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Unformatted text preview: en (xk , yk ), generate (xk+1 , yk+1 ) by (xk+1 , yk+1 ) = Tik+1 (xk , yk ), where
{i1 , i2 , ...} are chosen randomly from among {1, ..., n}.
Then the sequence
Σ = {(xi , yi ), i = 1, 2, 3 . . . }
lies entirely in A, and in fact A is the closure of Σ.
The ﬁnal reason, then, is this. So far as we know Stewart [147] ﬁrst posed
this question. What happens if the Ti are selected in a nonrandom order? This
perfectly natural question gives rise to driven IFS, one of the main topics of
Ch 4. Of course, this question is natural, or even makes sense, only after the
random IFS algorithm is familiar.
So, let’s look into the details of this method. That A is the closure of Σ
means A is closed and for every point (x′ , y ′ ) ∈ A, and for every ǫ > 0, there
is some (xi , yi ) ∈ Σ with d((x′ , y ′ ), (xi , yi )) < ǫ. Let us see why this is so. We
outline the steps of the argument now. Recall proofs are near the end of this
section.
First, the ﬁxed points of the Ti are good starting points because they belong
to A.
Lemma 2.4.1 The ﬁxed point (x0 , y0 ) of Ti belongs to A.
Next, we note that if we start in A, applicatio...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.
 Fall '14
 AmandaFolsom
 Geometry, Fractal Geometry, Limits, The Land

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