Convergence of Deterministic IFS

To prove the uniqueness of A, suppose there is another nonempty, compact set C with T (C) = C. Then

h(A,C) = h(T (A),T (C)) ≤ r⋅h(A, C) (by Prop. 2).

Now r = max{r1, ..., rN} < 1, so

h(A,C) ≤ r⋅h(A,C)

requires h(A,C) = 0. That is, C = A.

Return to the proof of the theorem.