Hausdorff distance and union of sets

Question

Let $X$ be a metric space; $A_1$, $A_2$, $B_1$, $B_2$ be non-empty subsets in $X$. Let $d(\cdot,\cdot)$ be the Hausdorff distance between sets in $X$. Then $$ d (A_1 \cup A_2 , B_1 \cup B_2) \leq \max \{ d(A_1,B_1), d(A_2,B_2)\}. $$

Answer 1

For a subset $A \subset X$ and a real $\epsilon>0$, let $\cup_{\epsilon}A$ denotes $\{x \in X \mid d(x,A)< \epsilon\}$.

Let $r =\max (d(A_1,B_1),d(A_2,B_2))$. Then $A_i \subset \cup_rB_i$ and $B_i \subset \cup_r A_i$. Therefore, $$\cup_r(B_1 \cup B_2) = \cup_rB_1 \cup \cup_r B_2 \supset A_1 \cup A_2$$ and in the same way $$\cup_r(A_1 \cup A_2) \supset B_1 \cup B_2.$$ Thus, $d(A_1 \cup A_2, B_1 \cup B_2) \leq r$.