfilled and f® is not onto.

Assume (3). It follows that

fzg+2(n- 1)(APAQ B) = f 2 k + 2(n- 1)(AQBRB),

and since PAQ ¢ QBR, (2) is fulfilled a n d f ~ is not onto. The proof is completed. Let x, y e X(SQ. Then x and y are negatively [positively] asymptotic provided

there exists N • I such that x i = y~ for i < N{ i > N} ; x and y are bilaterally asymptotic provided they are both negatively and positively asymptotic.

Let x ~ X(SQ, let m • I 1 and let i • L The m-block xixi+ 1" " "xi+,,- 1 will be denoted by B(x, i, m).

Let x, y ~ X(5 a) and let m • 11. Then x and y are totally m-separated provided that B(x, i, m) v~ B(y, i, m) (i E I); x and y are negatively m-separated provided there exists an integer N such that B(x( i, m) ~ B(y, i, m) for all i < N; x and y are positively m-separated provided there exists N • / s u c h that B(x, i, m) ~ B(y, i, m) for all i > N.

9.3 THEOREM. Let n • I2 , f e F(S¢, n) and let fo~ be onto. Let there exist y, z ~ X(~9°), y ~ z, such that fo~(y) = foo(z). Then y and z must be either positively ( n - 1)-separated or negatively ( n - 1)-separated; possibly both.

346 G.A. HEDLUND

Proof. Assume there exist y, z ~ X(S a) with y # z andf~(y) = fo~(z). Suppose that y and z are neither positively (n-1)-separated nor negatively ( n - 1 ) - separated. Since y # z, there exists k e I such that Yk # Zk. Choose M ~ 11 such that M > Ikl and M > n - 1 . Since y and z are not positively ( n - 1)-separated, there exists an integer i> M such that B(y, i+ 1, n - 1 ) = B(z, i+ 1, n -1 ) . Let this ( n - 1)-block be C. Since y and z are not negatively (n - 1)-separated, there exists an integer j < - M - n such that B(y, j + i, n - 1) = B(z, j + 1, n - 1). Let this ( n - 1)-block be D. Let P = Ys+, ' "Y i and let Q = z j+ , ' " z i . Then P # Q and Ys+ t ' "Y i+ , -x = D e c , zs+z . . . z i+ ,_ t = DQC. Since fo~(Y) =f~o(z), it follows thatfi_s(DPC) = f~_j(DQC). By Lemma 9.2, this implies that f® is not onto, contrary to hypothesis.


9.4 COROLLARY. Let n ~12 and let f ~ F(SP, n). I f there exist y, z ~X(A a) with y # z, y and z doubly asymptotic and f®(y) = .['~(z), then foo is not onto.

9.5 THEOREM. Let n >_ I 2, let f e F( Sa, n) and let f ~ be onto. Let there exist y, z e X( S e) with y # z and such that fo~(y) = f~(z). Then one and only one of the following statements is true.

(1) y and z are positively asymptotic and negatively ( n - 1)-separated. (2) y and z are negatively asymptotic and positively ( n - 1)-separated. (3) y and z are both positively and negatively ( n - 1)-separated.

Proof. It follows from Theorem 9.3 that i ff~ is onto, y # z andf~o(y ) = f®(z), then y and z must be either positively (n-1)-separated or negatively ( n - 1 ) - separated. Suppose that y and z are negatively ( n - 1)-separated. We show that either (1) or (3) is valid.

For suppose that y and z are neither positively asymptotic nor positively ( n - 1)-separated. Then, given any integer K, there exists k > K such that Yk # ZR, and given any integer M, there exists m > M such that B(y, rn+ 1, n - 1 ) = B(z, r e + l , n - l ) . Thus there exist integers, q, r, s such that q + n < r < s , B(y, q+ 1, n - 1 ) = B(z, q+ 1, n -1 ) , B(y, s+ 1, n - 1 ) = B(z, s+ 1, n - 1 ) and y, # z,. Let

C = B ( y , q + l , n - 1 ) = B ( z , q + l , n - 1 ) , D = B(y, s + l , n - l ) = B ( z , s + l , n - 1 ) , P = Y~+~'"Ys, Q = zq+,...z~.

Since q + n < r < s andyr = zr, we have P # Q.

Since fo~(y) = fo~(z), we havefs_q(CPD) =f~_q(CQD), and it follows from Lemma 9.2 thatfoo is not onto, contrary to hypothesis. Thus i f y and z are negatively (n-1)-separated, then either (1) or (3) is valid.

I f y and z are not negatively (n - 1)-separated, they must be positively (n - 1)- separated, and a similar proof shows that either (2) or (3) is valid.

Since the statements (1), (2), (3) are mutually exclusive, the proof is completed.

Let x, y ~ X(SP). Then x and y are positively [negatively] separated (under ~) provided that inf.~i~ d(cr"(x), o~(y)) [inf.~1~ d(~-"(x), ~-"(y))]>O; x and y are separated (under or) provided x and y are both positively and negatively separated.

9.6 COROLLARY. Let ~o e E ( ~ ) and suppose there exist y, z ~ X(SP) such that y ~ z and ~(y) = q)(z). Then either (1) y and z are positively asymptotic and negatively separated, or (2) y and z are negatively asymptotic and positively separated, or (3) y and z are separated.

Proof. Use Theorem 3.4 and Theorem 9.5. Let n ~ 12 and le t fE F(,9 °, n). Let m, k, q e I 1 with m >n - 1. LetM be a collec-

tion of blocks over 5 a. Then M has property P(m, k, q , f ) provided the following conditions are fulfilled.

(1) Each member of ~ is an m-block. (2) card ~ = k +q. (3) The collection d of initial ( n - 1)-blocks of the members o f ~ is identical

with the collection of terminal ( n - 1)-blocks of ~ and card d = k. (4) There exists an ( m - n + l ) - b l o c k D such that fm-n+i(B) = D for each

B e ~ .

9.7 LEMMA. Let n e 12, f e F( 9 o, n) and let there exist a collection ~ o f blocks over 5 ~ with property P(m, k, q, f ) . Then there exist a positive integer r and a collection ~ * of blocks with property P(2m - n + 1, k, q + r, f ) .

Proof. Let d be the collection of initial ( n - 1)-blocks of members of N', which, by assumption, is identical with the collection of terminal ( n - 1)-blocks o f ~ .

Let ~ * be the collection of all ( 2 m - n + 1)-blocks QAR, where QA e ~ and A R e ~ . Then A is an ( n - 1)-block and A ~ d . We show that M* has the desired property.

By definition, each member o f ~ * is a ( 2 m - n + 1)-block, so that M* satisfies condition (1).

Let ~ = {Bil i = 1, 2 , " ' , k+q} . Since card 6~ = k + q and card d = k < k + q, there must be two members of M with the same initial ( n - 1)-block, and we can assume that B 1 and B E have this property. Let B ~ = A~R ~, where A i is an ( n - 1)-block; thus A i e ~ and, in particular, A 1 = A 2.

Some member of M must have terminal (n-1) -b lock C ~ = A ~. Write this member as Q1C1 and let the remaining member o f .~ be given by {Q*C*[ i = 2, • . . , k+q} , where each C i is an ( n - 1)-block.

Let 1 k+q . In particular we can choose j(1) = 1. But Q1C1R 2 = Q1A2R2 is also a member of ~ * and thus card ~ * > k + q . Let r = card ~ * - k - q . Then r > 0 and card ~ * = k + q + r , and ~ * satisfies condition (2).

Since any member of ~ * is of the form QAR, where QA E ~ , A R ~ ~ , any initial ( n - 1)-block of a member of ~ * is in s~¢ and any terminal (n - ) -b lock of a member o f ~ * is in~¢. But if QA is any member of M, then there exists a block R such that A R ~ ~ , so that the collection of initial (n - 1)-blocks of the members o f ~ * is exactly the setz~¢. Similarly, if A R ~ ~ , there exists a block Q such that QA e ~ , so that the collection of terminal (n - 1)-blocks of the members of 9~* is also exactly the set~¢. Thus ~ * satisfies condition (3).

By hypothesis, there exists an ( m - n + 1)-block D such that fro-,+ 1 ( B ) = D

348 G.A. HEDLUND

for each B ~ ~ . Let C e ~* . Then C = QAR, where QA e ~ and A R e ~ , with A an (n-1)-b lock; hence f2m-2,+2(C) = DD. Thus condition (4) is fulfilled.

The proof is completed. The following theorem is a generalization of Lemma 9.2.

9.8 THEOREM. Let n ~ I 2 and let f ~ F( 5 ~, n). I f there exists a collection o f blocks over 5 ° with property P(m, k, p, f ) , then foo is not onto.

Proof. If f® is onto, it follows from Theorem 5.4 that card fq- I(B) = S "- 1 for every q-block B and every positive integer q. But if there exists a collection of blocks with property P(m, k, p, f ) , then by repeated application of Lemma 9.7 we can determine a q-block B such that card fq- I(B) > S"- 1. We infer tha t f~ is not onto.


10. Inverse of Recurrent Points are (n-1) -Separated

I f f ~ F ( ~ , n), n ~ I2, and f~ is onto, then the inverses of a recurrent point not only cannot be asymptotic but exhibit the much stronger property of being totally ( n - 1)-separated.

10.1 LEMMA. Let y, z e X ( ~ ) , let n e I2 and let y and z be positively [negatively] (n-1)-separated. Let there exist a sequence o f integers no< nl < " " [ " " < n 1 <n0] such that limi~o~ni(y) = u and lim~.~oo~ni(z) = v [limi_.~o~"i(y) = u and limi_,o~"~(x) = v]. Then u and v are totally ( n - 1)-separated.

Proof. Assume that y and z are positively ( n - 1)-separated. Then there exists an integer J such that if j_> J, then B(y, j , n - l ) = Y j Y j + I " " Y j + , - z and B(z, A n - 1) = zjzj+ 1"" "Zj+n- 2 are distinct.

Let no<n1 < "'" and let limi_,o~n~(y) = u, lim~_,®e"'(z) = v. Let k e L We show that B(u, k, n - l ) = Uk ' ' 'Uk+ , - z and B(v, k, n - l ) = Vk" "'Vk+,-2 are distinct.

Let M e I 1 with M > I k [ + n - 2 . There exists K e I 1 such that if i > K, the central (2m+ 1)-block of o"'(y) is identical with the central (2m+ 1)-block of u and the central (2m+ 1)-block of ~n'(z) is identical with the central (2M+ 1)- block of v. This implies that for i > K,

B(u, k, n - 1) = u k" • "Uk+m_ 2 ~ - " Y,~+k" "'Yn~+k+n-2 = B(y, n i + k , n - I),

B(v, k, n - 1 ) = Vk" " "Vk+,_ 2 = Z,,+k" " "Z,i+k+,_ ~ = B(z, n~+ k, n - 1 ) .

Now choose i > K such that n ~ - M > J . Then n ~ + k > n ~ - M > J , B(y, h i + k , n - l ) ~ B(z, n~+k, n - l ) , and hence B(u, k, n - l ) # B(v, k, n - l ) . It follows that u and v are totally (n-1)-separated.

The proof in case y and z are negatively (n - 1)-separated is similar.

10.2 COROLLARY. Let k e / 2 , let y~s X(£#) (i = 1, 2 , . . . , k), let n E 12

and let y i and y j be positively [negatively] (n - 1)-separated provided 1 < i < j < k. Let there exist a sequence o f integers n o < n x < " " [" "" < n - l < n o ] such that l imm_.~nm(y i) = U i [limm_._~¢"m(y i) = u i] (i = 1, 2 , ' ' ' , k). I f 1 <_ i < k <_ j , then u ~ and u j are totally (n - 1)-separated.

10.3 THEOREM. Let n e I 2 and let f e F(S#, n). Let x ~ X ( 5 : ) and let there exist y, z e X (S f ) , y ~ z, such that f ~ ( y ) = x = f ~ ( z ) and y and z are positively

[negatively] (n-- O-separated. Let u ~ to(x) [u ~ ~(x)]. Then there exist v E oJ(y) [~(y)], w e to(z) [~(z)] such that v and w are totally (n - 1)-separated and f , ( v ) = u

= f o~( w) .

Proof. Assume t h a t f , ( y ) = x = foo(z) and that y and z are positively (n - 1)- separated. Let u e to(x). Then there exists a sequence n o < n 1 < . - . o f integers such that l imi+~a" ' (x ) = u. Since X ( 5 : ) is a compac t metric space, we can assume, by choice o f a subsequence, that limi_.~er"'(y) = v and limi_.~o@'(z ) = w. Then v E to(y) and w ~ to(z). It follows f rom L e m m a 10.1 that v and w are totally ( n - 1)-separated. N o w

fob(v) = fo~(lim @'(y)] = lim cr",(y) = lim ~"'(foo(Y)) = lim ~"'(x) = u.

