3

Is there a well-known example (preferably in dimensions 2 or higher) of two homeomorphic spaces: (1) a metric space with the Euclidean metric and (2) a metric space that is not Euclidean?

user156619
  • 285
  • 1
  • 7
  • So you want an example of a metric space that is homeomorphic to $\mathbb R^n$, but with a different metric? – Benjamin Sep 08 '14 at 17:43
  • Not necessarily, but that would do. – user156619 Sep 08 '14 at 17:45
  • What is required to make a metric "non-Euclidean" ? Are you thinking about Euclidean vs. non-Euclidean geometry? – hardmath Sep 08 '14 at 17:46
  • Yes. A non-Euclidean metric would be just a distance function that is not the Euclidean metric $d_E$. $d_E$ is obviously just the metric between points $x$ and $y$ which, if $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$, is the length of the line segment between them:

    $$d_E(x,y)=\left(\sum_{i=1}^n \left(x_i-y_i\right)^2\right)^{1/2}.$$

     – user156619 Sep 08 '14 at 17:51
  • 1
    Example: the metric $d(x,y) = \sum_{i=1}^n |x_i - y_i|$. – Lee Mosher Sep 08 '14 at 17:53
  • 3
    The metric $$d_p(x,y)=\left(\sum_{i=1}^n|x_i-y_i|^p\right)^{1/p}$$ induces the same topology on $\Bbb R^n$ for any $p>0$. –  Sep 08 '14 at 17:55

4 Answers4

4

Consider the metric spaces $\mathbb R^2$ with the Euclidean metric $d$ and $\mathbb R^2$ with the "max metric" $d'$ defined by $$d'(x,y) = \max\{|x_1-y_1|,|x_2-y_2|\}.$$ The open balls in the Euclidean metric space are open discs, whereas the open "balls" of the max metric space are open squares.

To show that these metric spaces are homeomorphic, we need to show that every open disc composed of open squares and vice versa. This is straightforward.

I am fairly certain that this can be generalized to $\mathbb R^n$ as well.

Benjamin
  • 1,044
  • @JasonDeVito No, you are correct. Thanks for catching that, I've edited the answer to correct that! – Benjamin Sep 09 '14 at 12:22
4

Even though the question has been already answered, let me point out two non-trivial generalisations to the infinite-dimensional case:

Theorem (M. Kadets). All infinite-dimensional separable Banach spaces are homeomorphic.

There is even a more surprising result:

Theorem (Cz. Bessaga). Every infinite-dimensional Hilbert space is diffeomorphic to its unit sphere.

References:

Cz. Bessaga, Every infinite-dimensional Hilbert space is diffeomorphic with its unit sphere. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys., 14 (1966), 27–31.

M. I. Kadets, Proof of the topological equivalence of all separable infinite-dimensional Banach spaces, Functional Analysis and Its Applications, 1 (1967), 53–62.

Community
  • 1
Tomasz Kania
  • 16,361
3

To generalize on the fine examples given by Lee Mosher and Mike Miller, a finite dimensional vector space $\mathbb{R}^n$ has the same topology under any vector space norm. The Euclidean metric is the Euclidean norm. Lee's example is the 1-norm, and Mike's is the $p$-norm, for $1\le p \lt \infty$. We could add to these the $\infty$-norm (max-norm) that is described in Benjamin's Answer.

Community
  • 1
hardmath
  • 37,015
1

Stereographic projection is a homeomorphism $S^n∖{p}→R^n$.Look at here.

A.B.
  • 92
  • Perhaps you should describe the new example as "punctured" unit 2-sphere, since a point must be removed to get the homeomorphism. – hardmath Sep 08 '14 at 18:09
  • Yes, Indeed it is locally Euclidean. – A.B. Sep 08 '14 at 18:15
  • I'm not your downvoter (so I can't remove it), but this is an interesting example because the metric for the unit sphere is bounded (and for the plane, unbounded), so it differs from the family of equivalent norms described in my CW Answer. – hardmath Sep 08 '14 at 18:19
  • Manifolds are good examples. As you know a manifold is a topological space that is locally Euclidean. – A.B. Sep 08 '14 at 18:22