Let $(X,d)$ be a compact metric space and $K(X)$ the class of its compact subsets. Recall, that the Hausdorf distance between $A,B\in K(X)$ is given by
$$ d_{H}(A,B):=\max\big\{ \sup_{x\in A}\inf_{y\in B}d(x,y) , \sup_{y\in A}\inf_{x\in B}d(x,y) \big\} .$$
It seems quite "intuitive" (in view of the continuity of the distances), that if $\max\{d_{H}(A,A'),d_{H}(B,B')\}\leq \varepsilon$, for certain $\varepsilon>0$ and $A,A',B,B'\in K(X)$, then $|d_{H}(A,B)-d_{H}(A',B')|$ need to be "small". Formally, the question is the following: There is a continuous function $\theta: [0,\infty)\longrightarrow [0,\infty)$, with $\theta(t)=0$ iff $r=0$ such that
$$|d_{H}(A,B)-d_{H}(A',B')|\leq \theta (\varepsilon),$$
whenever $\max\{d_{H}(A,A'),d_{H}(B,B')\}\leq \varepsilon$? Can we show an (explicit) expression for such functino $\theta$?
Many thans in advance foy your comments and suggestions.
Yes, this works with $\theta(t)=2t$.
Since the Hausdorff distance is a metric (on compact subsets), the triangle inequality holds for $d_H$. Let $A,A',B,B'$ be given as compact subsets with $\max(d_H(A,A'),d_H(B,B'))\leq \varepsilon$. Using the triangle inequality we have $$ d_H(A,B) \leq d_H(A,A')+d_H(A',B')+d_H(B',B) \leq 2\varepsilon + d_H(A',B') $$ and $$ d_H(A',B') \leq d_H(A',A)+d_H(A,B)+d_H(B,B') \leq 2\varepsilon + d_H(A,B). $$
These imply $$ -2\varepsilon \leq d_H(A,B)-d_H(A',B') \leq 2\varepsilon $$ and therefore $$ | d_H(A,B)-d_H(A',B') | \leq 2\varepsilon. $$