A question from differential geometry I read a statement in a paper by J. Manton says that: 
Consider a circle. It is a manifold. It locally looks like $\mathbb{R}$ whereas globally does not. Mathematically, the circle cannot look like an open subset of $\mathbb{R}$; in particular, there does not exist a homemorphism mapping an open subset of $\mathbb{R}$ onto the circle. 
My point is about the terminology homemorphism in above explanation. As I checked the wiki for its definition I noticed a point says: the homeomorphism is a continuous stretching and bending of the object into a new shape. 
Since I am new in this area I would like to check whether my understanding is correct. So, I have questions:


*

*Does it mean, in this example, an open subset of $\mathbb{R}$ (a straight line without ending points cannot be stretched and bent to make a circle which is a closed set)?

*If we go through the mathematical definition of homemorphism mapping (which must be one-to-one+onto+continuous+$f^{-1}$ is continuous), there is no such $f$ between a line segment without ending points and a circle because of violating the conditions of being onto and continuous. Is it right? 

*In the above statement, it is stated that does not exist a homemorphism mapping from an open subset of $\mathbb{R}$ onto the circle. Why here it is stated that "from an open subset of $\mathbb{R}$"? 
Any comment/explanation/further perspective is appreciated. 
 A: 1- Yes that's exactly it. The point is that you're missing the endpoints; and thus you lack some sort of closure in the open set. The exact thing that you're missing is compactness, if you know what this means (there are other ways to prove that such a homeomorphism can't exist)
2- No, it depends on the mapping. There are many onto continuous maps $(0,1)\to S^1$ ($S^1$ is the circle here). The problem for those is that they won't be injective. 
As noted in the comments, adding one endpoint ( considering $[0,1)$ for instance) is not enough, though with this you can get a bijective continuous mapping $[0,1)\to S^1$. For this, the problem is that the reciprocal ($f^{-1}$) will not be continuous.
3- I don't understand the issue here : there is no way to homeomorphically send an open subset of $\mathbb{R}$ onto $S^1$ (and it's the same for $\mathbb{R}^n$, more generally - but for $n\geq 2$, the intuition should be better). Again, the reason is that open subsets of $\mathbb{R}$ don't have the same topological properties as $S^1$. 
A very important point here is the following heuristic : 

Homeomorphic spaces have the same topological properties

The problem with this heuristic is that it's not easy to define "topological properties", knowing what they are comes from practice; and at some point you'll be able to know it on the spot and say "no these spaces aren't homeomorphic because this one has property P and the other one does not" and in your mind, you will know that it wouldn't take more than a line or two to prove this. 
For instance "being compact" is a topological property, while "containing $1$" is not. Here we may use (for instance) the property that $S^1$ is compact, whereas any open subset of $\mathbb{R}$ is not to show that there can't be a homeomorphism between them. 
Edit: For 3 you were actually asking about the mention of "open". In fact, it turns out that this is true for any subset of $\mathbb{R}$ (proof below); but the mention of open is what interested the author. 
Indeed, manifolds are defined by how they look like locally, and the idea of locality is precisely what open sets are meant to represent. If $x$ is a point of the open set $U$ it means that all points close enough to $x$ are also in $U$.
Being a manifold means (in particular) that if you take any point $x$, there is a region surrounding it, containing all the points close enough to it, that looks like $\mathbb{R}^n$. In particular, the subsets of $\mathbb{R}^n$ that really look like $\mathbb{R}^n$ are its open subsets.
To see why these are the ones we choose, consider other subsets: do we really want to say that the Cantor set "looks like" $\mathbb{R}$ ? 
Actually there is a (non trivial) theorem that states that any subset of $\mathbb{R}^n$ that looks like $\mathbb{R}^n$ must be open. That's an a posteriori justification of the choice of open sets, but the initial choice was because we thought that these have the best properties, and they are what we want to generalize. 
For instance, differentiation only makes sense if you look like an open subset of $\mathbb{R}^n$ (with the caveat of manifolds with boundary, but that's mostly true)
Proof that a subset of $\mathbb{R}$ can't be homeomorphic to $S^1$: 
If $F$ is such a subset, then $F$ is connected because $S^1$ is, so it's an interval. It's also compact, because $S^1$ is, so it's of the form $[a,b]$. But now we know that these aren't homeomorphic to $S^1$ : if $f$ is a homeomorphism $[a,b]\to S^1$, then the restriction of $f$ is a homeomorphism from $[a,b]\setminus\{x\}\to S^1\setminus\{f(x)\}$ for $x$ the middle point of $[a,b]$. But the space on the left is now disconnected, while the space on the right isn't (there are other ways to see it: for instance $Homeo(S^1)$ acts transitively on $S^1$, while $Homeo([a,b])$ doesn't on $[a,b]$)
