4
$\begingroup$

I'm looking for the metrisable compact $M$ so that there is no isometry $f:M \rightarrow \mathbb{R}^{n}$ with $im(f) = K$ $-$ compact subspace.

First, isometry preserves completeness, separability, and boundedness. Then, let's see how do the compact subspaces of $\mathbb{R}^{n}$ look like. First, they are complete (this works not only in the case of $\mathbb{R}^{n}$ but is the cases of metrisable space), then they are bounded (Heine-Borel states that compact set is bounded and closed). Moreover, any compact space in metric space is separable.

So, it seems reasonable to find a metric space that is compact but not bounded. But, since a metric space is compact iff it's closed and totally bounded (which implies boundedness) then it's impossible.

So, the basic approach did not help much. How to construct such space or to prove that it doesn't exist?

Any help would be much appreciated.

$\endgroup$
  • 1
    $\begingroup$ Consider a suitable compact subset of the normed space of all sequences in $\ell_2$. $\endgroup$ – Ángel Valencia Nov 30 '15 at 9:06
  • 4
    $\begingroup$ The Hilbert cube will work, though I don’t offhand know of any way to prove this that doesn’t use dimension theory. $\endgroup$ – Brian M. Scott Nov 30 '15 at 9:18
  • 2
    $\begingroup$ Please look here: mathoverflow.net/questions/12394/… $\endgroup$ – MotylaNogaTomkaMazura Nov 30 '15 at 9:45
  • 1
    $\begingroup$ If $M$ is only a metrisable space and not a metric space, then what does it mean for $f:M\to\mathbb R^n$ to be an isometry? $\endgroup$ – bof Nov 30 '15 at 10:59
  • $\begingroup$ @bof - you are correct. I've deleted my comment. $\endgroup$ – levap Nov 30 '15 at 22:05
5
$\begingroup$

An example of a metric space that does not isometrically embed into any $\mathbb{R}^n$ is the circle with the path metric. (One can simplify this further to a $4$-point space formed by $4$ equally spaced points on the circle). The reason is that in $\mathbb{R}^n$, midpoints are unique.

A metrizable space doesn't have a metric yet, so it's meaningless to talk about its isometric embedding. One can interpret the desired property as "there is no isometric embedding into $\mathbb{R}^n$ for any metric on the space that induces its topology". But this is equivalent to saying "there is no topological embedding into $\mathbb{R}^n$ [i.e., homeomorphism onto its image]", so again "isometry" is a red herring.

A compact metrizable space of finite topological dimension embeds into some $\mathbb{R}^n$ (the Menger-Nöbeling theorem). So an example has to be infinite-dimensional in the topological sense, which leads to the Hilbert cube, mentioned by Brian M. Scott in a comment. Suppose that $f:C\to\mathbb{R}^n$ is a topological embedding of the Hilbert cube $C$. The cube contains a homeomorphic image of $\mathbb{R}^{n+1}$, and therefore contains uncountably many disjoint homeomorphic images of $\mathbb{R}^n$. Each of these must be mapped to an open subset of $\mathbb{R}^n$, due to the invariance of domain. But there cannot be uncountably many disjoint open subsets of $\mathbb{R}^n$, since each contains a point with rational coordinates.

$\endgroup$
  • $\begingroup$ Wonderful answer, thanks. Firstly, i found the example from the link given by MotylaNogaTomkaMazura useful (it describes a square with the path metric), but your proof regarding Hilbert cube made the picture complete. $\endgroup$ – hyperkahler Jan 4 '16 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.