"Let $F(U)$ be the set of all sections on $U$. $F(U)$ always contains at least one element, namely the zero section: the function $s$ that maps every element $x$ of $U$ to the zero element of the vector space $\pi^{−1}(\{x\})$."
The vector bundle is a map $\pi: E \to M$ and $U \subset M$ is an open set. A section is a continuous map $s: U \to E$ such that $\pi \circ s = id$.
Consider the sphere $M = S^2$ with the total space $E$ the sphere with the vector space spanned by its normal at each point and $\pi$ the projection. If $U = B(x_0,\varepsilon)$ is an open ball on $S^2$ at $x_0$, why is $s: U \to E$, $x \mapsto x$ not a section? And is there an example of a vector bundle where we don't have $x \in \pi^{-1}(\{x\})$? Thanks for the help.
$x$ is an element of $M$ and not of $E$. I guess what you really mean is $s(x) = (x,0)$. I would see $E$ as a subspace of $S^2 \times \mathbb R^3$, namely $$E = \{(p,v): p\in S^2, v\perp T_pS^2\}$$ rather than a subspace of $\mathbb R^3$. $E$ can be embedded in $\mathbb R^3$, but this embedding won't be natural in any sense of the word (addition in fibers cannot be the same as addition in $\mathbb R^3$, if you know what I mean). But, on the other hand, we could identify $S^2$ with the zero section in $E$ and then your expression makes perfect sense. So it depends on what you mean by $s(x) = x$ – Sam Oct 22 '12 at 8:22