Let z denote the collection of all two-sided sequence of 0, s and 1, s , and
define h : c*  z by
h(v)  the tow  sided sequence corresponding to v , for v in c*
(1)
Then h is well-defined, and identifies c* with z, as Lemma (2.2) asserts.
Lemma :-
h : c*  z is one-to-one and onto.
Proof: If v and w are in
c * and h(v)  h( w) , then because h(v ) and h (w) have
the same forward(backward) sequence , they lie on the same vertical
(horizontal) line in T . Therefore v = w , so that h is one-to-one .
Assume that x  ...x3 x 2 x1.x0 x1x2 ... is in z. For n  0 , let
J n  v in c0  c1 : h(v)  ...z3 z 2 z1.z0 z1z2 ... and z0 z1 z2 ...zn  x0 x1 x2 ...xn 
And
J  n  v in c0  c1 : h(v)  ...z3 z 2 z1.z0 z1z2 ... and z n ...z3 z 2 z1  x n ...x3 x 2 x1
Then J n and J  n are closed for all n. Because n o J n is a single vertical line in T
, if follows that  J n is a unique point v* . By construction, h(v* )  x, so that h
all n
is onto.
The map h has the property that
If h(v)  ...z3 z 2 z1.z0 z1z2 ..., then h(M (v))  ...z 2 z1z0 .z1z2 z3...
This means that the sequence associated with M(v) is the sequence associated
with respect to the decimal point. This fact , along with the association
between point of c* and two-sided sequence of 0, s and 1, s bears immediate
fruit. In particular, the two doubly-repeated sequences.
... 0.0... and 1 . 1 ...
Correspond to fixed points of M you can check that the fixed point p on the
border between B and c0 corresponds to ...0. 0... , Next, the sequence ...10.10...
and ...01.01... comprise a 2-cycle for M . using these sequences as models , one
can exhibit two-sided sequences corresponding to n-cycles for any positive
integer n. from the definition of ...z3 z 2 z1.z0 z1 z2..., one can even indicate where in
T members of such a cycle lie .
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Next we will introduce a distance on the set z of two-sided sequences . this
distance will make it possible for us to chow that c* and z are homeomorphic
Let
x  ...x3 x 2 x1.x0 x1x2 ... and z  ...z3 z 2 z1.z0 z1z2 ...
Then we define the distance x  z between x and z by the formula .


xz 
xk  zk
2
k  
(2)
k
The distance is a metric on the space of two-sided sequences. If
xk  zk for k  n, then x  z  1
moreover, if x  z  1
2n
2n 1
.
, Then x k  z k for k  n  1. thus the distance between x
and z is small provided that the central blocks of x and z are identical .
Theorem :-
h : c*  z is a homeomorphism .
Proof :- by lemma 2.3, h is one-to-one and onto . therefore we need only show
that h and h 1 are continuous.
Let   0 and choose n so large that 1 n 1   next , let v and w be in c* ,
2
with
h(v)  x  ...x3 x 2 x1.x0 x1 x2 ... and h(w)  z  ...z3 z 2 z1.z0 z1 z2 ...
If v  w  1 3n 1, then v and w lie in the same vertical strip of width 1 3n 1 , so
that xk  zk for k  0,1,2,...n. similarly , that is a 1  0 such that if v  w  1 then
v and w lie in the same horizontal strip at the nth stage , which means that
xk  zk for k  1,2,...,n. Now choose   0 such that   1 3n 1 and   1 , It
follows that if v  w   , then xk  zk for k  n , and thus
xz 


k  n 1
xk  zk
2
k

1
2n 1

Consequently h is continuous. The proof that h 1 is continuous follows by a
similar argument.
Therefore h : c*  z is a homeomorphism.
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The left shift map  : z  z is defined as the name suggests: if
z  ...z3 z 2 z1.z0 z1 z2 ... , then  ( z )  ...z3 z 2 z1 z0 .z1 z2 ...
Thus  shifts the entries the left one place with respect to the decimal point, or
equivalently, shifts the decimal point one place to the right. That  is a
homeomorphism on z can be proved in a straightforward manner . moreover,
we can show that  is strongly chaotic, which by definition means that .
i. Its domain has a dense set of periodic points
ii. It has sensitive dependence on initial conditions
iii. It is transitive (that is, it has an element with dense orbit)
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