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Topology




                    Notes          32.1.3 Besicovitch Covering Lemma

                                   There exists N = N(d)³1 such that for any cover by balls we can choose a sub-cover { }, such that
                                                                                                      i
                                   any point x lies in at most N balls.
                                   Thus we can bound

                                                                      1        N
                                                H d + (X)  å diam (  ) d +    å   (B )   .
                                                                i          i
                                                         i            C  i     C
                                                                 N
                                   Letting 0 we have that  H d + (X)   . Since C > 0 can be chosen arbitrarily large we deduce
                                                                 C
                                   that  H d + (X) = 0. In particular, dim (X)  d +for all> 0. Since > 0 is arbitrary, we deduce
                                                               H
                                   that dim (X)d.
                                         H
                                   The proof of the first statement is similar, except that a replace X by a set of full measure for
                                   which  d (x) d .
                                         

                                          Example 1: If L : X X  is a surjective Lipschitz map i.e. C > 0 such that
                                                        1   2
                                                   |L(x) – L(y)| C|x – y|,
                                   then dim (X )dim (X ).
                                          H  1     H  2

                                          Example 2: If L : X X  is a bijective bi-Lipschitz map i.e.C > 0 such that
                                                        1   2
                                                   ( )  |x – y| |L(x) – L(y)|C|x – y|,
                                                    1
                                                     C
                                   then dim (X ) = dim (X ).
                                          H  1     H  2
                                   Solution: For part 1, consider an open cover  for X  with dim ( )for all . Then the
                                                                            1          i           i
                                   images  = {L( ) : } are a cover for X  with dim (L( ))L  for all . Thus, from the
                                                                     2            i   
                                              
                                                                                                    
                                                                                             
                                   definitions, H (X ) H (X ) .³    In particular, letting0 we see that H (X )³H (X ). Finally,
                                              L   2    1                                     1       2
                                   from the definitions dim (X )dim (X ).
                                                       H  1     H  2
                                                                                              –1
                                   For part 2, we can apply the first part a second time with  replaced by L .
                                   32.1.4 Bernoulli’s  Measures
                                          Example 3: For an iterated function scheme T , ……, T  :  we can denote as before
                                                                             1      k
                                                        {
                                                      = x = (x ) ¥  : x  {1,  ,k }}
                                                              m m = 0
                                                                    m
                                   with the Tychonoff product topology. The shift map  : is a local homeomorphism defined
                                   by (x)  = x  . The kth level cylinder is defined by,
                                         m  m+1
                                                    [x ,  ,x k–1 ]  = { (i ) ¥    : i = x for 0   m   k -  1 }
                                                                           m
                                                                               m
                                                      0
                                                                  m m = 0
                                   (i.e., all sequence which begin with x , …, x ). We denote by W  = {(x , …, x ]} the set of all kth
                                                                0    k–1             k    0    k–1
                                   level cylinders (of which there are precisely k ).
                                                                        n
                                   Notation: For a sequence i   and a symbol r{1, …, k} we denote by k ( i ) = card{0mk–1 :
                                                                                            r
                                   i  = r} the number of occurrences of r in the first k terms of i .
                                   m
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