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

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.

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 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 $A_1$.
• If you wish to avoid sequences, every open ball around $Y$ in the $d_\Delta$ metric will contain my $Y_n$ for $n\ge n_0$, but small open balls around $Y$ in $d_H$ will not contain any of them, so the open balls can't be the same. – Kusma Sep 5 '18 at 16:21