The disk bundle of a vector bundle $p: E \to B$ is defined as :
$D(E)= \{e \in E : \|e\| \leq 1\} $ (Assuming there a norm on the space of course!) and the sphere bundle is defined to be
$S(E) =\{e \in E : \|e\|=1\}$.
I was wondering if someone could explain the following in a little more detail:
When the rank of $E$ is $0$, hence, $E=B\times ℝ^0 \cong B$,
$D(E) = E \cong B$ and $S(E) = \emptyset$.
I don't see how these follow from $E \cong B$.
Or alternatively if someone had a good reference for this?
To see what's going on, you just need to look at the fiber, which is a copy of $\mathbb{R}^0 = \{0\}$. In $\mathbb{R}^0$, there is only one vector with length less than or equal to one, namely $0$, so the closed unit disc of $\mathbb{R}^0$ is $\{0\} = \mathbb{R}^0$. On the other hand, there is no vector of length one, so the unit sphere in $\mathbb{R}^0$ is empty.