I have been studying Viro's textbook "Elementary Topology" and in the process of solving it I came across the following problem (in the textbook itself it has the number $4.Px$): Prove that on the set of convex polygons the metric $d_\triangle$ is equivalent to the Hausdorff metric.
The metric $d_\triangle$ is defined on the set of bounded polygons in the plane and maps a pair of elements of this set to the area of their symmetric difference. The Hausdorff metric is defined on the set of bounded closed subsets of an arbitrary metric space and looks like this: $d_\rho(A,B)=\max \lbrace \underset{a\in A}{\sup}\rho(a,B), \underset{b\in B}{\sup}\rho(b,A) \rbrace $ for bounded closed subsets $A,B$ of the metric space $(X,\rho)$.
To be honest, I don't have any ideas for a solution. The mechanism of applying the classical approach of proving the equivalence of two metrics (namely, demonstrating the inclusion of the first topology in the other and of the second topology in the first) seems unclear to me here, for it is not even fully visual or at least intuitive to see what the open ball represents on each of these metrics.
Best Answer
This question turns out to be very hard. The above link by @AlexRavsky confirms this feeling. The nice answer by @Kusma explains all the key ideas, except the last part, which he said that he "can't give an even semi-formal proof". Let me give a stab.
So we want to show this. Suppose $P$ and $Q_n$ are convex polygons such that $d_\Delta(P,Q_n)\to 0$ but $d_\rho(P,Q_n)\geq \epsilon_0>0$, then we have a contradiction.
My argument now depends on how $d_\rho(P,Q_n)=\max\{\sup_{p\in P}\rho(p, Q_n), \sup_{q_n\in Q_n}\rho(q_n, P)\}$ is achieved, that is, whether there are points in $P\backslash Q_n$ or $Q_n\backslash P$. Since there are only two cases, taking a subsequence puts us in one of the two situations.
$d_\rho(P,Q_n)=\sup_{p\in P}\rho(p, Q_n)\geq \epsilon_0$. That is, there is some point in $P$ that has a fixed distance from $Q_n$. Then this $p$ is a vertex of $P$, and the circular sector in $P$ with vertex $p$ and radius $\epsilon_0$ doesn't intersect $Q_n$, which is convex. If $\alpha$ is the smallest angle of $P$ (in radian), then $$ d_\Delta(P, Q_n)\geq \frac{1}{2}\alpha \epsilon_0^2, $$ and this contradicts $d_\Delta(P,Q_n)\to 0$.
$d_\rho(P,Q_n)=\sup_{q_n\in Q_n}\rho(q_n, P)\geq \epsilon_0$. That is, there is some point $q_n\in Q_n$ that has a fixed distance from $P$. Then by projection, $\exists p_n\in P$ such that $\rho(p_n, q_n)=\rho(P, q_n)\geq \epsilon_0$. Then $p_n\in \partial P$. Suppose first $p_n$ is not a vertex, and so lies on the inside of an edge. Then $\overline{q_np_n}$ is perpendicular to this edge. Let $\ell$ be the smallest side-length of $P$. Since $d_\Delta(P,Q_n)\to 0$, there exists two points $r_n, s_n$ on the edge that belongs to $Q_n$ with $\rho(r_n, s_n)\geq \frac{\ell}{2}$. (The disks around the vertices of this edge must contain points in $Q_n$, since otherwise the $d_\Delta(P,Q_n)$ is big.) Then $Q_n$ would have a triangle $\Delta(q_n,r_n,s_n)$ (again since it is convex) whose area is $$ \geq \frac{1}{2}\frac{\ell}2 \epsilon_0. $$ This is a fixed number, so contradicts that $d_\Delta(P,Q_n)\to 0$. If the projection $p_n$ is a vertex of $P_n$, by arguing with the angles of $P$, we can still argue that there exists an edge of $P$ such that the height of $q_n$ above this edge is a positive fixed number in terms of $\epsilon_0$ and $P$. Concretely, there exists one edge of $P$ such that the height of $q_n$ above it is $$\geq \epsilon_0\min_{\beta}\sin\frac{\beta}{2},$$ where the minimum is taken over all interior angles $\beta$ of $P$. Then the above argument of a triangle with its base on this edge again applies.