“Area metric” and “Hausdorff metric” are not equivalent on all closed polygons, but equivalent on convex closed polygons

Question

Suppose $X$ is the set of all closed polygons, $d_\Delta$ is the “area metric” defined by the area sum of the symmetric difference of two closed polygons, and $d_H$ is the Hausdorff metric on $X$. How should I prove that $d_\Delta$ and $d_H$ do not generate the same topology on $X$? Also why do they generate the same topology on the subset of convex polygons? I tried to visualize how a typical open ball in both metrics looks but this seems rather impossible.

Answer 1

I can show you a sequence $Y_n$ that converges to $Y=[-1,0]^2$ in the area metric, but not in the Hausdorff metric. Near the points $(\frac k n,0)$, $k=1,\dots, n-1$, consider a triangular spike of height $1$ and area $\frac{1}{n^2}$, for example the triangle with vertices $(\frac k n-\frac{1}{n^2},0), (\frac k n,1), (\frac k n+\frac{1}{n^2},0)$. Define $Y_n$ as the union of $Y$ and these triangles. Then the distance between $Y_n$ and $Y$ in the area metric tends to zero, but the distance in the Hausdorff metric does not.

For the positive result, try to consider two convex polygons and try to bound the difference of their areas in terms of the Hausdorff distance and vice versa. Even if the metrics are not equivalent, that may be an easier approach to showing the topologies are the same than considering open balls.

For example, if you have two sets whose Hausdorff distance is small, you have $A_1\subset B_\epsilon(A_2)$ and vice versa. This means that $A_1{\small\Delta} A_2\subset B_\epsilon(A_1)\setminus A_1\cup B_\epsilon(A_2)\setminus A_2$, and the area of that set is bounded by $C\epsilon$ times the perimeter of $A_1$.

On the other hand, if two convex polygons have positive Hausdorff distance, you need to show their area distance can't be arbitrarily small. I can visualise why that should be true, but I can't give you an even semi-formal proof at the moment.