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I have a set $E \subset X$ within a metric space ($X, d$). I want to prove that it is isomorphic to $\mathbb{R}^{n \times n}$, in the sense that there exists a continuous bijection between the two. Because $E$ is a fairly complicated set, it would be a huge pain to actually find an exact bijection, so instead I hope to identify a sufficient suite of conditions that I can test $E$ for that will suffice to show that the two are isomorphic.

Is there some sort of known method for doing this? Or will I have to find the exact function?

Thanks.

Austin Mohr
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GMB
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  • What are $E$ and $X$? – Jonas Meyer Jan 21 '13 at 03:54
  • It need not be the case in general, i.e. $\Bbb Q$ is not homeomorphic/isomorphic to $\mathbb{R}$. – Clayton Jan 21 '13 at 03:55
  • @JonasMeyer $X$ is $\mathbb{R}^{n \times n} \times \mathbb{R}^n$ and $E$ is a strange set within that that would take a fair amount of effort to explain. I hope to find a test suite of conditions that applies to any $E$ we might choose. – GMB Jan 21 '13 at 03:59
  • @Clayton I don't follow: I don't believe there is a continuous bijection between $\mathbb{Q}$ and $\mathbb{R}$. – GMB Jan 21 '13 at 04:00
  • @LevDub: There isn't; that is what I was getting at. If $E\subseteq\mathbb{R}$ and $E=\mathbb{Q}$, then there doesn't exist what you're asking for. You'll be better off putting in the question what it is you have and want exactly. – Clayton Jan 21 '13 at 04:03
  • @Clayton Ah, I understand now. My question is "does there exist a condition (or set of conditions) such that $E$ and $\mathbb{R}^{n \times n}$ are isomorphic iff $E$ satisfies the condition(s)?" So it would defeat the purpose to specify $E$. – GMB Jan 21 '13 at 04:10
  • Are you sure that you only want to show that there is a continuous bijection between the two? This does not in general imply that they are homeomorphic (or isometrically isomorphic, for that matter). – bradhd Jan 21 '13 at 04:12
  • @Brad that's interesting, I didn't know that. What's an example of a continuous bijection with a non-continuous inverse? – GMB Jan 21 '13 at 04:16
  • Would a better title for your post be "what properties characterize $\Bbb R^n$ among metric spaces"? – MJD Jan 21 '13 at 04:17
  • @LevDub The map $f:[0,2\pi)\to S^1\subset\mathbb{C}$ defined by $f(t) = e^{it}$ is continuous and bijective, but its inverse is not continuous at the point $1$. – bradhd Jan 21 '13 at 04:18
  • Note that your notion of "isomorphism" is not the same as the notion of isomorphism in the category of metric spaces, which is customary called "homeomorphism". – Matemáticos Chibchas Jan 21 '13 at 04:36
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    A continuous bijection of manifolds is an homeomorpism, as a consequence of the theorem of Invariance of Domain. It is important here that there be no boundaries (although this extends to manifolds with compact boundaries, iirc) – Mariano Suárez-Álvarez Jan 21 '13 at 04:44

2 Answers2

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The necessary and sufficient set of conditions for $E$ to be homeomorphic to $\mathbb R^{m}$ (in your situation $m=n^2$):

More details here.

Community
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Another less general, but easier solution is that if $E \subset \mathbb{R}^n$ is a star-shaped (or less generally convex) open set, then $E$ is homeomorphic to $\mathbb{R}^n$.

Note that the converse is not true. The idea of the proof is to expand the shape radially in such a way that the map stays continuous. More details can be found here.

Teo Miklethun
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