# Subsets of a metric space in which Hausdorff semi-distance is symmetric

These are the definition of Hausdorff distance and Hausdorff semi-distance for subsets of a metric space $X$.

‎‎Hausdorff semi-distance of two subsets ‎$‎A‎, B‎ \subset X$ is defined as below: ‎$‎d(A ‎\mid ‎B‎)‎ ‎:= ‎\sup ‎\lbrace ‎d(x,B) : x \in A ‎\rbrace‎ ‎‎$‎‎‏‎, where ‎$‎d(x, B‎)‎ ‎:= ‎\inf ‎\lbrace ‎d(x,y) : y \in B ‎\rbrace‎ ‎‎$.

Also Hausdorff distance for the subsets‎ $‎A‎, B‎ \subset X$‎‎‏‎ is defined as follow : ‎$d_H (A,B) := \max‎\lbrace ‎d(A\mid B), d(B \mid A) ‎\rbrace‎ ‎$‎‎ .

I want to know that are there any sufficient conditions on $A$ and $B$ which imply that $d(A\mid B) = d(B \mid A)$‎ ?

I guess that for convex sets the equality hold, but I can't prove my intuition. Thanks in advance for any help/comment.

No, equality rarely holds, and convexity does not help. For one thing, notice that $d(A|B)=0$ whenever $A\subset B$. So, if $A$ is a proper subset of $B$, the one-sided distances $d(A|B)$ and $d(B|A)$ are not equal.
For another example, take two disjoint closed balls $A=\mathbb B(c_1,r_1)$ and $B=\mathbb B(c_2,r_2)$. Then $d(A|B) = \|c_1-c_2\|+r_2-r_1$ and $d(B|A) = \|c_1-c_2\|+r_1-r_2$. So, the distances agree only when $r_1=r_2$.
A geometrically natural special case when $d(A|B)=d(B|A)$ is when the sets are parallel (or equidistant), meaning there is $d$ such that $d(a,B)=d$ for all $a\in A$ and $d(b,A)=d$ for all $b\in B$. These sets arise as level set of submetries $f:X\to Y$, which are maps between metric spaces that satisfy $f(\mathbb B(c,r))=\mathbb B(f(c),r)$ for all $c\in X$ and all $r>0$. (A special case of a submetry is quotient map by a group of isometries.)