Similarly, foo(w) = u. The p r o o f when y and z are negatively ( n - 1)-separated is similar.

10.4 C O R O L L A R Y . Let n ~ 12 and let f e F ( 5 : , n). Let x e X ( S # ) and let there exist {yi I 1 < i < k, f i e X(~9°) } such that foo(y i) = x (i = 1, 2 , " ", k) and i f 1 <_ i < j <_ k, then yl and y i are positively [negatively] ( n - 1)-separated. Let u s to(x) [u s e(x)]. Then there exists a set

[vii 1 <_ i < k, v i ~ w(yi)] [[vii 1 _< i _< k, v i ~ 0c(yi)]]

such that i f 1 < i < j < k, then v i and v i are totally ( n - 1)-separated and foo(v i) = u , l < i < k .

The following theorem is due to L. R. Welch.

10.5 T H E O R E M . Let n ~ /2, let f e F ( 5 : , n) and let fo~ be onto. Let x ~ X ( 5:), let x be recurrent and let there exist y, z ~ X ( SF), y ~ z, such that foo (y) = x = foo (z). Then y and z are totally ( n - 1)-separated.

Proof. Iff~o is onto, it follows f rom Theorem 5.5 that c a r d f ~ ' ( x ) < oo for each x ~ X(5 : ) . Let x be recurrent ; that is, x is bo th positively and negatively recurrent, and let there exist y, z ~ X(5¢), with y # z, such thatfo~(y) = x = foo(z). Let cardfo~ X(x) = K and letfo~ X(x) = {yi[ i = 1, 2 , . . . , K}. Since y, z ~ f ~ l(x), K _ > 2 .

F r o m Theorem 9.3, y and z are either positively (n - 1)-separated or negatively (n -1 ) - s epa ra t ed . Suppose y and z are positively ( n - D - s e p a r a t e d . Since x is positively recurrent, x e co(x), and it follows f rom Theorem 10.3 that there exist u, v e X ( 5 : ) such that foo (u) = x = foo (v) and u and v are totally (n - 1)-separated. The same result follows when y and z are negatively ( n - D-separated.

Let M be the maximal number of mutual ly totally ( n - D-separa ted members of fo~m(x). I t has been shown that M >_ 2, and we can assume that {yi[ i - - 1, 2," • ", M} is such a set. I f M = K, the s ta tement of the theorem is true. We assume tha t M < K and show that this leads to a contradict ion.

L e t j be an integer such that M < j <_ K. We show that yJ is positively asymptotic to some y~, 1 _< i _< M, and negatively asymptot ic to some yk, 1 < k <_ M, k # i .

For suppose yJ is not positively asymptot ic to any member of {y~," • . , ym}. Then according to Theorem 9.5, yJ is positively ( n - 1 ) - s e p a r a t e d f rom each member of { y l , . . ", yM}. Since x is positively recurrent, x ~ to(x), and it follows f rom Corol lary 10.4 that there are M + 1 members o f f ~ l(x) which are mutual ly

350 G.A. HFa3Ltn~I)

totally (n - 1)-separated. Since this is not the case, we infer that yJ is positively asymptotic to some y ' , 1 _< / _< M.

The p roo f that yJ is negatively asymptotic to some yk, 1 < k < M, is similar. It follows f rom Corollary 9.4 that i # k.

Let B(y i, q, n - l ) i i . = YqYq+l"" Yq+n-2, where 1 __ N ( j ) , and hence B(y i, q, n - 1) = B(y j, q, n - 1) provided q >_ N ( j ) . Let N = sup {N( j ) I j = M + 1,-" ", K}. Then i f q > N , the collection o f ( n - 1)-blocks {B(y i, q, n - 1)l i = 1, 2 , . . . K } is identical with the collection {B(y ~, q , / / -1 ) [ i = 1, 2,. • . , M}.

Similarly, there exists an integer Q such that i fq _< Q, the collection o f ( / / - 1)- blocks {B(fi, q, n - l ) [ i = 1, 2 , . . - , K} is identical with the collection {B(y', q, n - l ) [ i = 1, 2 , . . - , M}.

We show that for some m > N, the set {B(y i, m, n - 1)1 1 <_ i _< M} is identical with the set {B(y ~, Q, n - l ) l 1 _< i _< M}. By hypothesis, x is recurrent, and hence positively recurrent, and it follows that ~rQ(x) is positively recurrent, so that there exists a sequence of integers nl <I/2 < " " such that limj_~oo~J(x) = eQ(x). We can assume, by choosing a subsequence if necessary, that lims_, ooo~(y ~) = z ~ (i = 1, 2 , - . - , M) . F rom Corollary 10.2, we infer that if 1 < i < j < M, then z i and z j are totally ( / / - 0 - s e p a r a t e d . Fur thermore , for 1 < i _< M,

f®(z ~) = fo~( lim o"J(yi)) = limf~o(~"S(y~)) = lim ~"~(f~o(Y')) \j~oo j ~ j-~oo

---- lira #'fix) ---- crQ(X). j-,oo

But foo(~e(fi)) = ~e(x) (1 < i < M) and the set {aQ(yi)[ 1 _< i _< K} must be a maximal collection of totally ( n - 1)-separated inverses of crQ(x), so that the collection {z~l 1 < i < M} must be identical with the collection {~e(y~)] 1 N and let m = nj. Then {B(y i, m, / /-1)11 <_ i_< M} = {B(o"J(y~),O,n-1)l 1 < i < M } = {B(z~,O,n-1)l l 0, it follows f rom Theorem 9.8 that f ~ is not onto, contrary to hypothesis. We infer that K = M and the p roo f of the theorem is completed.

10.6 C O R O L L A R Y . Let ~ e E ( ~ ) , let x be recurrent and let there exist y, z e X ( A a) with y ¢ z such that 9(Y) = x = rp(z). The/ /y and z are separated.

Proof. According to Theorem 3.4, there exist m e I , / /E 11 and f ~ F (A ° , / / ) such that ~ = d~f~. Since ~ e E(~°) , ~ is onto; hence f ~ is onto. I f / / = 1, then, by Theorem 4.1, f ~ is a homeomorph ism of X ( ~ ) onto X(~9 °) and the same is true of ~, so that there cannot exist y, z s X(A °) with y # z such that ~0(y) = x = ~(z). T h u s / / s 12.

We assume that x is recurrent and that ~0(y) = x = ~(z) with y ¢ z. Then e~f~(y) = x = omf~(y) and hence f~ (y ) = ~-m(x) = f~(z) . But if x is recurrent, so is e-re(x), and it follows f rom Theorem 10.5 that y and z are totally ( / / - 1)- separated. This implies that y and z are separated.

Endomorphisms and Automorphisms of the Shift Dynamical System 35 1

11. Almost all Points Have the Same Number of Inverses

It has been proved by A. M. Gleason and L. R. Welch, using quite dissimilar methods, that iffo~ is onto, there exists an integer k such that c a r d f ~ l ( x ) = k for every bilaterally transitive x E X(50). Our proof of this fact has somewhat the flavor of that of Welch.

11.1 THEOREM. Let n E 12, let f s F(50. n) and let f ~ be onto. Let x e X(50), let x be bilaterally transitive and let c a r d f ~ l(x) = M (x). Then cardfo~ i(y) > M(x)

for every y ~ X(50) and f ~ l ( y ) has at least M(x) members which are mutually totally (n - 1)-separated.

Proof. L e t f o ~ ( x ) = {x~l 1 < i < M(x)}. Since x is bilaterally transitive, x is recurrent, and it follows from Theorem 10.5 that if 1 < i < j < M(x), then x ~ and x j are totally (n-1)-separated, hence, in particular, positively ( n - 1 ) - separated. Let y ~ X(50). Since x is positively transitive, y E co(x), and it follows from Corollary 10.4 that there exists as set [yi[ 1 M(x) and the statement of the theorem is proved.

11.2 THEOREM. Let n ~ lz, let f ~ F(50, n) and let f ~o be onto. Then there exists an integer M ( f ) such that i f x ~ BT(50), then c a r d f ~ l(x) = M ( f ) .

Proof. Let z ~ BT(50) and let M ( f ) = card f~- l(z). Let x ~ BT(50) and let M(x) = card f ~ l ( x ) . It follows from Theorem 11.1 that M(x) > M ( f ) and M ( f ) > M(x) , and thus M(x) = cardf~- l (x) = M ( f ) . The proof is completed.

11.3 Remark. Let n E I2, let f ~ F(50, n) and let f~o be onto. For x E X(50), let M (x) be the maximum number of members o f f ~ l(x) which are mutually totally (n-1)-separated. Then M ( f ) = infx~x(z)M(x ).

Proof. This follows from Theorem 11.1 and 11.2. The integer M ( f ) can be characterized in terms of the inverses of finite blocks. Let m ~ 1 1, let k E/1 with k >_ m, and let A = aaa2" • "ak and B = bib2" • "bk

be k-blocks over 50. Then A and B are m-separated provided that for 1 < i < k

- m + 1, aiai+ 1 • • "ai+m_ 1 ~ bibi+ 1 • • .bi+m_ 1. A set ~ of k-blocks is m-separated provided any two members of ~ are m-

separated. In particular, if card M = 1, then M is m-separated. Let n ~ I 2 , f ~ F(50, n) withf~ o onto, and let B be an m-block over 50. Then

f~-l(B) is a collection of ( m + n - 1)-blocks and card fm-i(B) = S "-1. The set f~-l(B) contains an (n-1)-separated subset (any single element off~-I(B)) ; let

be the collection of ( n - 1)-separated subsets of f~- l (B) , Then supG~ ~ card G is finite and we will denote this integer by M ( B , f ) . Clearly 1 < M ( B , f ) < S ~- i.

11.4 THEOREM. Let n ~ I 2 , f E F(50, n) and let foo be onto. L e t ~ denote the collection of all finite blocks over 5 °. Then

M ( f ) = inf M ( B , f ) .

Proof. Let B e ~ and let B be of length k. Let x e X(50) be bilaterally transitive.

352 G.A. HEDLUND

According to Theorem 11.2, c a rd f~ l(x) = M ( f ) . Since x is bilaterally transitive, x is recurrent and, from Theorem 10.5, the members off~-l(x) are mutually totally ( n - 1)-separated. Le t f~ l (x ) = (YJl 1 <_ j <_ M ( f ) } . But B appears in x, so that we have B = Xi+lXi+2"''Xi+k for some i e I . Since f®(yJ)= x, i <_ j <_ M ( f ) , it follows that

J J . A(Yi+lYi+2 "'YJ+k+n-1) = Xi+j" "Xi+k,

• "" J efk-X(B), 1 < _ M ( f ) . Since yJ and y" are ( n - 1)- so that Y~+t" Yi+k+n-1 J < separated i f j # p, the set {y~+ 1"" "Y~+k+n- 1l 1 <_ j < M ( f ) } is (n-- 1)-separated and hence M(B, f ) >_ M ( f ) for all B s ~ .

Suppose M(B, f ) > M ( f ) for all B s ~ . That is, corresponding to any m ~ 11 and any m-block B e ~ , there exists an ( n - 1)-separated set ~'(B) of (m+n - 1)- blocks with card ~ ( B ) = M ( f ) + 1 and ~(B) = f~- ~(B).

Let x be bilaterally transitive and for p E 11 let Bp = x _ / . . X o . . . x , . Let

~(Bp) = {y_p( j ) . . . yo ( j ) . . . y ,+n_ l ( j ) I 1 < j <_ M ( f ) + l }

be a set with the properties listed above. For p~I1 and 1 _<j _< M ( f ) + l , let y ( j , p ) s X ( ~ ) be such that [y(j,p)]~ =

Yi(J) ( - P <- i <_ p + n - 1 ) and let f~ (y ( j , p ) )= x(j, p ) ( j = 1, 2 , . . . , M ( f ) + 1). Then [x(j, p)]i = x~ ( - p < i <p ) , so that for each j(1 _<j < M ( f ) + l , limp_.~ox(j, p) = x. Since X(5 f) is compact, there exists a sequence Pl <P2 < " '" such that limk-, ooY(j, Pk) = y J(1 < j < M ( f ) + 1). Then

f o~(Y J) = f o~ [~im y(j, Pk)]l = k-*®lim f ®(Y(J' Pk)) = k-~o~lim x(j, Pk) = X.

Since the set g(Bp) is (n-1)-separated, it follows that the members of {yJ[ 1 <_ j <_ M ( p ) + 1} are mutually totally (n-1)-separated. This contradicts Theorem 11.2. We conclude that for some B, M(B, f ) = M ( f ) .

The proof of the theorem is completed. In the preceding theorems, the restriction has been imposed t h a t f s F(~9 ~, n)

with n > 2. But in view of Theorem 4.1, the above theorems are still valid if n = 1. We note that i f f e F(SP, n) and foo is onto, the M ( f ) = 1.

