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 $\mathbb{R}^n$.
- 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.
- 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?
"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.