I have the following property I suspect about the Hausdorff distance between convex sets in $\mathbb{R}^d$, and I thought that perhaps someone can find a fault with this idea if it is wrong.
Given nonempty convex compact sets $A,B\in \mathbb{R^d}$, their Hausdorff distance is given by $d_H(A,B)=\max \Big\{ \underset{a\in \partial A}{\sup} d(a,B), \underset{b\in \partial B}{\sup} d(b,A) \Big\}$. When $d=1$ and $A=[\alpha,\beta]$ and $B=[\gamma,\delta]$, then $d_H(A,B)=\max \big\{ \vert \alpha-\gamma\vert , \vert \alpha -\delta\vert , \vert\beta – \gamma \vert, \vert \beta- \delta \vert \big\}$.
While the Hausdorff distance is given by $d_H(A,B)= \max \Big\{ \underset{a\in A}{\sup}d(a,B), \underset{b\in B}{\sup} d(b,A) \Big\}$. In the $1$-dimensional case this follows just by looking at all the cases for $a,b,c,d$. I think that in $\mathbb{R}^d$ this should follow by the hyperplane separation theorem and since $A=\partial A \sqcup int(A)$. For example if $a\notin B$ and $a\in int(A)$, then there is a hyperplane separating $\{ a \}$ and $B$. We then take a vector of length $\epsilon$ perpindicular to the separating hyperplane and the openess implies that there exists some $a'\in A$ such that $d(a',B)>d(a,B)$. Therefore, $\underset{a\in A}{\sup}d(a,B)=\underset{a\in \partial A}{\sup}d(a,B)$.
Does this argument seem correct? If not, does the statement seem correct?
Best Answer
Since I stumbled across the answer, I will say that the answer to this question is yes and one can strengthen it to a general metric space by demanding closedness and boundedness instead of compactness. This is proven very clearly in this paper.
I hope that if someone else stumbles upon this, the reference I gave will help them.