Convergence of Deterministic IFS

Finally, let B be any compact subset of the plane. Then because A = T (A), we have

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

Iterating this argument, we obtain

h(T k(B), A) ≤ rk⋅h(B, A)

Because r < 1, we see

limk→∞ h(T k(B), A) = 0.

Return to Convergence of determinisitc IFS.