By virtue of Theorem 3.4, the preceding theorems apply to any continuous map of X(S~) onto X(5 ¢) which commutes with the shift.

11.5 THEOREM. Let ~o E E(Na). Then there exists a positive integer M(~o) such that i f x e X(SP) and is bilaterally transitive, then card ~o- l(x) = M(rp). Furthermore, i f y e X( Sf), then card ~o- l(y) >_ M(~) and the set q)- X(y) contains M (~o) members which are mutually separated.

11.6 THEOREM. (L. R. Welch) Let n e 12, let f e F ( ~ , n) and let foo be onto. Let x be positively [negatively] transitive and let f~o(Y) = x = fo~(z) with y ¢ z. Then y and z are negatively [positively] (n - 1)-separated.

Proof. We assume that x is positively transitive, that f~(y) = x = f~(z) with y ~ z and that y and z are not negatively (n - 1)-separated. It follows from Theorem 9.5 that y and z are negatively asymptotic and positively ( n - 1)-separated.

Since y and z are negatively asymptotic, there exists k e I such that y, = x, (i < k+n) . Define u~ (i < k - l ) such that u i e 6 ¢ and such that the left ray

• " ' uk- 2Uk- X contains every finite block over 5#. This is possible since the totality of all finite blocks over 5 ° is countable. Let

y ' = . . " Uk_ 2Uk_ lYkYk + x • • . ,

2 ' ~-- " " " U k _ 2 U k _ l Z k T . k + l " ' "

. . . . U k - 2 U k - l Y k Y k + 1" " " Y k + n Z k + n - l Z k + n + 2 " " " ,

X r : • . " V k _ 2 V k _ x X k X k + l . • •

where v~ = f ( u ~ . . . u i + , _ 0 (i < k - n ) ,

v k - ,+ x = f ( u k - , + x " .u~_ xY~),

Vk-,+ 2 = f ( U k - , + 2" " "Uk- lYkYk+ X),

Vk-1 = f ( U k - x Y k ' ' ' Y k + , - 2 ) .

Thenf~(y ' ) = x ' = f ~ ( z ' ) , y ' and z' are negatively asymptotic and positively (n - 1)-separated. Since y ' is negatively transitive, the set {a-"(y')[ n e I 1 } is dense in X(5#), and hence the set {~-"(x')I n ~ I x } is dense in X(Se), so that x' is negatively transitive• Since x is positively transitive and x'~ = x~ (i > k), it follows that x ' is positively transitive. Thus x' is bilaterally transitive. But it then follows from Theorem 10.5 that y ' and z' are totally ( n - 1)-separated, which is not the case. We infer that y and z must be negatively ( n - 1)-separated.

The proof of the second reading is similar.

12. Invariance o f Propert ies Under cp-1 W h e n ~ e E ( 5 # )

We have already seen (Theorem 9.1) that if q~ ~ E(5#), ~(y) = x and x is periodic, then y is periodic. We show that the corresponding result is valid when x is almost periodic or recurrent• It is not true in all cases.

12.1 Remark. There ex i s t s q~ e E ( 5 : ) and x , y ~ X ( 5#) such that q~(y) = x and

x is regularly a lmos t per iodic [isochronous] while y is not.

P r o o f Use 12.39, 12.54, 12.55 of [10]. Let X = X(5#), and for k e 11 let X (k) denote the cartesian product X t × X2 x

• . . x Xk, where X~ = X(5 #) (i = 1, 2 , . . . , k). Let X (k) be provided with the product topology. Then X (k) is a compact metrizable space•

Let (X (k), ~b) be the space product of the family of transformation groups ((Xi, ~)l i < i < k). That is, ~b is the homeomorphism of X (k) onto X (k) defined by ~b(x 1, x2, . . . , x k) = (a(xX), ~(x2), ' ' ' , ~(xk)). The homeomorphism ~b is the produc t t rans format ion on X (k) induced by ~.

12.2 THEOREM. L e t n ~ 12, let f e F ( 5#, n) and let x ~ X ( 5: ) . L e t yX, y2, . . ", yk e f ~ I(X) with k > 1 and f o r 1 < i < j < k , let y t a n d y j be pos i t ive ly [negatively] ( n - 1 ) - s e p a r a t e d . Then there ex i s t s a = (a x, a2, . . . , a k) e X (k) such that:

(1) a is a lmos t per iodic under ~ ; (2) a' ~ w(y ' ) {=(y')} (i = 1, 2,. • -, k); (3) a i is a lmost per iodic under o;

(4) i f 1 <_ i < j < k , then a ~ and a ~ are total ly ( n - 1)-separated; (5) there ex i s t s c ~ oo(x) [~(x)] such that f ~ ( a ~) = e (i = 1, 2 , . . . , k) .

354 G.A. HEDLUND

If, in addition to the s ta ted hypotheses, it is assumed that x is almost periodic, then the fo l lowing addit ional condition can be fulfilled.

(6) f~o(a t) = x (i = 1, 2 , " . , k).

Proof. We assume that the members o f {y~[ 1 < i < k } are mutually positively (n - 1)-separated.

Let u = (y l , y 2 , . . ", yk) ~ x(k) and let ~o(u) be the co-limiting set of u under ~b. Since X <k) is compact , oJ(u) is non-vacuous. Moreover , co(u) is closed and invariant under ~b and thus oJ(u) contains an orbit-closure M which is minimal under ~b. Let a = (a ~, aZ. • . . , a k) E M . Since M is minimal under ~b, it follows that a is almost periodic under ~b, which implies that each a ~ (i = 1, 2 , . . . , k) is almost periodic under ~. Thus (1) and (3) are proved.

Since a ~ M c ~o(u), there exists a sequence of integers no < nl < n2" ' " such that limj_.oo~b"J(u) = a and hence limj_,o~o~J(y i) = a i (1 < i < k). This implies a i ~ oJ(y i) (1 _< i _< k) and (2) is proved.

Statement (4) follows f rom Corollary 10.2. As above, limj_.®cr"J(y t) = a i (1 _< i _< k), and since f® is continuous, it

follows that limj_.~ofo~[cr"J(yt)] = fo~(ai), 1 _< i < k. But f~o commutes with ~, andf~o(y i) = x so that limj_+~o fo~[o~(y~)] = limj..~o~"~[f~o(yi)] = lim~_.o~o~ffx)= foo(a i) (1 < i < k). Let c = fo~(a 1) = limj~o~"ffx). Then c ~ o~(x) andfoo(a i) = c, 1 < i < k. This proves (5).

Now we assume that x is almost periodic under or. Then the orbit-closure of x, l ' (x) , is a minimal set, so that oJ(x) = _r'(x). There exists a sequence of integers no<nx < " " such that l imj_~"~(c ) = x, and we can assume the sequence so chosen that l i m j _ ~ " ~ ( a i) = ai, (1 < i < k) and thus limj_~o~b"ffa) = a , , where a , = (al,, a2,, . . . , ~ ) . Since a belongs to the minimal set M, a , e M and a , is almost periodic under ~b. It follows that each a~, is almost periodic under

i (1 < i < k). Since a , ~ M c co(u), a , ~ o~(fi) (1 < i < k). Since the members of {a~t 1 _< i < k} are mutually totally ( n - 1)-separated, the same is true of the members of {a~,l 1 _< i _< k}. Sincef~o commutes with ~, we have

x = lim o~(c) = lim enJ(foo(ai)) j-*oo j~oo

= limfo~(~r"~(a')) = fo~(a~) (1 < i < k). j -* oo

The p roo f of the first reading is completed. The p roo f of the second reading is similar.

12.3 T H E O R E M . Let n ~ 12, let f E F ( 5 ¢, n) and let foo be onto. Le t x ~ X(Aa), let x be almost periodic under ~ and let f ~ l(x) = {ill i = 1, 2,- • -, k } with k > 1. Then the point (y l, y2, . . ., yk) o f the product space X (k) is almost periodic under ~b.

Proof. Since x is almost periodic, x is recurrent, and it follows from Theorem 10.5 that the members of {yi I i<_ i < k } are mutually totally (n-1) -separa ted . Applying Theorem 12.2, we infer the existence of a = (a 1, a2, • • . , a k) ~ X (k) with properties (I) through (6) of that theorem. But then the set {ail 1 _< i < k} must be identical with the set {ill 1 _< i < k}. It follows that (y l , y 2 , . . . , yk) is almost periodic under or.

12.4 T H E O R E M . Let n E 11, let f e F( ~ , n) and let f ~ be onto. Let x ~ X ( aa), let x be almost periodic and let y e f ~ l(x). Then y is almost periodic.

Proof. I f card f~ - l ( x ) = 1, the theorem follows f rom Theorem 8.8. In particular, i f n = 1, then by Theorem 4.1, c a r d f ~ - l ( x ) = 1, so that we can assume tha t n > 2 and c a r d f ~ - l ( x ) > 1. But now the stated result follows f rom Theorem 12.3.

12.5 C O R O L L A R Y . Let q~ ~ E( 5~), let ~o(y) = x and let x be almost periodic. Then y is almost periodic.

Proof. Use Theorems 3.4 and 12.4.

12.6 T H E O R E M . Let n E I2, let f ~ F( 5 a, n) and let f ~ be onto. Let x ~ X(og a) be recurrent and let c a r d f ~ l(x) = k > 1. L e t f ~ l(x) = {yi[ 1 < i < k}. Then the point (y~, y2,. . ., yk) ~ x¢k) is recurrent under the product transformation ~b on X (k) induced by ~.

Proof. Sincef~o is onto and x is recurrent, it follows f rom Theorem 10.5 that the members of {Yil 1 _< i < k} are mutual ly totally ( n - 1)-separated. Since x is positively recurrent, there exists a sequence of integers m0 < ml < " ' " such that limj_.®~m;(x) = x. We can assume, replacing this sequence by a suitably chosen subsequence if necessary, that

(1) lira o " ' ( y i) = z / (i = 1, 2 , . . . , k). j-+m

Since the members of {yi[ 1 < i _< k} are mutual ly totally ( n - 1)-separated, the same is true of the members of {z il i < i < k}. We also have

f®(z') = f®(lim o-"(yi))j= limf®(o'ffyi))

= lim g"'(f®(Yi)) = lim flmJ(x) = X .

Thus the collection {zil 1 < i < k} is identical with the collection {yi[ 1 < i < k}, and there exists a pe rmuta t ion ~r of the set o f integers (1, 2,. • . , k) such that z ~ = y~(i). Then

(2) lim crm~(y ~) = y~(O (i = 1, 2 , ' " , k). j--+~

Let ~ and/3 be permuta t ions of the integers (1,2, . • . , k). We say that an ordered set o f q-blocks (B'(1), • " . B "(k)) over 5 a appears arbitrarily far to the right in the ordered set (yp(~),.. ", yp(k)) provided there exists a sequence o f integers

no < n~ < • • • such tha t v#(i)v#(i) . (~) "'Y~,j+q-1 = B~(° ( i = 1 , 2 , " " k ; j = O , 1 , . " ) . J n j J n j + l

Let Bi(m, N) i • i = Y m - N " " "Y~,'" " Y m + N (i = 1, 2 , . . . , k).

F r o m (2) we infer that any ordered set (B"(~)(m, N) , " . ' , B"(k)(m, N) ) appears arbitrari ly far to the right in (yl , y 2 , . . . , yk). But (2) implies that

(3) lira o" f fy ~ ) ) = y~'(o (i = 1, 2 , . - . , k), j~cv

and hence any ordered set

(B"~(l)(m, N)," " , B"~(k)(m, N) )

356 G.A. HEDLUND

appears arbitrarily far to the right in ( y l , . . . , yk). Proceeding inductively, we infer that any ordered set

(B"~°)(m, N), . . ., B"~<k)(m, N))

appears arbitrarily far to the right in ( f l , . . . , yk).

But there exists an integer i such that rr i is the identity. Thus any ordered set (Bl(m, N ) , . . . , Bk(m, N)) appears arbitrarily far to the right in ( y l , . . . , yk). It follows that the point ( y l , . . ", yk) of X <k) is positively recurrent under ~b.

The proof of negative recurrence is similar and will be omitted.

12.7 THEOREM. Let n ~ 11, let f ~ F( A a, n) and let f o~ be onto. Let x, y ~ X ( S#), let x be recurrent and let foo(y) = x. Then y is recurrent.

Proof. If card f ~ l ( x ) = l, the theorem follows from Corollary 8.7. If card fo~ l(x)> 1, then n > 2 and the result follows from Theorem 12.6.

12.8 COROLLARY. Let ? ~ E(Aa), let x, y ~ X(S~, let x be recurrent and let qo(y) = x. Then y is recurrent.

Proof. Use Theorems 3.4 and 12.7.

