Let $A,B \subset \mathbb R^n$ be two closed and non-empty sets. We define the Hausdorff distance by:
$\delta_H(A,B):=\max \big [ \sup_{X\in B} d(X,A),\sup_{Y\in A} d(Y,B) \big ]$
where $d(X,A):=\inf_{Y\in A} \vert X-Y \vert$
Now let $C:=\{x\in A: dist(x, A^\complement) \ge \theta \gt 0 \}$ and suppose is compact.
QUESTION: If $C \subset B$ then can we claim that $\delta_H(B,A) \le \delta_H(C,A)=\theta$ ?
In my defense, I 'm not very familiar to the Hausdorff distance and all my approach is rather intuitive. I would appreciate if somebody could give me some details on whether this assumption is valid or not.
Thank you in advance!
First of all, you would want both $A$ and $B$ to be bounded if you want the Hausdorff distance to actually be a metric (otherwise, the distance might be $\infty$, as the following also illustrates). I'm assuming $A$ compact throughout:
The answer is no. Consider $B=B[0,r]$ (the closed ball of radius $r$. Assume $r>0$ large enough that $A\subseteq B(0,r/2)$. \
Then, $x=r(1,0,0,...0)\in B$ and for any $y\in A,$ we have $||x-y||\geq ||x||-||y||\geq r/2$. Hence, $d(x,A)\geq r/2$ and hence, $\delta_H(A,B)\geq r/2.$
So, making $r$ very large, I can make the distance arbitrarily large, even though every compact subset of $A$ is a subset of $B$.