Convergence of Deterministic IFS

Here we prove the Lemma (Ae)f ⊆ A(e+f), illustrated by this sequence of inclusions

Take any point x in (Ae)f. Then there is a point y in Ae with d(x,y) ≤ f.

Now because y belongs to Ae, there is a point z in A with d(y,z) ≤ e.

Then by the triangle inequality for Euclidean distance,

d(x,z) ≤ d(x,y) + d(y,z) ≤ e + f.

That is, for any point x in (Ae)f, there is a point z in A with d(x,z) ≤ e + f.

This shows (Ae)f ⊆ A(e+f).

Return to Triangle inequality.