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Unformatted text preview: y compact set S
diam(Ti (S )) ≤ ri · diam(S )
(see Exercise 2.4.3) and so for any sequence i1 , i2 , ..., ik ,
diam(Tik ...Ti1 (S )) ≤ rik · ... · ri1 · diam(S ) ≤ (max ri )k · diam(S ) (2.11) To show A is the closure of Σ, we must show for every point (x, y ) ∈ A, and for
every ǫ > 0, there is some point (xj , yj ) ∈ Σ with d((x, y ), (xj , yj )) < ǫ. So take
k large enough that (max ri )k · diam(A) < ǫ. Now because
n Aik ...i1 A=
ik ,...,i1 =1 if (x, y ) ∈ A, then (x, y ) ∈ Ai′ ...i′ for some i′ , ..., i′ . We must show some point
of Σ lies in Ai′ ...i′ .
This relies on the randomness of the sequence producing Σ. Suppose the
point (x0 , y0 ) is the ﬁxed point of T1 . Then (x0 , y0 ) ∈ A1 . Next, (x1 , y1 ) =
Ti1 (x0 , y0 ) ∈ Ai1 1 . Next, (x2 , y2 ) = Ti2 (x1 , y1 ) ∈ Ai2 i1 1 . Continuing, we see
each successive application of Tij has this eﬀect on the address of the region containing (xj , yj ): the address digits are shifted to the right and ij is inserted in the
left-most location. T...
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