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Let's say a metric space $X$ has property $P$ if $X$ is not isometric to any of its proper subsets. I'd like to know what this property is called in the literature and whether there's a nice characterization of spaces having this property.

What I know so far:

  1. Compactness implies property P, so compactness is a sufficient condition,
  2. The set $ \mathbb{Z}$ of integers ( with Euclidean metric) is not compact but satisfies $P$, so compactness is not necessary.

Any necessary or sufficient conditions ( weaker than compactness) are also appreciated.

Thanks in advance.

Q-rious
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  • In your title you speak of 'isomorphic' and in your question of 'isometric'. – drhab Jul 05 '14 at 11:32
  • Sorry, I've corrected the title. – Q-rious Jul 05 '14 at 11:33
  • You may find http://math.stackexchange.com/questions/579380/isometry-of-a-metric-space-with-proper-subset interesting. – Gerry Myerson Jul 05 '14 at 12:42
  • @GerryMyerson: Thanks for the pointer, but I'd already seen that. The example provided is indeed interesting, though I've been unable to use it to make any progress on my problem. – Q-rious Jul 05 '14 at 13:32
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    Generally, if an object is not (somehow) isomorphic to any of its proper subsets, it is called (somehow) co-Hopfian. This term came up in group theory, where a group is co-Hopfian if it not isomorphic to any of its proper subgroups. But it has since been applied to metric spaces: one speaks of quasisymmetrically co-Hopfian metric spaces or quasi-isometrically co-Hopfian ones. So, the property you describe would be isometrically co-Hopfian. However, it seems that this exact term is yet to be used by anyone, you can be first. –  Jul 05 '14 at 18:27
  • I think there can't be a condition other than a tautological one. Think of $\mathbb N$ with the standard metric: it is not isometrically co-Hopfian. But if we redefine the metric so that $d(42,43)=1.001$, then it becomes isometrically co-Hopfian. What mathematical gadget would detect such a difference? –  Jul 06 '14 at 21:28
  • @Thisismuchhealthier. : thanks for the comments. :) At first I was tempted to say that studying the symmetries of the space could detect that kind of difference, but as far as I know automorphisms are what we use to define the symmetries of an object. So I agree with you it's probably not possible to come up with any interesting characterisations of these spaces. – Q-rious Jul 07 '14 at 05:05
  • @Thisismuchhealthier.: Do you know if a Banach space is "co-Hopfian" if and only if it is finite-dimensional? I believe, an answer one way or another would be a fantastic answer to OP's question. (If you do not know, I will post it on Mathoverflow where Bill Johnson might know the answer.) – Moishe Kohan Jul 07 '14 at 05:18
  • @Thisismuchhealthier.: On a second thought, I see even symmetries won't help. Take $\mathbb{N}$ and ${1}$ for example. They have the trivial group as their automorphism group, yet one is isometrically co-Hopfian and the other is not. – Q-rious Jul 07 '14 at 05:31
  • @studiosus Gowers constructed an infinite-dimensional Banach space that is isomorphically co-Hopfian (which is a stronger property). –  Jul 07 '14 at 06:04

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