Lemma 1 For all compact sets A, B, C, and D in the plane, h(A ∪ B, C ∪ D) ≤ max{h(A,C),h(B,D)}. |
Proof Let d = h(A,C) and e = h(B,D). Then |
A ⊆ Cd, C ⊆ Ad, |
B ⊆ De, D ⊆ Be. |
Writing f = max{d, e}, we see |
A ⊆ Cf, C ⊆ Af, |
B ⊆ Df, D ⊆ Bf. |
Consequently, |
A ∪ B ⊆ Cf ∪ Df |
It is an easy exercise to show that |
Cf ∪ Df ⊆ (C ∪ D)f. |
Combining these last two inclusions we see |
A ∪ B ⊆ (C ∪ D)f. |
A similar argument gives |
C ∪ D ⊆ (A ∪ B)f. |
Consequently, |
h(A ∪ B, C ∪ D) ≤ f = max{h(A,C),h(B,D)}. |
Return to Hausdorff contraction factor.