Convergence of Deterministic IFS

Take B a compact set large enough that
Ti(B) ⊆ B for i = 1, ..., N.
This is always possible because for large enough B, the effect of translation will be negligible compared with contraction.
Then
B ⊇ T1(B) ∪ ... ∪ TN(B) = T (B)
Applying T to this inclusion gives
T (B) ⊇ T (T (B)) = T 2(B)
Continuing to apply T and combining the inclusions gives
B ⊇ T (B) ⊇ T 2(B) ⊇ T 3(B) ...
and so
B, T (B), T 2(B), T 3(B), ...
is a nested sequence of nonempty compact sets. Then
A = ∩T k(B)
is a nonempty, compact set.

Return to the proof of the theorem.