12.9 LEMMA. Let B be a block over 6:. Let n ~ 1 I. For each q > n - 1 let there be defined a partition ~ q of ~ q( A a) into disjoint subsets, each having S"- 1 members. Then there exists an integer Q such that i f q >_ Q, some D~ ~ ~q has the property that B appears in every member of Dq.

Proof. We suppose the lemma false. That is, there exists a sequence of integers ql < q2 < " " such that B does not appear in every member of Dq, for any Dq, ~ ~q,. Then the number of ql-blocks in which B does not appear is at least as great as the number of elements of the partition ~q,, which is Sq'/S "- 1. Let N(B, k) denote the number of k-blocks over S~ in which B does appear. Then

and thus S q' - N(B, qi) > Sq'/S"- 1

N(B, qi) 1 - - < l-g -i

s q i - -

which implies N(B, qi) 1

lim sup Sq ~ < 1 - S,_---- q . i~Qo

This contradicts Lemma 5.7.

12.10 THEOREM. Let n ~ 11, let f E F(5: , n), let foo be onto, let x ~ X( 5#) and let y e f ~o l(x). I f x is transitive [negatively transitive] [positively transitive] [bilaterally transitive], then y is transitive [negatively transitive] [positively transitive] [bilaterally transitive].

Proof. Let x be positively transitive and let B be an arbitrary k-block. Since foo is onto, it follows from Theorem 5.4 that card fm- I(A) = S "- 1 for all A ~ Mm(S:) and for all m > 1. This implies that the collection {f~- I(A)I A ~' , , (5a)} defines a partition 9 , ,+ ,_ 10fMm+,-1(5 a) with each element of the partition containing S"- 1 blocks. From Lemma 12.9 we infer that for m sufficiently large, there exists an m-block A such that B appears in each member offml(A).


Since x is positively transitive, there exists a positive integer i such that xi+ lxi+ 2" " "Xi+m = A. Let ye f ~ l(x). Then Yi+ lYi+ 2" ' "Yi+m+,- 1 eft2"1(A) and thus B appears in y~+ lY~ + 2" " "Y~+,, ÷,- 1. Since B was arbitrary, it follows that y is positively transitive.

The proof that, if x is negatively transitive and y e f ~ ' (x), then y is negatively transitive, is similar.

Suppose x is transitive. By Theorem 7.16, x is either negatively transitive or positively transitive; hence, i fy e f ~ t(x), y must be either negatively transitive or positively transitive. But either of these conditions implies that y is transitive.

Suppose x is bilaterally transitive. Then x is both negatively and positively transitive. If y e f ~ l ( x ) , then y is both negatively and positively transitive, and hence bilaterally transitive.


12.11 COROLLARY. Let go e E( Aa), let x, y e X( 6 a) and let go(y) = x. I f x is transitive [negatively transitive] [positively transitive] [bilaterally transitive], then y is transitive [negatively transitive] [positively transitive] [bilaterally transitive].

Proof Use Theorems 3.4 and 12.10.

13. Compositions

Let m, n e 11 and l e t f e FG9 °, m), g e F(SP, n). Then gm maps ~ , , +,_ 1 (~) into ~m(~9 °) a n d f m a p s N'm(~) into ~1(~9°), so that fgm is a well-defined mapping of ~m+n- 1(~9~) into ~,(S~). The mappingfg,, will also be denoted byfg. In general,

f g # g f

13.1 Remark. Let f ~ F(6 a, m), g e F( 6 a, n). Then (fg)~ = f~og~. Proof Let x e X(~9 o) and let (fg)oo(x) = y. Let i ~ L Then Yi = fgm(xix~+ 1"'"

Xi+m+n-2) = f[g(x i ' ' 'Xi+n-1) '" "g(Xt+m-a'" "Xi+m+,-2)]. Let goo(x) = u; thus ul = g ( x i ' " x i + , - t ) . Letfo~(u) = v. Then vi = f ( u i " "

ui+~-l) = f [ g ( x i " "Xi+n-1)'''g(Xi+m-l'''Xi+m+n-2)]. Thus Yi = vi ( / E l ) ; hence y = v, ( fg)~(x)=f~og~(x) for all x e X(Sg), which implies that (fg)® =

f~go~.

13.2 THEOREM. Let go e dP( 5 a) and let Y be a closed invariant subset of X(Aa). I f go(Y) = X(5°), then Y = X(Aa).

Proof If go(Y) = X(Sg), then go(X(Aa)) = X(5~), so that go is onto. Let x e X(5 p) be transitive. There exists y e Y such that go(y) = x. According to Corollary 12.11, y is transitive, so that the orbit o f y under ~ is dense in X(Aa). Since Yis invariant, the orbit o fy lies in Y, so that Yis dense in X(S t') and Y = Y = X(~e).

13.3 THEOREM. Let % ~b E @( 6P). Then go~ (~ followed by go) is onto i f and only i f both go and ~b are onto.

Proof If both go and ~b are onto, then, clearly go is onto. Suppose that ~o~b is onto. If go is not onto, and thus go(X(SP)) ¢ X(Aa), then,

since ~(X(Aa)) = go(~(X(~)) c cp(X(6a)), it follows that go¢ is not onto. Thus ~o must be onto.

Let Y = ~(X(S~')). Then Y is a closed invariant subset of X(A'~). Since go~b is

358 G.A. I-IEDLUND

onto, X(S~') = 9~b(X(Aa)) = ~0(Y). It follows from Theorem 13.2 that Y = X(Sa), and thus @(X(Aa)) = X(6 e) and ~b is onto.


13.4 COROLLARY. Let m, n e 11, let f e F( 6 a, m) and let g ~ F(~9 ~, n). Then ( fg)~ = f~g~ is onto if and only i f f ~ and g~ are both onto.

Let % ~b ~ E(Aa). From Theorem 11.2 there exist positive integers M(~o) and M(~b) such that if x ~ BT(6a), then card ~o- l(x) = M(~o) and card if- l(x) = M(~b). Now ~off e E(Aa), so that M(~0ff) exists.

13.5 THEOREM. Let % ~b ~ E(S~). Then M ( ~ ) = M(~o)M(~b). Proof. Let % ff ~ E(6a). Let x ~ BT(Aa). Then card f0- X(x) = M(~o) and, accord-

ing to Corollary 12.11, each member of ~o-l(x) is in BT(S~). Let ~o-l(x) = {y(i) I 1 < i < M(~o)} and thus y(i) ~ y ( j ) unless i = j . Again, for each i(1 < i < M(~0)), card ~ - l ( y ( i ) ) = M(~b), and from Corollary 12.11, each member of ~b-l(y(i)) is in BT(Se). For 1 < i < M(~0), let ~b-l(y(i)) = {y(i , j) I 1 < j < M(~b) } and thus y(i, j ) ~ y(i, k) unless j = k.

Now y( i , j ) = y(m, n) only i f / = m a n d j = n. For if y( i , j ) = y(m, n), then y(i) = ~b(y(i, j )) = ~b(y(m, n)) = y(m), which implies i = m. But y(i, j ) = y(i, n) impliesj = n.

Let Y = {y(i,j)l 1 <_ i < M(~o), 1 < j < M(~b)}. Then card Y = M(~)M(~b) and Y = ~b-l(~0-~(x)) = (~0~b)-l(x).

Since x e BT(Ae), M ( ~ ) = card (f0~b)- ~(x) = card Y = M(~o)M(~) and the theorem is proved.

13.6 COROLLARY. Let m, n ~ I~ and let f s F( Se, m), g ~ F( A a, n), with both f ~ and g~ onto maps. Then M ( f g ) = M ( f ) M ( g ) .

14. Maximal Compatible Extensions

With any on tof~ defined b y f ~ F(Sf, n), there has been associated an integer M ( f ) which is multiplicative under composition. We associate with any o n to f~ two integers R ( f ) , L ( f ) which are also multiplicative under composition. These were first defined by L. R. Welch.

Throughout this paragraph, the symbol set ~ will be fixed with card A a > 1 and references to ~9 ° will, in general, be omitted.

Let m ~ 11 and let A be an m-block. A right [left] extension of.4 is a block AB [BA] where B is an arbitrary block. If B is a p-block, then, by definition, .4B [B.4] is a right [left] p-extension of .4.

Let n E 12, let f 6 F (S ~', n), let m s 1,_1, let .4 be an m-block and let .4B, .4C be right p-extensions of .4. Let q = m +p - n + 1. By definition, .4B and .4C are compatible with f provided f~(AB) = fq(.4C). Similarly, the left p-extensions BA and CA, of A are compatible with f provided f~(BA) = FQ(C.4).

Let n e I2, l e t f e F (6 a, n) let m e I,_ 1 and let A be an m-block. Le tp e 11 and let M be a set of p-blocks. Define AM = {AB[ B~M}; thus AM is a set of right p-extensions of A. By definition, the set AM is compatible with f provided each pair of members of AM is compatible withf. If the set AM is compatible w i t h f and q = m + p - n + 1, then there exists a q-block D such that fq(AB) = D for all B~M.

The definition of a set MA of let p-extensions of A, compatible with f , is analogous.

14.1 Remark. Let n ~ 12, let f ~ F( 6 a, n) and let A be an m-block with m > n - 1. Let p ~ 11 and let AM [MA] be a set of right [left] p-extensions of A compatible with f . Iffoo is onto, then card AM < S n- 1 [card A < S n- 1].

Proof. This follows from Theorem 5.4.

14.2 THEOREM. Let n ~ 12, let f ~ F( 6 a, n) and let foo be onto. Let m ~ In- 1, let A ~ M,~( SP), let p ~ 11 and let AM [MA] be a set of right [left] p-extensions of A compatible with f . Let Mr(n--1) [Mi(n--l)] be the collection of terminal [initial] (n-1)-blocks of AM [MA]. Then card Mr (n -1 ) = card AM = card M [card M i ( n - 1) = card MA = card M].

Proof. We consider the case of right extensions. Clearly, Mr (n - 1) < card AM = card M. Suppose card M t ( n - 1 ) < c a r d AM. Then there are distinct members AB and AB' of AM with the same terminal (n - 1)-block D. I fp _< n - 1, then AB = AB', so that p > n - 1 . Let C = ala2""a ._1. Then AB = CPD, AB' = CP'D, where P and P ' are distinct [m+p-2(n-1)]-blocks. Since AB and AB' are right p-extensions of A compatible with f , f~(CPD) = fq(CP'D), where q = m + p , n + 1. But according to Lemma 9.2, it would follow thatfoo is not onto, contrary to hypothesis. Thus card Mr(n - 1) = card AM.

The proof in the case of left extensions is similar. Let n ~ 12, l e t f e F(6 a, n) withf~ onto, let m ~ I n_ 1 and let A be an m-block.

Let p ~ 11 and let AM be a set of right p-extensions of A compatible with f . It follows from Remark 14.1 that card AM < S n- 1. For given p, the collection of all sets of right p-extensions of A compatible with f has finite cardinality and there is one M(A, p , f ) of maximum cardinality. Let R(A, p , f ) = card M(A, p , f ) . Now R(A, p, f ) < S n- 1 for all p e 11, and we let R(A, f ) = supp~11R(A, p, f ) . Then R(A, f ) < S n- 1, and for some p ~ 11, there exists a set AM of right p- extensions of A compatible with f such that card AM = R(A, f ) .

14.3 Remark. Let n E 12 and let f e F(SP, n) with foo onto. Let A and C be blocks each of length at least n - 1. Then R ( A , f ) = R(C , f ) .

Proof. Let AM be a set of rightp-extensions of A, compatible with f , such that card AM = R(A, f ) . Let q be the length of A. Then CAM is a set of right (q +p)- extensions of C, compatible with f , so that R(C, f ) >_ R(A, f ) . By interchanging the roles of A and C, we infer that R(A , f ) > R(C , f ) and thus R(A , f ) = (R(C,f) .

Let R ( f ) = R(A, f ) , where A is any block of length at least n - 1. The integers L(A, p, f ) , L(A, f ) and L(.f) are defined analogously for left

extensions. The following basic theorem is due to L. R. Welch.

14.4 THEOREM. Let n ~ 12, let f e F ( 6 a, n) and let f oo be onto. Let m ~ I n_ 1, let A be an m-block and let AM [MA] be a set of right [left] p-extensions of A, compatible with f , such that card AM = R ( f ) [card MA = L(f)] . Let i = m + p - n + 1. There exists an i-block D such thatf~(AB) = D (B E M) [fi(BA) = D (B ~ M)]. Let E be an arbitrary q-block, q ~ 11. Then there exists a set A ~ [C~A] of right [left] (p + q)-extensions of A, compatible with f , such that, i f j = i+ q, then f j (AC) = DE (C c go [fj(CA) = ED (C ~ gO] and card A ~ = R ( f ) [card C~A = L(f) ] . Further-

3 6 0 G . A . H E D L U N D

more, i f C e ~ a n d ~ = CLC2" " "%+q, then c1"'" c v = B for some B ~M [Cq+l" • ' Cp+q = B for some B EM].

