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Unformatted text preview: j ≥ k , we can take k = j and apply the triangle inequality:
d((x′ , y ′ ), (wj , zj )) ≤ d((x′ , y ′ ), (xj , yj )) + d((xj , yj ), (wj , zj )) < δ/2 + δ/2 = δ
That is, (wj , zj ) ∈ C and consequently
0 < d((x, y ), (wj , zj )) < diam(B ) < ǫ
Because ǫ > 0 was arbitrary, this shows (x, y ) is a limit point of OI + (w0 , z0 ).
(The case j < k is Exercise 2.4.4.))
To see Λ(OI + (w0 , z0 )) ⊆ A, note by Theorem 2.4.1 that for every inﬁnite
sequence I ′ and for every (x0 , y0 ) ∈ A, that OI ′ + (x0 , y0 ) ⊂ A. For any limit
point (x′′ , y ′′ ) of OI + (w0 , z0 ), there is a subsequence (wpi , zpi ) → (x′′ , y ′′ ). By
eq (2.12), for this same subsequence (xpi , ypi ) → (x′′ , y ′′ ). That is, every limit
point of OI + (w0 , z0 ) is a limit point of a sequence of points in A. Because A is
closed, it contains all its limit points. That is, Λ(OI + (w0 , z0 )) ⊆ A. 2.4. CONVERGENCE OF RANDOM IFS 61 Proof of Proposition 2.4.1. Because i1 → · · · → in is forbidden, we know
int(Ain ...i1 ) = ∅. Now fo...
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