# Local Euclidean Spaces

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?
• What you called an "open disk in $\Bbb R^3$" isn't homeomorphic to an open ball in $\Bbb R^2$. – Lord Shark the Unknown Jan 29 '19 at 19:05
• How is that? (Maybe the name "open disk in $\mathbb{R}^3$" is wrong. I mean the set I wrote above.) – GottlobtFrege Jan 29 '19 at 19:50
• What you wrote is homeomorphic to a closed disc in $\Bbb R^2$, which is compact, and so not homeomorphic to an open disc. – Lord Shark the Unknown Jan 29 '19 at 19:56
• Ah, yes. That was a typo. Edited. – GottlobtFrege Jan 29 '19 at 20:03
• 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 "$\forall x\exists U\exists n \ldots$" – Neal Jan 29 '19 at 23:04

"Locally Euclidean of dimension $$n$$" implies that $$n$$ is invariant for all points of the space.