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I'm confused about some properties of local Euclidean spaces. I'm working with the definition:

A topological space, $X$, is locally Euclidean of dimension $n$, iff $\forall x\in X. \exists U$ a neighborhood of $x$ such that $U$ is homeomorphic to an open ball of $\mathbb{R}^n$.

  1. If we're looking at an open disk in $\mathbb{R}^3$, i.e. $\{(x,y,z)\in \mathbb{R}^3|x^2+y^2< 1, z=0\}$, then this has to be homeomorphic to an open ball in $\mathbb{R}^2$, right? How about a non-connected union of say a circle ($S^1$) and a disk as before in $\mathbb{R}^3$. Is this locally Euclidean but just for varying $n$ in this case? Which means my definition doesn't hold for the space $X$ but for each connected part.
  2. Is locally euclidean a property of open sets? Or are we looking at the subset topology in each case? If we have a closed rectangle in $\mathbb{R}^2$, how could we map a neighborhood to an open set in $\mathbb{R}^2$. Or am I missing something here?
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  • $\begingroup$ What you called an "open disk in $\Bbb R^3$" isn't homeomorphic to an open ball in $\Bbb R^2$. $\endgroup$ – Lord Shark the Unknown Jan 29 at 19:05
  • $\begingroup$ How is that? (Maybe the name "open disk in $\mathbb{R}^3$" is wrong. I mean the set I wrote above.) $\endgroup$ – GottlobtFrege Jan 29 at 19:50
  • $\begingroup$ What you wrote is homeomorphic to a closed disc in $\Bbb R^2$, which is compact, and so not homeomorphic to an open disc. $\endgroup$ – Lord Shark the Unknown Jan 29 at 19:56
  • $\begingroup$ Ah, yes. That was a typo. Edited. $\endgroup$ – GottlobtFrege Jan 29 at 20:03
  • $\begingroup$ The trick here is that "$n$" is specified before and independently of $x$, so you should read this as "$\exists n\forall x \exists U \ldots$" instead of "$\forall x\exists U\exists n \ldots$" $\endgroup$ – Neal Jan 29 at 23:04
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"Locally Euclidean of dimension $n$" implies that $n$ is invariant for all points of the space.

The closed rectangle is not locally Euclidean in that definition because the boundary points do not have a neighbourhood (in the subspace topology always) that is homeomorphic to an open set of a Euclidean space. That is why the notion of "manifolds with boundary" was introduced. Open subsets of Euclidean spaces are trivially locally Euclidean of the same dimension.

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  • $\begingroup$ Therefore the set I described in point 1 (non-connected union of line and disk) is not locally euclidean? But the connected parts themselves are? $\endgroup$ – GottlobtFrege Jan 30 at 11:09
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    $\begingroup$ @GottlobtFrege indeed. Not locally Euclidean of 1 or 2 dimensions that is. $\endgroup$ – Henno Brandsma Jan 30 at 12:33

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