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

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 $$\mathbb{R}^3$$?”