Proof. First reading. The existence of an / -b lock D such that f i (AB) = D, B e M, is implied by the compatibility of the set AM withf . Let D = did2"'" dl, E = ei+:" • "ei+q. L e t M ' = M6:. Since card AM = R ( f ) and cardS/ ' = S, we have card AM' = S R ( f ) . Now fi+ I(AB') ~ DS: (B' ~ M'). For s ~ 6 a, let M~ = {B'[ B' ~M' , f i+I(AB' ) = Ds}. Then AM~ is a set of right ( p + 1)-extensions of A for each s ~ 6~, compatible with f ; hence

(1) card AM'~ <_ R(F) (s ~S#).

But U s~g = M~, so that

(2) ~ card AM's = ~ card M's = card M' = SR( f ) . s ~ S ~ s ~

It follows from (1) and (2) that

(3) card AM', = R ( f ) (s ~ 5e).

Choose s = eg÷ 1. Then AM's is a set of right (p + l)-extensions of A, compatible with f , f i+l(AB') = Dei+l (B' eM') , card AM~ = R( f ) , and if B' ~M~, then B' = Bt for some B ~ M and some t ~ 6:.

This proves the theorem if E is a 1-block. I f we assume the theorem is true for E a j-block, the same argument shows that the theorem is true for a ( j + 1)- block, and hence the theorem is true for any q-block E.

The proof of the second reading is similar. Let n ~12 and l e t f ~ F(6 : , n) withfo~ onto. Let d E 11 and let D be a d-block.

A connected maximal f-covering of D is a set cg'AM of (d+n-1)-blocks , where A is an m-block, m >_ n - 1 , AM is a set of right p-extensions of A, compatible with f , such that card AM = R(f), ~ A is a set of left q-extensions of A, compatible with f , such that card c~A = L ( f ), and ~ A M c f 7 X(D) • Necessarily, d + n - 1 =

m + p + q .

14.5 Remark. Let n ~ 12, let f E F( 6 a, n) with f® onto and let k ~ 11. Then there exists a d-block D, d > k, which has a connected maximal f-covering.

Proof. Let m > n - 1 + k and let A be an m-block. Let AM be a set of right p-extensions of .4, compatible with f , such that card M = card AM = R( f ) . Let ~ A be a set of left q-extensions of A, compatible with f , such that ~ = card ~ A = L ( f ) . Let d = m + p + q - n + l > k. There exists a d-block D such that fd(CAB) = D for every C ~ ~ and every B e M. Thus ~ A M is a connected maximal f-covering of D.

14.6 LEMMA. Let n E 12 and let f ~ F( 5#, n) with fo~ onto. Let CgAM be a connected maximal f-covering of the k-block D. Let E~fk-I (D) . Then either E ~ ~ A M or E is ( n - 1)-separated from each member o f ~ A M .

Proof. Let A be of length m _> n - 1, l e tp be the length of the members of M and let q be the length of the members of c~. Suppose that E ~fk- ~(D), E ¢ C~AM, and there exist C ~ cg, B e M such that E and CAB are not ( n - 1)-separated. Let

E = e I • • " e q e q + l " • " e q + m e q + m + 1 " " "eq+ra+p,

CAB = c 1 • • • Cqaq+ ~" • . a q + m b ~ + m + 1" " " b q + m + p

= h~...h~hq+ 1" " " h q + m h q + m + 1" " " h q + m + t .

and thus c l " " c q ~cg, aq+l"" "ag+,, = A, bq+, .+l- . .bq+, ,+p eM. Since E and C A B are not ( n - 1)-separated, there exists an integer i, 1 < i+ 1 __ n - 1, either q < i or i + n - 1 < q + m . We will consider only the first case, q < i. The completion of the proof in the other case is similar.

Let D = dl"" "dk, where k = q + m + p - n + 1. Consider the set gahg+l.. • hi+,_ 1 = C~hq+l. . .hiei+l . . "ei+n_ 1. Then

f , (Ch~+l"" "hi+,_ 0 = d 1 • " 'd, ( C ~ g 3 and

f i ( e l ' " ' e i+ , -1 ) = dt"" . d i .

Thus the members of gahq+x.. .hi+,_ 1 together with e l . . 'ei+n_ 1 constitute a set of left/-extensions of ei+ 1" " "ei÷,- 1 compatible withf . Since card Cghq+ 1" " " h i + , - , = L ( f ) , it follows that el"" "ei+,-1 eCghq+x "" "hi+,-1 and hence e l " " eq+n_ 1 = C'a~+ 1 • • "aq+n_l, where C' ecg.

Let t = m + p - n + 1. Then

f , ( a q + i ' " a ~ + m B ) = d q + l " " 4 (B ~%z~) and

ft(eq+ x " " eq+,.+,) = ft(aq+ l" "aq+meq+m+ l ' "eq+m+p).

Thus the members of AM together with eq+ 1" " "eq+m+p constitute a set of right (m + p - n + 1)-extensions of aq + 1" " "aq + n - - 1 ~--- e~ + 1" " "eq +._ x compatible w i t h f Since card A M = R ( f ) , it follows that eq+l""e~+, .+p = AB' , where B '~M. Now E = e l " ' e ~ + , . + p = C ' A B ' ~ ~ A M , contrary to hypothesis.

The statement of the lemma follows.

14.7 T H E O R E M . Let n ~ 1 z and let f s F ( 5 a, n) with f ~ onto. Let k ~ 11. There exists a d-block D, d >_ k, and M * ~ 11 such that

fe-X(D) = {CgiA,M,[i = 1, 2 , . . - , M*},

where for each i, 1 < i < M* , CiAiMi is a connected maximal f-covering o f D. Moreover, i f 1 k, with a connected

maximal.f-covering CglA1M 1. Iffd71(D1) = CglA1M,, the theorem is proved by setting d = dl, D = D, and M* = 1.

We assume that f ~ l ( D , ) ~ CglA1M 1. Then there exists A 2 Efa7I(DI) such that A 2 ¢ CglAxM I. According to Lemma 14.6, A 2 is ( n - 1)-separated from each member ofCglAx~l. Let m, be the length of A 1, p~ the length of members o f ~ , and qt the length of members of cgl.

Let A2M 2 be a set of right pz-extensions of A2, compatible with f , and with card A2M 2 = R ( f ) . Let CgzA2 be a set of left qz-extensions of A2, compatible with f , and with card C~2A2 = L ( f ) . Let d 2 = q2 +q l + m l +Pl + P z - n + 1. Then there exists a dz-block D2 such thatfa~(C2AzB2) = D2 for all C2 ~c~z, B2 6M2.

According to Theorem 14.4, there exist a set A ~ M ~ of right (p ~ +pz)-extensions of A 1, compatible with f , with card A 1M; = R(f ) , and a set C~[A ~ of left (q~ + q2)- extensions of A 1, compatible with f , with card ~ A x = L ( f ) , such that

f a , ( ~ ; A 1 M ; ) = Dz for all C[ scg;, B[ e ~ [ .

362 G.A. HEDLUND

Each of the sets c~'A 1M ~, (~2A2~,,~2 is a connected maximal f-covering of D2. Since A2 is (n - 1)-separated from each member of C~IA1M 1, the sets C~[Al~' 1 and (~2A2~.~2 are disjoint, and by Lemma 14.6, each member of (~2A2~2 is (n--1)-separated from each member of C~AiM ~. If fd~'l(D2) = C~A1MI' u (~2A2~2, the statement of the theorem is proved by setting d = d 2, D = D z and M* = 2.

If fd~(D2) ~ C ~ A l ~ ~ t,.)(~2A2~2, then by the same construction we can obtain (~3A3~3, ~ h 2 , . ~ , c~ 'A i~ ' , each a connected maximal/=covering of a d3-block D a and with members of different sets (n-1)-separated. Iffd~l(D3) is the union of these three sets, the theorem is proved by setting d = d3, D = D 3 and M* = 3.

Iffa~l(D3) is not the union of these sets, the process can be continued. But since cardf~- X(B) = S"- 1 for every m-block B and every m EIi, the process must terminate and the conclusion of the theorem is obtained.

In Section 11 it was shown that to any f s F(SP, n), where n s /2 and f~ is onto, there corresponds an integer M ( f ) such that if x is bilaterally transitive, then c a r d f ~ l ( x ) = M ( f ) .

14.8 THEOREM. Let n E 12, let f • F(A a, n) and let f ~ be onto. Then the integer M* of Theorem 14.7 is M ( f ) .

Proof. Let D be the k-block with the properties affirmed by Theorem 14.7. It follows from Theorem 11.4 that there exists an e-block E such that M ( e , f ) = M ( f ) . That is, fe-x(E) contains a subset consisting of ( n - 1)-separated members of cardinality M ( f ) and does not contain such a subset of greater cardinality.

By applying Theorem 14.4 to each of the sets C~iAiM ~, i = 1, 2 , . . . , M*, we obtain a collection {C~iAiM'il i = 1, 2,. -. , M*} of maximal connected f-coverings

I IM*c~ ~ . - ~ ' = UM*lc6~iAi~i S "-1, we have of DE, and since card u / = l /,1; ~ card _ = f ~ I ( D E ) = I I M* c~,~ ~ , w i = l i,.i i. For any i, no two members of C~iAiM~ are ( n -1 ) - separated, while if i # j, each member of CKiAiM'i is ( n - 1)-separated from each member of C~jAj~j. This implies that M* = M(DE, f ) and M(E, f ) >_ M*; thus M* <_ M ( f ) . But M* = M(DE, f ) > M ( f ) , by Theorem 11.4. Thus M* = M ( f ) and the theorem is proved.

I f f e F(S~, 1) a n d f ~ is onto, then, by Theorem 4.1, f~o is one-to-one and card f ml (B ) = 1 for every m-block B and every m ~ I1. In view of this, we define L ( f ) = M ( f ) = R ( f ) = 1.

14.9 THEOREM. Let n • Ii, let f • F(SP, n) and let foo be onto. Then L ( f ) M ( f ) R ( f ) = S"- 1.

Proof. I f n = 1, then L ( f ) = M ( f ) = R ( f ) = 1 and sn-1 = 1. We assume that n e 12. By Theorems 14.7 and 14.8, there exists a d-block D

such that fd-l(D) = {(~iAi~i[ i = 1, 2 , ' " , M ( f ) } ,

where for each i, 1 < i < M ( f ) , CdiAi~ i is a connected maximal f-covering of D, and if 1 < i < j < M ( f ) , then each member of CgiAiMi is (n-1)-separated from each member of C~jAj~j. Since for each i, 1 < i < M ( f ) , card cgi = L ( f ) and card ~ = R ( f ) , it follows that card ~M_t{)~Ai~ i = L ( f ) M ( f ) R ( f ) . By Theorem 5.4, card fd-l(D) = sn-1. This proves the theorem.

15. L(fg) = L( f )L (g ) and R(fg) = R( f )R(g )

Let m, n e I t and l e t f e F(5~, m), g e F(Se, n), with bo th f~ and g~ onto maps. By Corollary 13.6, M ( f g ) = M ( f ) M ( g ) . We show that this multiplicative property holds for L and R.

15.1 T H E O R E M . Let m, n ~ 11 and let f ~ F( Sf, m), g e F( 5 ¢, n), with both f oo and g~ onto maps. Then L(fg) = L( f )L (g ) and R(fg) = R( f )R(g) .

Proof. We show first that R(fg) > R( f )R(g) . Let A be a block over 5 ~ of length a with a > m + n, and if n > 1, let AM be a

set of right b-extensions of A, compatible with g, such that card AM = R(g). I f n = 1, the set N is any single block of length b. Then there exists a block C of length c = a + b - n + 1 such that gc(AB) = C for all B e N.

I f m > 1, let Cff be a set of right e-extensions of C, compatible with f , such that card Cff = R( f ) . I f m = 1, the set ~ is a single block of length e. Let ~ = {E~I 1 1, there exists a set AN, of right (b + e)-extensions of A, compatible with g, such that card AN, = R(g) and gc+e(AB~) = CEi for all Bi eNi . I f 1 < i < j < R( f ) , then N / n N j = ~. For if B i = Bj, where Bi ~N, and Bj ~Nj , then CE, = ge+e(ABi) = g~+~(ABj) = CEj. This implies E~ = Ej, which is not the case. I f n = 1, the set N , still exists with the same properties, but now, due to Theorem 5.4, it consists of a single block.

Let N* = Wi=a/|R(S)Ni" Then card AM* = R(f )R(g) . Let B* EN*. Then B e N i for some i, 1 _ R( f )R(g) .

A similar argument shows that L(fg) >_ L( f )L(g) . From Theorem 14.9, we have

(1) L ( f ) M ( f ) R ( f ) = S " - 1, L(g) M(g) R(g) = S"- a.

