Talk:Hausdorff distance

Is the figure wrong?
The figure doesn't seem to fit the definition of Hausdorff distance.

To get sup_x inf_y d(x,y), you want to choose an x in X that is maximally distant from Y. That would not be a point right on the near boundary of X, but on the opposite side (which is not pictured in the figure.) For example, take the left line segment in the figure, the one supposedly exhibiting sup_x inf_y d(x,y). Take the point in X at the end of it. Now move that point a ways up into the interior of X. Now you have an x in X with a  larger value of inf_y d(x,y). —Preceding unsigned comment added by 128.226.120.167 (talk) 16:17, 30 October 2007 (UTC)


 * If we suppose that X should be the complement of the green part (instead of X=the green part, as indicated in the picture), then perhaps this figure will work. --Kompik 19:28, 3 December 2007 (UTC)


 * The figure is indeed entirely wrong and needs to be replaced as soon as possible as it is no doubt causing a lot of confusion.


 * Kompik suggests that assigning X as the complement of the green part would make it work, and this is true for X. However, this would only solve the problem for the directed Hausdorff distance from X to Y (in essence, we should demonstrate that the Hausdorff distance is in fact the maximum of both directed distances). Also, the local maximum to the right remains confusing and as would be that the figures would be touching. Instead, I suggest an illustration with points instead of blobs: illustrate that the maximum distance of all paths between each point in X to a nearest point in Y is the Hausdorff distance. In other words: each point in X is at most $$ d_{\mathrm H}(X,Y) $$ from Y.


 * This page has an excellent demonstration that could perhaps be animated in a GIF file. &mdash; Stimpytalk 09:48, 3 January 2008 (UTC)


 * That link seems to have stopped working (404 Not Found). --Lambiam 19:51, 11 April 2008 (UTC)


 * The answer is yes. I try a new version of this figure. Rocchini (talk) 12:53, 6 May 2008 (UTC)


 * I think the figure is correct as long as we take X and Y as being the green and blue lines respectively. The confusion arises if we try to interpret X and Y as being the regions enclosed by these lines, in which case the Hausdorff distnace from each point of X to the set Y is 0, as X is contained in Y. And if we assume that Y is the region outside of the blue line then we don't even have a Hausdorff distance between X and Y because Y is no longer a compact set. I have added a caption to the image to try to clarify the correct interpretation. Gandalf61 (talk) 11:22, 19 May 2008 (UTC)

Is distance in the remark wrong?
The remark is not that important. If $$d(X,X) > \epsilon$$, then it is true that $$X \subseteq Y_\epsilon$$. It is just like expecting y to belong to B(x), the ball with radius 1 centered at x, just because d(x,y) = 1. André Caldas (talk) 20:03, 12 May 2015 (UTC)

Perhaps I am misunderstanding something here, but shouldn't the distance between the two sets be 2 rather than 1? — Preceding unsigned comment added by Bowiki (talk • contribs) 15:44, 15 November 2012 (UTC)

- No. 2 is the supremum of d(x,y), not the Hausdorff distance. André Caldas (talk) 20:03, 12 May 2015 (UTC)

According to my understanding the distance between the sets is 1+ε, where ε<δ, ε>0 and δ>0 is an arbitrary small number. Thus, the example in the remark doesn't prove anything. --130.75.51.124 (talk) 13:25, 9 January 2013 (UTC)

- No. There is no ε in the definition of X and Y. So, the distance cannot deppend on ε. André Caldas (talk) 20:03, 12 May 2015 (UTC)

-Please can someone explain to me why 1∉ Y_1 ? I take the sequence y_n=-1/n∈Y and evaluate inf|1-y_n|=1 thus 1∈Y_1. Please correct me. — Preceding unsigned comment added by Stikawi (talk • contribs) 16:47, 13 November 2019 (UTC)

The remark is wrong for the following reason: the distance map $$d(\cdot, X)$$ is Lipschitz-continuous, hence the $$\varepsilon$$-fattening $$X_\varepsilon = \{y \in M \vert d(y, X) \leq \varepsilon\}$$ is closed as the continuous preimage of a closed set; next, it is easy to prove $$X_\delta = \bigcap_{\varepsilon > \delta} X_\varepsilon$$ for any $$\delta \geq 0$$. Now set $$\delta := d_H(X, Y)$$, then the definition implies $$Y \subset X_\varepsilon$$ for all $$\varepsilon > \delta$$, and hence $$Y \subset \bigcap_{\varepsilon > \delta} X_\varepsilon = X_\delta$$; symmetry yields $$X \subset Y_\delta$$.

Also Stikawi is right: the correct 1-fattenings in the example are $$X_1 = [-1, 2]$$ and $$Y_1 = [-2, 1]$$. 129.22.1.29 (talk) 15:06, 6 October 2023 (UTC) M.L.Juhos

Re the motivation for the definition
Getting a symmetric function is not the only point for going from d to d_H. d(X,Y)=0 iff X\subset Y as well, so that d_H(X,Y)=0 iff X=Y as required. Would someone make a nice edit? 89.180.15.198 (talk) 10:37, 7 July 2008 (UTC)


 * ✅. --Lambiam 02:10, 12 July 2008 (UTC)

Hi there, I don't understand the notation '[1,3,6,7]' in this section. Shall I change it to {1,3,6,7}, and then also change all the other square brackets to curly brackets? Thanks! — Preceding unsigned comment added by The man who was Friday (talk • contribs) 14:39, 29 March 2012 (UTC)

Link to localized pages
Is it possible to have links to localized pages when appropriate? For instance, this article have corresponding Russian page Метрика Хаусдорфа and Italian page Distanza di Hausdorff. I am sure that lots of math articles have variants in other languages.

Arkadi kagan (talk) 07:28, 17 May 2009 (UTC)


 * The link exists already. When you look the article page, it appears normally on the left, far down the Wikipedia image.  You can also see it (and possibly modify) in the "source" of the article page, by clicking on "edit this page" and moving to the very bottom on the page contents. --Bdmy (talk) 07:44, 17 May 2009 (UTC)

Definition of natural number
It might be wise to state in the Remark on the Definition that this article adheres to the notation $$\mathbb{N}$$ as the set of strictly positive integers, in contrast to the positive integers which would result in a Hausdorff distance of $$\infty$$. Both definitions are found in literature, see, so it seems necessary to disambiguate. ikingut 15:30, 12 June 2014 (UTC) — Preceding unsigned comment added by Rphlypo (talk • contribs)

External links modified
Hello fellow Wikipedians,

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False claim under "Properties"
"If X is a subset of Y then d(X,Z) >= d(Y,Z)" is just false for general Z. E.g., take Z=X; then this says d(X,X) >= d(Y,Z); the LHS is zero and this is true only when Y happens to be "the same as" X (more precisely: same closure). I'm not just removing this because (1) right now I'm away from my usual computer and am not logged in, and I don't want to make significant changes without accountability, and (2) maybe there's some not-wrong thing that this was intended to say. (I think _probably_ not.) So for now I'm just leaving this comment here and also on the talk page of the user who added that bit (https://en.wikipedia.org/w/index.php?title=User_talk%3ABongobums#Hausdorff_distance_property). Gareth McCaughan — Preceding unsigned comment added by 15.90.166.11 (talk) 13:06, 5 January 2024 (UTC) I've now removed that claim. Gareth McCaughan (talk) 02:30, 8 January 2024 (UTC)


 * Yes, you are right. Thank you for pointing that out! Next time I will check my claims more thoroughly before I post them on the wiki. Bongobums (talk) Bongobums (talk) 08:33, 24 January 2024 (UTC)