As a bonus question in an exam we were asked to find compact metrix spaces $X,Y$ and $Z$ such that $d_{GH}(X,Y)=d_{GH}(X,Z)=d_{GH}(Y,Z)>0$.

The proposed answer is to take $\{0\},\{-1,1\}$ and $\{-1,0,1\}$. And the distances can be easily calculated by trying all appropriate correspondances and calculating distortions.

However I proposed the following three sets ( there are only two radii and they are $R>r$)

enter image description here

set $X$ is the big closed ball of radius $R$.

Set $Y$ is the small closed ball of radius $r$.

Set $Z$ is a closed ball of radius $r$ union two perpendicular line segments of length $2R$ that intersect in the center of the ball.

The conjecture is that all distances are $R-r$. To achieve this distance simply place the figures concentrictly. (I think we may need that $\frac{r}{R}$ is big)

In order to get lower bounds for the distances involving $Y$ one simply uses the bound $d_{GH}(A,B)\geq |\text{Diam}(A)-\text{Diam}(B)|/2$.

But I am stuck calculating the distance between $X$ and $Z$. One approach is to use contradiction and try to use distortions. If we take $x_1$ and $x_2$ diametrally opposite then if $z_1\sim x_1$ and $x_2\sim z_2$ we must have $d(z_1,z_2)>r$ and so at least one of $z_1$ and $z_2$ is outside the small ball, but Im stuck after this.

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    $\begingroup$ I would call this "Hausdorff distance". As would Gromov. His name is attached to the topology using that distance. $\endgroup$
    – Igor Rivin
    Dec 15, 2017 at 16:54
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    $\begingroup$ @IgorRivin No. The topology of Hausdorff distance (between subsets of a metric space) owes nothing to Gromov and is due to Hausdorff too. Gromov's invention is also a distance, but between different metric spaces, instead of subsets of a given metric space. It's inspired of Hausdorff distance, and hence is known as Gromov-Hausdorff distance. $\endgroup$
    – YCor
    Dec 15, 2017 at 17:57
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    $\begingroup$ I don't understand the OP's question. What are "these three sets"? are $X$ and $Y$ meant to be closed disks? of which radius? is $Z$ the union of a closed disc and a circle? A picture can help clarifying a description, but should not replace it (especially when the "formal" description is easy and the picture confusing) $\endgroup$
    – YCor
    Dec 15, 2017 at 18:01
  • $\begingroup$ Yes, $X$ and $Y$ are closed disks and $Z$ is the union of $Y$ and two perpendicular segments of length $2R$. $\endgroup$
    – Gorka
    Dec 15, 2017 at 19:52

1 Answer 1


Assume $R=1$, $r=1-\varepsilon$ and $\varepsilon$ is small enuf.

Consider the maximal regular hexagon in $X$. If $d_{GH}(X,Z)< \varepsilon$ then there is a hexagon in $Z$ with sides at least $>1-2\cdot\varepsilon$ and main diagonals $>2-2\cdot\varepsilon$. The latter leads to a contradiction.


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