Sincef~ and g~ are both onto, it follows from Corollary 13.4 that (fg)~ is onto, and since fg e F(SP, m + n - 1 ) , we have

(2) L(g f )M( fg )R( fg ) = S m +,-2

From (1) and (2) we obtain

(3) L ( f ) L(g) M ( f ) M(g)R( f )R(g) = L(fg) M( fg ) R(fg).

We have shown that M ( f g ) = M ( f ) M(g), and from this and (3) it follows that

(4) L ( f ) L(g) R( f )R(g ) = L( fg ) R(fg).

From (4) and the inequalities L ( f g ) > L ( f ) L ( g ) , R(fg)>_ R ( f ) R ( g ) , we infer that

(5) L(fg) = L ( f ) L(g), R(fg) = R ( f ) R(g).

364 G.A. HEDLUND

16. Cross-Sections of the Mappings f~

Let X and Y be topological spaces and let f : X---~Y be continuous. Then g: Y--~X is a cross-section o f f provided g is continuous and f(g(y)) = y (y ~ Y). It is clear that if f has a cross-section, t h e n f m a p s X onto Y.

A. Nerode raised the question as to conditions on fo~, f ~ F ( S e , n), which are equivalent to the existence of a cross-section. O. S. Rothaus has given a definitive answer by proving the following theorem in case card S~ _- 2.

16.1 THEOREM. (O. S. Rothaus) Let n ~ 12 and let f E F(5 t', n) with foo onto. Then the following statements are equivalent.

(1) There exists tz E 11 such that foo is an exactly tz-to-one map of X(5¢) onto x(Y3.

(2) foo is open. (3) f~o has a cross-section. (4) For each x ~ X(6a), any two distinct members o f f Zo l(x) are totally ( n - 1)-

separated.

The proof given by Rothaus is valid for card S~ > 2. In his proof that (4) implies (1), he makes crucial use of a matrix semigroup,

associated with f ~ F(SP, n), and the algebraic structure of this semigroup as developed by A. M. Gleason. We show that (4) implies (1) without making use of this semi-group, though with somewhat more labor, but the basic ideas are due to Rothaus.

I f f e F(SP, 1) and f® is onto, then, by Theorem 4.1,f~ is a homeomorphism of X(6 a) onto X(Se), and hence f® has a cross-section namelyfo~ 1, and f~o is open. Thus it is no restriction to assume that n > 2.

We prove three theorems, which, together with a theorem of E. A. Michael [19], imply Theorem 16.1.

16.2 LEMMA. Let n ~ I2 and let f e F( 5~, n) with f oo onto. Let there exist a sequence {x(i)[ i E I1 } of points Of X( Sf) converging to x such that for each i E It, fo~ X(x(i)) is a set of mutually totally (n-1)-separated bisequences, each set of cardinality k. Then f ~ l ( x ) contains a set of mutually totally (n-1)-separated bisequences of cardinality k, each of which is a limit point of a subsequenee of the set U i E,1 f ~ l(x(i)).

Proof. Let fo~ X(x(i)) = {y~l),..., y~k)} (i ~ I1). We can choose a subsequence i(m) of I1 such that each of the sequences tyi(m),S ,,(J) , m e It} is convergent for each j = 1,. •. , k. Let limm_.~oy~[~ )) = y(J). Sincefoo(y~[~))) = x(i(m)) and limi~ o~x(i) = x, it follows that f®(y(J)) = x.

Since "(J) and "(P) ~i(,,) ~(,,) are totally ( n - 1)-separated provided j ~ p, hence at a distance apart exceeding a fixed positive number 8, it follows that y(J) # y(P) if

~inc~ , . o ) ~.~.q ,.(p) j ~ P. ~ ,, 3"i(m) . . . . ,~i(m) are totally ( n - 1)-separated providedj ~ p, it follows that y(J) and y(P) are totally ( n - 1)-separated i f j ~ p.

The conclusion of the lemma follows.

16.3 THEOREM. Let n e I2, let f e F(Sg, n)andlet foo be an exactly tz-to-one map fo r some tz e 11. Then the following statements are valid.

(1) f ~ is onto. (2) /z = M ( f ) . (3) For each x e X(Sa), distinct members o f f £ X ( x) are totally (n - 1)-separated. (4) f~o is open.

Proof. It follows from Theorem 5.12 that iff~o is an exactly t~-to-one map, then f® is onto.

Let x ~ BT(Sa). By Theorem 11.2, cardfo~ l(x) = M ( f ) , so that t~ = M ( f ) . Suppose that y, z e X(5¢), y # z and f~ (y ) = u = foo(z). By Theorem 11.1,

f ~ l ( u ) contains M ( f ) members which are mutually ,totally (n-1)-separated. Since M ( f ) = tz, it follows that distinct members o f f~1 (u ) are totally ( n -1 ) - separated.

To prove (4), it is sufficient to show that i fy e X(Sa) and Uis a neighborhood of y, then f ~ ( U ) is a neighborhood of x = fo~(Y).

Suppose that there exist y s X(Se) and U, a neighborhood of y, such that V = fo~(U) is not a neighborhood of x = fo~(Y). Then there exists a sequence of points {x(i)[ i e I 1 ) converging to x such that U n f ~ o l ( x ( i ) ) = o ( i e 11). But for each i, card fo~ l(x(i)) = M ( f ) , and the members o f f ~ l(x(i)) are mutually totally (n - 1)-separated. It follows from Lemma 16.2 thatfo~ x (x) contains a set Y of mutually totally (n-1)-separated members such that card Y = M ( f ) , and each member of Y is a limit point of a subsequence of the set U ~ r l f ~ t ( x ( i ) ) • Since u n f ~ l ( x ( i ) ) = ~ (ieI1), it follows that y ¢ Y. But then card f ~ l ( x ) > M ( f ) , which is not the case. Thus f® is open.


16.4 LEMMA. Let n eI2, let f e F ( 5 a, n) and let f® be onto. Le t there exist y, z e X ( 5 ¢) such that y ~ z , f ~ ( y ) = x = f ®(z) and y and z are not totally (n - 1)- separated. Then there exist u, v, w e X(SP), u ~ v, such that one o f the following statements is valid.

(1) u and v are positively (n - 1)-separated and negatively asymptotic; :t and v are negatively transitive; f~o(u) = w = f®(v).

(2) u and v are negatively ( n - 1)-separated and positively asymptotic; u and v are positively transitive, f~o(u) = w = foo(v).

Proof. It follows from Theorem 9.3 that y and z are either positively ( n - 1)- separated or negatively ( n - 1)-separated.

We show that if y and z are positively (n- l ) -separated, then (1) is valid. Since y and z are not totally (n - 1)-separated, there exists i e I such that y~. • • Yi+n-2 = Zi ' ' 'Z i+n-2" Define u, v e X ( 5 a) as follows. For j > i, let uj = yj, vj = z~. For j < i, choose u s so that u is negatively transitive. For j < i, let vj = uj. Then u ¢ v, u and v are positively ( n - 1)-separated and negatively asymptotic and f ~ (u ) = f~ (v ) . Let w = f~ (u ) and then (1) is valid.

I f y and z are negatively ( n - 1)-separated, a similar argument shows that (2) is valid.

366 G.A. HEDLUND

16.5 LEM_MA. Let n • 12, let f • F(A a, n) with f ~ onto and let f~o have a cross- section g. Let Y = g(X( A*)). Then Y is open and closed, g: X(,9~)--~ Y is a homeomorphism and f ~ I Y is a homeomorphism o f Y onto X( S#).

Proof. Since f~og is the identity map, each of the mappings g and fo01Y is one-to-one and f~[ Y maps Y onto X(5~). Since X(A") is compact and g is continuous, it follows that Y = g(X(A°)) is compact and hence dosed, and since g is one-to-one, g is a homeomorphism ofX(A '~) onto Yandf~o I Yis a homeomorphism of Yonto X(S~). It remains to show that Yis open.

Let y • Y. We show that Y is a neighborhood of y and thus Y is open. Let x = g - l (y ) and let U be an open and dosed neighborhood of y with diam U< (n + 1)-1. Let V be an open and closed neighborhood of x such that g(V) c U. Let v • V. Then g(v) • U and since f~o(g(v)) = v, g(v) • f ~ l(v), so that g(v) • U n

f ~ l(v). If the members of f ~ l(v) are mutually totally (n-1)-separated, then since diam U< (n + I)-1, U contains at most one member offo~ l(v), and hence g(v) = v f g l ( v ) .

Nowfoo(y) = foo(g(x)) = x; let W be an open and closed neighborhood of y such that W c U andf~o(W) c V. By Remark 7.15, the set BT(A p) of bilaterally transitive points is dense in X(S'), so that the set W n BT(G*)is dense in IV. Let z • Wc~ BT(S~) and let v = f~o(z)e V. Then from 12.10, v • BT(S ~) and from Theorems 11.1 and 11.2, it follows that the members of f~ l (v ) are mutually totally (n-D-separated; hence g ( v ) = U n f ~ l ( v ) . Now z • W c U and z e f ~ l(v), so that z = g(v) and W r~ BT(3~) c g(V). But g(V) is closed and since it contains the set W r~ BT(S~) which is dense in W, it follows that W c g(V) c y. Thus Y is a neighborhood of y and Y is open.


16.6 THEOREM. Let n e 12, let f • F( A'*, n) and let f ~ have a cross-section g. Then for each xeX(G'~), the members of fZol(x) are mutually totally ( n - l ) separated.

Proof. We assume that n • /2 and that there exist y, z • X(5 p) with y ¢ z, f~ (y ) = x = f~o(z), and such that y and z are not totally ( n - D-separated. By Lemma 16.4 there exist u, v, w E X(SP) with u ~ v, f~(u) = w = foo(z) and such that either statement (1) or statement (2) of Lemma 16.4 is valid. We assume that (1) is valid.

The set Y = g(X(S/~)) is non-vacuous and, by Lemma 16.5, Y is open. Since u is negatively transitive, there exists a sequence . . . < k _ 1 < k o such that limi~ook'(u) = y and since u and v are negatively asymptotic, it follows that limi_~oook'(v) = y. Since Yis open, there exists k • / s u c h that <rk(u) • Y, ~k(v) e Y.

If (2) is valid, the same conclusion is reached by a similar argument. Now f~o(ak(u)) = ak(fo~(U)) = ak(w) and fo~(ak(v)) = crk(f~(v)) = crk(w), and

footY is not a homeomorphism of Y onto X(A'~), contrary to Lemma 16.5. The proof is completed.

Definition. Let n e 11 and let f E F ( ~ , n). A right f-branch of length k is a pair of ( n - 1 + k)-blocks over ~ ,

a - n + l " " a - i b o ' " b k - 1 , a - n + l " " a - i b ~ ' " b ~ - l , such that bo # b~ and

A ( a - , + 1"" .a_ l b0"" .b~_ 1) = A(a-n + 1 " " a_ l b~""" b~_ 1).

A left f-branch o f length k is a pair of ( n - 1 + k)-blocks over ~ ,

b-k+1"" "boax" " "a.-1, b'-k+x"" " b ~ a l " " a . - 1 ,

such that bo ~ b~ and

fk(b-k+1"" "boa1"" "a,-1) = fk(b'-k+l"" "b~at'" "a.- 1).

16.7 LEMMA. Let n ~ I 2, [et f e F ( ~ , n) and let foo have the property that for all x ~ X(Sa), the members o f f ~ t(x) are mutually totally ( n - 1)-separated. Then there exists an integer K ( f ) such that there exists no right or left f-branch of length exceeding K( f ) .

Proof. Suppose that, under the hypotheses of the lemma, corresponding to each k E 11, there exists a right f-branch

• " "ak-1 bo'" "b[-1, a~.+l"' "ak-i co k'" "c~- 1. a - n + l

Let x k be any member of X(SQ such that

xk_..+ 1 . . . xk t_ = a_.+ 1 . k " "ak_l bg'" "bk_l

and let yk be any member of X(5 e) such that

Y~.+I" "Y~-x a~.+l" k C k . . . C k_ • = " " a - 1 1

k (i< - -n+ 1). Then d(x k, yk) = 1, since Xo k ~ y~. and y~ = xi Let u k = foo(xk), v k = f~o(yk). Then if k>n , d(u k, v k) < (1 + k - n ) - 1 , so that

limk_, o~d(u k, v k) = 0. Since X(5 a) is a compact metric space, there exist a sequence of integers ko < kl < • • • and x, y, u, v ~ X(S/') such that limi~®x k' = x,lim~o~y k' = y, limi..~u k' = u and l i m ~ v k' = u. But thenfoo(x) = u, fo~(y) = v, d(x, y) = 1, Xo ~ Yo and u = v. Thus x and y are distinct members o f f ~ l(u) which are not totally ( n - 1)-separated, contrary to hypothesis. It follows that there exists KR E 11 such that there is no right f-branch of length exceeding KR.

