For each positive integer n, let E(n) be n-dimensional Euclidean space-with its standard metric-and let M(n) be the separable metric space whose "points" are all the non-empty compact subsets of E(n) and whose "distance" is the Hausdorff distance between each pair of these subsets. Let F(n) be the mapping of M(n) onto E(n) which maps every non-empty compact subset of E(n) to its centroid. Is F(n) continuous at all "points" of M(n)?
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asked Aug 24, 2016 at 20:08
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3$\begingroup$ How do you define the "centroid" of an arbitrary compact subset of $E(n)$? $\endgroup$– Noam D. ElkiesCommented Aug 24, 2016 at 20:34
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1 Answer
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Even if you restrict to convex sets, I don't think there's a definition of centroid that makes this work for $n > 1$.
For instance, in $\mathbb{R}^2$ consider rectangles of the form $[0, 1] \times [0, \epsilon]$ versus triangles with vertices $(0, 0), (1, 0), (0, \epsilon)$.
As $\epsilon \to 0$, both families of shapes converge to the same line segment, but their centroids converge to different points: $(1/2, 0)$ and $(1/3, 0)$ respectively.
The best bet to rescue the theorem might be to restrict to the space of convex sets with positive measure.
answered Aug 25, 2016 at 1:41
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$\begingroup$ Your question made me realize that there might be something wrong with my question. I am not sure that every non-empty compact subset of a Euclidean space has a centroid. Some definitions of "centroid " might not imply that it was a unique point. Where, for example, would be the centroid of an uncountable and totally disconnected compact set such as the Cantor middle third set -which has zero Lebesgue measure. A denumerably infinite compact set might pose an even worse problem. I see why you are advocating that one should only consider sets having positive measure. $\endgroup$ Commented Aug 25, 2016 at 19:41
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$\begingroup$ I Can slightly modify my question to remove this flaw. If K is any non-empty compact subset of E(n), I can substitute for the centroid of K, the so-called Tchebycheff center of K. This is the center of the smallest closed ball of E(n) that contains K as a subset.. Every non-empty compact subset of E(n) contains a unique Tchebycheff center. With this change, I would not be surprised if my question about continuity had a positive answer. At least I have not been able to find any counter-examples. $\endgroup$ Commented Aug 26, 2016 at 20:48