When is an open subset of an orientable manifold an orientable submanifold?

Question

I am trying to complete problem $4$ from the introductory chapter of Riemannian Geometry by do Carmo. The question asks us to show that the projective plane $P^2(\mathbb{R})$ is non-orientable, first by proving any open subset of an orientable manifold is an orientable submanifold.

do Carmo's definition for a manifold $M$ to be orientable is if it admits a differentiable structure $\left\{(U_\alpha,\mathbf{x_\alpha}) \right\}_{\alpha \in \Gamma}$ such that for every pair $\alpha,\beta$ with $\mathbf{x_\alpha}(U_\alpha)\cap\mathbf{x_\beta}(U_\beta)$ nonempty the differential of the change of coordinates, $d(\mathbf{x_\beta}^{-1} \circ \mathbf{x_\alpha})$, has positive determinant.

My approach to this was to say let open $V \subset M$ be given. Then $\left\{(U_\alpha\cap \mathbf{x_\alpha}^{-1}(V),\mathbf{x_\alpha}) \right\}_{\alpha \in \Gamma}$ is a differentiable structure on $V$ that satisfies the definition.

My confusion stems from the fact that an open Möbius band can embedded in $\mathbb{R}^3$, for example by the embedding $f:U\to \mathbb{R}^3$

$$f(x,y) = \left(\left(1+\frac{y}{2}\cos\frac{x}{2}\right)\cos{x},\;\left(1+\frac{y}{2}\cos\frac{x}{2}\right)\sin{x},\;\frac{y}{2}\sin\frac{x}{2}\right)$$

Where $U = \left\{(x,y)\in \mathbb{R}^2 : 0 \leq x < 2\pi,\;-1 < y < 1\right\}$ (taken from Wikipedia).

So I guess my confusion comes down to "why isn't the open Möbius band a submanifold of $\mathbb{R}^3$?"

Answer 1

A möbius strip is not an open subset of $\mathbb{R}^3$.

Answer 2

This is not in relation to your confusion. I just want to give you a characterisation of orientability that is often much easier wo work with:

An $n$ dimensional smooth manifold $M$ is orientable if and only if there ex. a non-vanishing differential of degree $n$ on $M$.

With this you should be able to prove that open subsets of orientable manifolds are orientable manifolds themselves.