I am studying spivak's Calculus on Manifolds by myself,now a sophomore college student majoring in Mathematics. In Spivak - Calculus on Manifolds, Comments and Errata,the author said "Page 118) In definition of orientation the W must be connected, otherwise no manifold would be orientable." Why is it necessary to require the open subset W in $\mathbf{R^n}$ must be connected ? I need some details to explain if the condition of "connected" were canceled,no manifold would be orientable.
(page 118)
It is often necessary to choose an orientation $\mu_x$ for each tangent space $M_x$ of a manifold $M$. Such choices are called consistent provided that for every coordinate system $f : W \to \mathbf R^n$ and $a, b \in W$ the relation $$ [f_{\ast}((e_1)_a), \ldots, f_{\ast}((e_k))_a] = \mu_{f(a)} $$ holds if and only if $$ [f_{\ast}((e_1)_b), \ldots, f_{\ast}((e_k)_b)] = \mu_{f(b)}.$$ Suppose orirentations $\mu_{x}$ have been chosen consistently.If $f : W \to \mathbb R^n$ is a coordinate system such that $$ [f_{\ast}((e_1)_a), \ldots, f_{\ast}((e_k))_a] = \mu_{f(a)} $$ for one, and hence for every $a\in W,$ then $f$ is called orientation-preserving. If $f$ is not orientation-preserving and $T :\mathbb R^k \to \mathbb R^n$ is a linear transformation with $\det T=-1,$ then $f\circ T$ is orientation-preserving. Therefore there is an orientation-preserving coordinate system around each point.
For the notation. With $v_p$ he denotes the tangent vector $v$ at the point $p$, for a function $f : A \to \mathbf R^m$ with $A \subseteq \mathbf R^n$ differentiable on $A$ we have $f_{\ast}(v_p) := (Df(p)(v))_{f(p)}$ and for a basis $v_1, \ldots, v_n$ he denotes by $[v_1, \ldots, v_n]$ the orientation of $v_1, \ldots, v_n$. Also a coordinate system is defined as:
(page 111)
$\textbf{5-2 Theorem}$
A subset $M$ of $\mathbf R^n$ is a $k$-dimensional manifold if and only if for each point $x \in M$ the following ''coordinate condition'' is satisfied:
(C) There is an open set $U$ containting $x$, an open set $W \subseteq \mathbf R^k$, and a $1-1$ differentiable function $f : W \to \mathbf R^n$ such that
(1) $f(W) = M \cap U$,
(2) $f'(y)$ has rank $k$ for each $y \in W$,
(3) $f^{-1} : f(W) \to W$ is continuous.
Such a function $f$ is called a coordinate system around $x$.
Here's a silly little example: on the manifold $\mathbb{R}$, consider the following coordinate system: $$W=(0,1)\cup (2,3)$$ $$U=(0,1)\cup (2,3)$$ $$f(y)=\begin{cases}1-y&0<y<1\\y&2<y<3\end{cases}$$ If we try to match this with the usual coordinates on $\mathbb{R}$, the orientation would be inconsistent; it flips between the two halves of $W$ using $f$, while it doesn't flip in the usual coordinates.
And that's the problem; every manifold has examples like this, where we can take two disconnected pieces out of a larger coordinate system and flip one of them around for an opposite orientation. If we allowed disconnected pieces, every orientation would be inconsistent in this way.