Similarly there exists KL ~ 11 such that there exists no left f-branch of length exceeding KL. If we let K ( f ) = max [KR, KL], the statement of the lemma is proved.

16.8 LEMMA. Let n ~ 12, let f e F(5¢, n) and let there exist an integer K ( f ) such that there exists no right or left f-branch of length exceeding K ( f ). Let E be a block of length e >_ n - 1 and let EM [ME] be a set o f right [left]p-extensions of E compatible with f . Then EM = E'M' [~E = M'E'], where E' is a block of length e' > e, E '~ " [M'E'] is a set o f right [left] p'-extensions of E' compatible with f , andp ' < K ( f ) .

Proof. First reading. We assume that E ~ is a set of right p-extensions of E compatible withf. If card E ~ = 1, the conclusion of the lemma is obvious. We assume card E ~ > 1.

Let EB, EC ~ E ~ with B # C. Let

EB = e l " " e ~ b l " " b p ,

EC--- e I . . . e ec 1...cp.

Let k be the least positive integer such that bk # ck. Since B # C, k exists and 1 < k < p. Suppose p - k > K ( f ) . Then the terminal ( p - k + n - 1)-blocks of EB and EC constitute a right f-branch of length exceeding K ( f ) , contrary to hypo-

368 G.A. HFX)LU~

thesis. Thus p - k < K ( f ) , and hence k > _ p - K ( f ) . It follows that the initial (e+k)-blocks of all members of E ~ are identical, where k > p - K ( f ) . Let this initial (e+k)-bloek be E'. Then E ~ = E ' ~ ' where E ' ,~ ' is a set of right p ' - extensions of E ' compatible with f , and p' = p - k < K ( f ) . This proves the first reading.


16.9 THEOREM. Let n e I2, let f e F( SP, n) and let f~o have the property that for all x e X(Aa), the members o f f a l ( x ) are mutually totally ( n - 1)-separated. Then f~o is onto and there exists M e 1, such that i f A = a I • • "am+n- 1 is any block o f length m + n - 1 > M and fro(A) = D, there exists a set al " " " a._ l ~ [~am+ 1"'" am+._1] o f right [left] m-extensions o r a l . . . a . _ 1 [am+ l . . "am+._i] such that:

(1) card ~ = R ( f ) [card ~ = L ( f ) ] ; (2) A ~a t ' " "an- l~ [,4 e~am+l ' "am+, -1] ; (3) fro(a1"" .a ,_ lB) = D (B ~ ) [fro(Sam+l"" "am+,-1) = O (B e~')].

Proof. First reading. We assume that the members offo~ l(x) are mutually totally ( n - 1)-separated for each x ~ X(Se). This implies that if y, z efoo(x), y # z, then d(y, z)> (n + 1)-1, and hence, since X(5 a) is compact, card f ~ l (x)< a3 for each x e X(Sa). By Theorem 5.12,f~o is onto.

Choose M e I1 such that M > K ( f ) + n - 1 , K ( f ) as in Lemma 16.7. Let E be a block of length e, e > n - 1. Let Ef t be a set of right q-extensions of E compatible w i t h f such that card E ~ = R ( f ) . Let p = e + q - n + 1. Then there exists a p-block H such thatfp(EG) = H (G e~) . Let f¢ = {Gil 1 M and let r = e + q + m. Then f ,(EG1A) = HS, where S is a block of length m + n - 1.

According to Theorem 14.4, there exists a set ECg of right ( q + m + n - 1 ) - extensions of E, compatible with f , such that card EC~ = R ( f ) andf , (EC) = H S for all C ec~. Now G1A ~¢g, for otherwise the set E ~ ' , where c~ ' = W u {G1A}, would be a set of right (q+m+n-1) -ex tens ions of E, compatible with f , and with card W' = R ( f ) + 1, which is impossible.

It follows from Lemma 16.8 that ECg = E'Cg ', where E ' is a block of length e' _> e, E'cg ' is a set of rightp'-extensions of E'compatible w i t h f a n d p ' _< K ( f ) . Since e + q + m + n - 1 = e ' + p ' , it follows that e ' = e + q + m + n - l - p ' > e + q + m + n - 1 - K ( f ) > e + q + M - K ( f ) > e + q + n - 1. Let k =- e ' - e + q and let A = al"" "am+,-1. Then k > n - 1 and since G1A eW, the terminal k-block of E' is al" • • ak. Let ~ = a,. • • ak~'. Then the set a I • • • a,_ 1 satisfies conditions (1), (2) and (3), which completes the proof of the first reading.


16.10 THEOREM. Let n e 12, let f e F( 5 a, n) and let f have the property that any two distinct members o f f ~ l(x) are totally (n - 1)-separated for all x e X( Sa). Then f~o is onto and foo is an exactly M ( f )-to-one map of X ( ~ e) onto X ( Sg).

Proof. By Theorem 16.9, foo is onto. From Theorem 11.1, card f ~ l ( x ) > M ( f ) ( x e X(S~)) and card f ~ l ( x ) =

M ( f ) (x e BT(Sa)). Suppose there exists x ~ X(5 a) such that card f ~ l(x) > M ( f ) . Let {yti)[ i = 1 , . - . , M ( f ) + 1} be distinct members o f f a l ( x ) . Choose m ~/1 such that m > M, where M is the constant which exists by Theorem 16.9.

Le tA, ) = y~i)...y~O_l (1 _< i _< M ( f ) + l ) . By Theorem 16.9, there exist collections rd~° and ~ t o of m-blocks over A '~ such that card c~(i) = L ( f ) , card ,~(~) = R ( f ) ,

Y~)m+~ . . . . . ( i ) , , ( i ) . . . . . (i) ~,(i) . . . . . ( i ) e ( ~ ( i ) A ( i ) ~ ( i ) , • YO ..rl . Y n - l . Y n Y m + n - 1

and all members of :d¢i)A<i)~ ¢i) have the same image D under f 2 m and hence D = X_m+l""Xm. Since the members o f f a l ( x ) are mutually totally (n-1) - separated, it follows that A <i) ¢ A <j) if i # j, and thus <d¢°A~°~<° n :d¢J)A¢i)~J) = ~ i f i # j . Thus

M ( f ) + 1

card U <g(°A(i)~<i) = L ( f ) [ M ( f ) + l ] R ( f ) , i=1

which is impossible since cardf~ m 1)(D) < S"-1 = L ( f ) g ( f ) R ( f ) . The proof is completed. Proof of Theorem 16.1. By Theorem 16.3, (1) implies (2). That (2) implies (3) follows from the following theorem of E. A. Michael [19].

THEOREM..4 continuous open map of a complete metric space onto a zero- dimensional paracompact space always has a cross-section.

By Theorem 16.6, (3) implies (4). It follows from Theorem 16.10 that (4) implies (1). This completes the proof of Theorem 16.1. Theorem 16.1 gives us information concerning the existence of cross-sections

of members of E(~9~).

16.11 THEOREM. Let ~ e E( Af). Then the following statements are equivalent.

(1) There exists i~ E 11 such that ~ is an exactly/~-to-one mapping of X ( ~ ) onto X(SQ.

(2) q~ is open. (3) ~ has a cross-section. (4) For each x e X( 5#), any two distinct members of qo-l(x) are separated.

Proof. Let q~ e E(5#). By Theorem 3.4, there exist m e/ , n e 11 a n d r e F(St', n) such that q~ = ~"fo~. Since ~ is onto, so is f®. I fn = 1, then, by Theorem 4.1,f~o is a homeomorphism, and hence ~0 is also, and statements (I), (2), (3) and (4) are valid. Thus we can assume that n e/2.

Assume (1). Since q~ = ~mf®, it follows that f~ is an exactly tL-to-one mapping of X(5:) onto X ( ~ ) . By Theorem 16.1,f~o is open, and hence ~o is open and (2) is proved.

Assume (2). But i f~ is open, thenf~ = ~-m~o is open, and by Theorem 16.1, f ~ has a cross-section g. Let h = a - m g . Then h: X(5:)--->X(5:) is continuous and q~h(x) = qxr-mg(x) = ~r-mq~g(x) = fo~g(x) (X e X ( 5:). Thus h is a cross-section of ~0 and (3) is proved.

Assume (3). Let h be a cross-section of % Then ~mh is a cross-section o f f~ and by Theorem 16.1, for each x e X(5:), any two distinct members o f f ~ ~(x) are totally ( n - 1)-separated. Suppose that ~o(u) = ~(v) and u # v. Thenf®(~"(u)) = fo~(oS"(v)), ~m(u) ~ ~m(v), and am(u) and Ore(V) are totally (n-1)-separated. This proves (4).

370 G.A. HEDLUND

Assume (4). Suppose that there exist y, z ~ X(S#) such that y ~ z, f®(y) = x = f~o(z) and y and z are not totally ( n - 1)-separated. By Lemma 16.4 there exist u, v e X(Se) such that u # v, u and v are asymptotic in one sense andf~(u) = w = foo(v). But then qo(u) = amf~(u) = a'(w) = ~foo(v) = ~0(v) and u and v are not separated, contrary to hypothesis. Thus, for each x e X(S#), any two distinct members offo~l(x) are totally (n-1)-separated. By Theorem 16.1, there exists tz e/1 such that f® is an exactly /z-to-one map of X(S:) onto X(SQ. Since qo = mf®, the same is true of ~, which implies (1).


17. Converse of Theorem 6.7

It was shown in Theorem 6.7 that i f f ~ F(S °, n) a n d f is permutive in both x 1 and x,, then c a r d f ~ X(x) = S n- 1 for all x ~ X(~ga). The converse of this has been proved by O. S. Rothaus.

17.1 LEMMA. Let n ~ I l, let f ~ F( 5 a, n) and let foo be onto. Then the following statements are equivalent.

(1) f is permutive in x 1 Ix,]. (2) L ( f ) = 1 JR(f) = 1].

Proof Forn = 1, the equivalence ofthe statements follows from the definition. We assume n > 1.

Let fbe permutive in x x. Suppose L ( f ) > 1. Then there exist x2"'" x, ~ M, _ 1 (5a) and distinct elements xl, x~ of 5 a such that f ( x l x2 " ' " x,) = f ( x ] x 2 " " x , ) , which implies t h a t f i s not permutive in xl. Thus L ( f ) = 1.

Suppose that L ( f ) = 1. Let x 2 " " x , ~M,_x(Sa). Then if xl, x'l ~5 : with xl # x~, it follows t h a t f ( x x x 2 " .x,) ¢ f(x~x2"" .x,) and hence f is permutive in Xl. This proves the first reading.


17.2 THEOREM. Let n ~ I x and let f ~ F( 5 a, n). Then the following statements are pairwise equivalent.

(1) f is permutive in both xl and x,. (2) f~o is onto and there exists x E BT(S#) such that c a r d f ~ l(x) = S "-1 (3) c a r d f £ l(x) = S"- 1 (x ~ X(Se)).

Proof Assume (1). By Theorem 6.6,f~o is onto and by Lemma 17.1, L ( f ) = R ( f ) = 1. Since, by Theorem 14.9, L ( f ) M ( f ) R ( f ) = 1, it follows that M ( f ) = S "-1. By Theorem 11.2, if x ~ BT(S#), then ca rd f~ l (x ) = M ( f ) = S "-x. This proves (2).

Assume (2). By Theorem 11.1, cardfo~l(y) _> S "-1 for every yeX(~9°). But by Theorem 5.5, card fo~ l(y) _< S"- 1 for every y ~ X(5:). This proves (3).

Assume (3). It follows from Theorem 5.12 tha t f~ is onto. By Theorem 11.2, M ( f ) = S" - 1. By Theorem 14.9, L ( f ) M ( f ) R ( f ) = S"- 1 ; hence L ( f ) R ( f ) = 1 and thus L ( f ) = R ( f ) = 1. From Lemma 17.1 we conclude that f is permutive in both xl and x, and hence (3) implies (1).


For any ontofo~, f ~ F(5 t', n), the maximum cardinality o f f £ l(x) as x ranges

over X(Ae) is S ~- ~. By the preceding theorem, if this maximum eardinality is attained for a point x in BT(Aa), t h e n f , is an exactly S ~- 1-to-one mapping. It is not known whether the same conclusion can be reached if it is assumed that c a r d f ~ l ( x ) = S n-1 for some x e X(Aa).

18. Roots of Powers of the Shift

Let p e I1 and let p = qk with q e 12. It is easy to construct ~ e O ( ~ ) such that ~ is not a power of ~ and ~ = uP. For, by Theorem 6.13, there exists ~b e ~ ( ~ ) such that ~q = ~o, the identity mapping on X(Af), and ~b ~ a °. Let

= k . Then ~ = ~qk~bq = ~P. If ~ = aJ, then ~b = a i - k , ~q = aq(j-k) = O ,

j = k and ¢ = ao, which is not the case. L. R. Welch has shown that in a sense the converse is true if S = c a r d ~ is a

prime. Namely, if q, p e I1, ~ = ~(~T) and ~q = o p, then q divides p.

