Can a mobius strip manifold exist in $3$ dimensions? Background and Question
I've always thought one can embed a manifold of dimensions $n$ into a higher dimension manifold of $R^{(n+1)}$. Is this true? If yes, what about the counter example below? If not, what is the minimum number of dimensions required?
Counter example
The Mobius band is formed as a quotient of $X = [0, 1] \times R$. A point $(0,y)$ is identified with the point $(1,-y)$. Thus:
$$Y = [ (x,y) , [(x,y),(1,-y)| x \in (0,1),y \in R]$$ 
Now we shall try to embed this in dimensions $R^3$. Imagine doing this first with a finite width mobius strip and then take width to infinity later.  However, this cannot be done when the mobius strip has infinite length. Due to the twist in $3$ dimensional space will have infinite width when extending and hence, it is guaranteed to self intersect and will not be a manifold anymore! (try visualizing it, I wasn't sure how to draw this)
 A: When we talk about whether manifold $M$ can be embedded into $\Bbb{R}^N$ for some $N$, we mean whether a homeomorphic (or diffeomorphic) copy of $M$ can be embedded into $\Bbb{R}^N$, rather than a specific parameterization of $M$. Since $\Bbb{R}$ is homeomorphic to the open interval $(0,1)$, your construction of the open Mobius band as a quotient of $[0,1]\times \Bbb{R}$ is homeomorphic to the analogous construction of the open Mobius band as a quotient of $[0,1]\times (0,1)$.  
The open Mobius band $M$ is embeddable into $\Bbb{R}^3$. One can see this by the standard cut and paste construction with a sheet of paper. Take a finite length sheet of paper, twist one time, and glue one end to the other. Voila, you have a Mobius band in $\Bbb{R}^3$, that looks something like this:

Examples of surface that cannot be embedded into $\Bbb{R}^3$ include the real projective plane and the Klein bottle.
As mentioned in the comments, the Whitney embedding theorem says that any smooth $n$-dimensional manifold $M$ can be smoothly embedded into $\Bbb{R}^{2n}$. 
