Suppose A and B are compact subsets of the plane. Then the Hausdorff distance between A and B is |
h(A, B) = min{ε: A ⊆ Bε and B ⊆ Aε}. |
For example, suppose A is the red unit square, and
B is the rectangle with base length |
What is h(A, B)? |
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We see |
A is contained in B1/4 and |
B is contained in A1/2. |
These are the smallest ε-thickenings that will work: |
for any ε < 1/4, Bε misses the left and right sides of A, and< |
for any ε < 1/2, Aε misses the top of B. |
So h(A, B) = 1/2. (Remember, the same ε must work for both A ⊆ Bε and B ⊆ Aε.) |
As usual, the only work required in proving h is a metric lies in showing the triangle inequality holds. That is made easier by first proving the lemma |
(Aε)δ ⊆ A(ε+δ). |
Exercise Find h(A, B), where A
is the union of the line segments
from |
Return to Convergence of determinisitc IFS.