18.1 THEOREM. Le t S = card A f be a prime. L e t n e 11, let f ~ F ( ~ , n)

and let there exis t q e la, p e I such that f ~ = uP. Then q divides p.

Proof. Let g = f q . Then g e F(~9 ~, r), where r = q n - q + 1. Since g ~ = ( f q ) ~

= f ~ = uP, it follows from Lemma 3.3 that 0 < p < r - 1 = q(n - 1). It is easily verified that L(g) = S p. Since g~ = aP is onto, so i s f~ ; hence L ( f ) , M ( f ) and R ( f ) exist. From Corollary 13.6 and Theorem 15.1 it follows that L ( f q) =

[L(f)]L Thus [L(f) ] q = S p. By Theorem 14.9, L ( f ) M ( f ) R ( f ) = S ~- 1. Since S is prime, there exists A e 11 such that L ( f ) = S ~. But then S p = [ L ( f ) ] q = S~q;

hence p = Aq. This proves the theorem.

18.2 COROLLARY. Le t S = card A a be a prime. L e t ~ ~ O(A a) and let there

exis t q ~ 11, p ~ I such that q~q = uP. Then q divides p.

Proof. Since aP is onto, it follows from Theorem 13.3 that ~ is onto, so that e E ( ~ ) . By Theorem 3.4, there exist k e I, n e 11 and f e F ( ~ , n) such that = ~ f~ . Sincefoo commutes with a, (akf~)~ = ~ f ~ and we h a y e r q = aP-~q.

From Theorem 18.1, q divides p - kq , hence q divides p. It follows from Theorem 6.13 that in the group A(A f) there exist many pairs

~, ¢, with neither the identity mapping, yet with ~b the identity mapping. The next theorem, due to Curtis, Hedlund and Lyndon, shows that when c a r d ~ = 2, the product of an foo and a g0o is the identity mapping only in exceptional and simple situations.

The identity mapping of ~9 ~ onto ~ will be denoted by ~. Then ~ ~ F ( ~ , 1) and a~ is the identity mapping of X ( ~ ) onto X ( ~ ) .

Let c a r d ~ = 2 and l e t ~ = {0, 1 }. The permutation o f ~ defined by 0-~1, 1---~0 will be denoted by ft. Then fle F ( ~ , 1), fl~ is a one-to-one mapping of X ( ~ ) onto X(~9 ~) and f12 = ~ .

18.3 THEOREM. Le t card ~ = 2. Le t m, n ~ 11 and let f e F(~9 ~, m), g

F ( ~ , n). Then the fo l lowing s tatements are valid.

(1) I f ( f g ) ~ = ~ , then either f ~ = o%, g ~ = ~oo or f ~ = fl~, g~ = fl~.

(2) / f (fg)o~ = [3o~, then either foo = ~ , g ~ = fl~ or foo = flo~, goo = ~oo.

Proof . Suppose that (fg)o~ = ao~. Since (fg)o~ = foogo~ and ~o~ is onto, it

372 G.A. HEDLUND

follows from Theorem 13.3 that both f00 and g00 are onto. Since ~00 is one-to-one, it follows that both f00 and g0o are one-to-one.

N o w f g is a mapping from ~,~ +,_ 1(5°) onto MI(S:) = So, and since (fg)00 = ~00, it follows that (fg)00(x) = ~00(x) = x (x E X(5~)), [(fg)0o(x)]i = [~00(x)]i = x, ( x e X ( S : ) , i E I ) and hence ( fg ) (x1" ' 'Xm+n-O = Xl for all Xl""Xm+,-x e ~m+,-1(S~). This implies that fg is permutive in Xl and, from Lemma 17.1, f is permutive in xl and g is permutive in Xl. Since f00 and g0o are one-to-one, it follows from Theorem 6.19 that there ex i s t f ' e F(S#, 1) and g' ~F(6 a, 1) such that f00 = f ~ and g00 = g~. But then, sincef~ and g~o are onto, f ~ = ~00 or/300 and g~ = c% or/300. Since ~0o/30o =/300 = fl00~00, it follows that either f00 = ~0o, g00 = e00 or f00 =/300, g00 =/300, and (1) is proved.

Now suppose that (fg)00 =/300. Then (fg)00(fg)00 = [f(gfg)]00 = t3 2 = ~ . It follows from (1) that either f00 = ~0o orf0o =/30o.

Iff0o = ~0o, then g0o = ~00g00 = f0og0o = (fg)0o = fl00. 2

If f00 =/30o, then g00 =/300g00 =/300f00g00 = fl0o(fg)0o = flz = ~00. The proof is completed. The analogue of the preceding theorem, for the case where card S# > 2, is not

valid, as shown by the example given in Section 6. For the case 50 = {0, 1, 2 }, a n f ~ F(oq', 2) was defined as follows:

f(00) = 0, f(01) = 2,

f ( 1 0 ) = 1, f ( 1 1 ) = 1,

f ( 2 0 ) = 2, f ( 2 1 ) = 0,

f ( o 2 ) = o,

f ( 1 2 ) = 1,

f ( 2 2 ) = 2.

It is easily verified tha t f~ is the identity mapping on X(Sa), yet, as remarked in Section 6, there does not ex i s t f ' ~ F(St', 1) such that f00 = fo~.

19. Polynomial Mappings

Let n s /1 and let P ( x 1 , ' " , x n ) be a polynomial in ( X l , " " ", Xn) over GF(oq'), where St' = {0, 1 , - . . , S - 1 } is a cyclic field of S elements. Then P defines a polynomial mapping fp of ~ , ( 5 : ) into ~ l ( S ,°) = 5 : and thus fp ~ F ( 5 a, n). Let ~(~9 °, n) denote the set of all polynomials, of degree at most S - 1, for each vari- able over GF(Sa), and let F~,(5 t', n) = {fpl P e ~ ( 50, n) }. In general, Fe(5: , n) F(5 : , n), but the equality holds if S = card 5 ° is a prime.

19.1 THEOREM. Let S = card 5 : be a prime and let n ~ I x. Then F~(S/', n) = F( 5 a, n).

Proof. Let n ~ 11 and let P and Q be polynomials in (Xx," • ", x.) over GF(Sa). We show that i f fp = f~, then P = Q or, equivalently, i f f p ( X l , " ", x,) = 0 for all Xa'" "x. e ~ , (Se) , then P is the zero polynomial (all coefficients are zero).

Let P = ~, a ( m l , ' . . , m,)x~l . . . . x~,".

m 1 " ' " mn e , ~ n ( , ~ )

The proof is by induction. Let n = 1 and let

8 - 1

P = X a(m) xm" m = 0


Suppose P(x) = 0 for x = 0, 1,. • . , S - 1. But then P is a polynomial o f degree S - I with S distinct zeros. It follows that a(m) = 0 for m = 0, 1,. • . , S - 1.

We assume that the result is true for n = k ~ 11 and prove that it is true for n = k + l . Let

P = ~, a ( m l , ' " , mg+l)X~l l" • ".~+11. m l " " ' Ink+ 1 E . ~ k ÷ 1( ,~ )

For any choice of x~ = xi (i = 1 , . . . , k), P is a polynomial in xk+ 1 with coefficient polynomials in ( x l , . . . , Xk) evaluated at ( x l , ' " , Xk)" By assumption, P(gl ," " ' , Xk, Xk+l) = 0 for xk÷~ = 0, I , . . . , S - 1, and thus, by the first par t of the proof , the coefficient polynomials vanish if x; = $~ (i = 1,. • . , k). Since the g~ are arbi t rary in GF(~a) , by the inductive hypothesis, each of the coefficient polynomials is a zero polynomial . Since each coefficient of P is a coefficient o f one of these polynomials , it follows that P is a zero polynomial . This implies that i f f e = f o , then P = Q.

It is easy to show that card ~ (S , ~, n) = S s~ and card F ( 6 a, n) = S s". Since different polynomials define different mappings , it follows tha t F~(Se, n) = F(,2T, n).

The p r o o f is completed.

20. Another Property of the Groups A(S~) and A(SP)/Y,(A a)

It has been shown, Corol lary 6.15, that every finite group is i somorphic to some subgroup of A(Aa)/E(SP). Thus A ( A a ) / E ( ~ ) contains elements of any given finite order. We show that A(6f)/E(~9 °) contains elements of infinite order. More particularly, we show that A(Ae)/E(6 ~) contains two elements, each of order two, whose produc t is of infinite order.

Let ~: X(Aa)---~X(6 a) be defined as follows. For any x s X ( S f) and i s I, [~(x)]i = xl unless xi = 0 or 1; if xi = 0, [~(x)]i = 1; if xl = 1, [~(x)]i = 0. Clearly, ~ s A(S~) and ~2 is the identity mapping.

Let /3: X(Af)--~X(._~) be defined as follows. For any x s X ( 6 a) and i ~ I, [fl(x)]i = xi unless x i_ xXiXi+ lxi+2 = 1001 or 1101 ; if x i - lXiXi+ lXi+z = 1001, [/3(x)]i = 1; if x i _ l x ~ x i + l x i + : = 1101, [/3(x)] i = 0. I t is easily verified that /3 s A(6~) and/32 is the identity mapping.

Let ~ = ~/3. Let x s X ( 6 e) be defined by

x i = 0 (i >_ 0),

x _ 7 x _ 6 " " x _ l = 1001001,

x _ 2 i _ l x _ 2 i = 10

F o r p s 11, let x ~2") ~ X(Sf ) be defined by

x~ 2p) = 0

x(2p) ,.(2p) = 01 - - 2 i . ~ - 2 i + 1

x ~ ) p - 7 . . . . -~-2p-1~2') = 1001001,

x(au) ~(2,) 10 _ 2 i _ 1 . ~ _ 2 i

(i > 4).

(i > 0),

(i = 1 , . . . , p ) ,

(i > p + 4 ) .

374 G.A. HEDLUND

For p E lx, let x (2p+1) E X(Sf) be defined by

x~ 2p+I) - 1 (i > 0),

x(2p+l)v(2p+l) - - (01) (i 1 , ' " , p + l ) , - 2 i " ~ - 2 i + 1 - -

x ( 2 P + 1) . . . . ( 2 p + 1) = 1001001, - 2 p - 9 "~'- 2 p - 3

X 2 p + 1 ) v ' ( 2 p + 1 ) 10 (i > p+5) . _ 2 i _ 1 . ~ _ 2 i =

More briefly, we write

x = L 100100 R,

where R is the right periodic sequence 0 0 . . . , L is the left periodic sequence • • • 101010 and the "do t" notation indicates that x is indexed at the first element of R.

For p E I1, let (01) 9 denote the (2p)-block 01010.. . 1 and let /~ denote the right periodic sequence 11. • .. Then

X (2p) = L 1001001 (01) p R,

x (2p+1) = L 1001001 (01) p+I/~.

It is easy to verify that ~02p(x) = x 2p and ~2p + X(x ) = x(2p + 1). But for p e 11, q~2p(x) = x c2p) ~ x and ~2P+l(x)= xC2p+l)~ x. Thus we have proved the following theorem, due to Curtis, Hedlund and Lyndon.

20.1 THEOREM. The group A( 5") contains two elements, each of order two, whose product is of infinite order.

Let G be the subgroup {~"l n ~ I} of A(5¢). Let G' = (gy(Se)[ g ~ G}. Then G' is the subgroup of A(Sg)/Z(Sg) generated by ~0E(5¢). If ~0Z(S g) were of finite order, there would exist n ~ 11 such that 9~"E(Se) -- E(5°), and hence there would exist k e I such that ~oncr k = cr °, or ~o" = or- k, and thus ~o"(x) = g- k(x). But ~o"(x) is either x (2p) or x c2p÷ x). Since the block 100100101 appears in both x (2p) and x(2p + 1), and this block does not appear in x, neither x t2p) nor x (2p + 1)is a trans- late of x, so that 9~"(x) ~ g-k(x). We conclude that q~Y~(Se) is an element of infinite order in A(S,°)/E(,.9").

Let ~' = 0~Z(5¢), 3' = flY(Sa). Then ~' and 3' are elements of A(SQ/Z(S¢), each is of order two and ~o' = ~0Z(de) = ~flE(SQ = ~'/3'. Thus we have the following extension of Theorem 20.1, also due to Curtis, Hedlund and Lyndon.

20.2 THEOREM. The group A(Se)/Z(5 a) contains two elements, each of order two, whose product is of infinite order.

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(Received 25 April 1969)

G. a. Hedlund -- Endomorphisms and Automorphisms of the Shift Dynamical System

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