I am trying to understand how the Hausdorff distance acts on complements but struggling to find any good resources. Is it true in general that if I have $3$ compact sets $A,B$ and $C$ that the following implication holds?

$$d_H(A,B)\leq r \implies d_H(C\setminus A, C\setminus B) \leq r$$

Where $d_H$ is the Hausdorff distance (assume the sets $C\setminus A$ and $C\setminus B$ are non-empty). Also, does anyone know of any good resources for further reading on the Hausdorff distance? Any with exercises would really help!


1 Answer 1


Update: I think I have a counterexample in $\mathbb{R}^2$:


$$A = \bigg(\bigg[\frac{1}{n},1\bigg]\times [0,1]\bigg) \cup ([0,1]\times [2,3])$$ $$B = ([0,1]\times [0,1]) \cup ([0,1]\times [2,3])$$ $$C = ([0,1]\times [0,1]) \cup ([0,1]\times [1,3])$$

We have that:

$$C\setminus A =\bigg(\bigg[0,\frac{1}{n}\bigg)\times [0,1]\bigg) \cup ([0,1]\times [1,2))$$ and:

$$C\setminus B =[0,1]\times [1,2)$$

Trivially we can make $d_H(A,B)$ as small as we like by increasing $n$. But for $d_H(C\setminus A,C\setminus B)$ will always be greater than or equal to $1$.


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