Suppose there is a vector bundle (smooth, with constant rank finite-dimensional fibres) over a (smooth, second-countable, Hausdorff, not necessarily connected) manifold $B$ of dimension $n$.
(a) Is it true, that the manifold B can be covered by a finite number of sets $U_1,\dots,U_N$ s.t. the vector bundle, restricted to $U_i$, is isomorphic to a trivial one for every $i=1,\dots,N$?
(b) If yes, can $N$ be taken to be $n+1$?
P.S. Some observations:
- It's proven in the book by Milnor and Stasheff, that every bundle allows a countable locally finite trivialization cover.
- Part (a) is obviously trivial for compact manifolds.
- It seems, that (b) is true if $B$ is an $n$-dimensional CW-complex. Proof: Denote with $B_k$ union of cells of dimension $0,\dots,k$. Prove by induction in $k$, that there are subsets $U_0,\dots,U_k$ of $B$, which cover $B_k$, s.t. the restriction of the bundle to each of them is trivializable. Start with case $k=0$: construct contractible neighbourhoods of each 0-cell, which do not intersect with each other. Take there union. Now to prove the claim for the next value of $k$ it is enough to construct a contractible non-intersecting neighbourhoods of each $X_\alpha=e_\alpha\setminus (U_0\cup \dots\cup U_{k-1})$. Call the desired neighbourhood with $V_\alpha$. First, note, that $X_\alpha$ is closed in $e_\alpha^k$ and doesn't intersect with its boundary $\partial e_\alpha^k$, so we can find its neighbourhood in $e_\alpha^k$, which doesn't intersect with $\partial e_\alpha^k$. This set is our candidate for $V_\alpha\cap B_k$. Extending it to an open set in $B_{k+1}$ can be done cell by cell: interpreting $e^{k+1}_\beta$ as a unit ball with the center in the origin, we can write every its point as $r\theta$, where $\theta\in S^k$ and $r\in [0,1]$. We include $r\theta$ in $V_\alpha\cap B_{k+1}$ iff $\theta$ is already there and $r>0.99$. Repeating this procedure we extend it to $B$.
Edit: Open sets $U_1,\dots, U_N$ are assumed to be open. I don't ask them to be connected.
Best Answer
The answer (to both questions (a) and (b)) is YES (assuming $B$ is a smooth manifold). A proof can be found on Walschap's book "Metric Structures in Differential geometry", p. 77, Lemma 7.1.
For the OP's convenience, here's a sketch of the proof. Choose an open cover of $B$ such that your vector bundle is trivial over each element. From general results in topology, this (and in fact any) cover of an $n$-dim manifold $B$ admits a refinement $\{ V_\alpha\}_{\alpha\in A}$ such that any point in $B$ belong to at most $n+1$ $V_\alpha$'s. Let $\{\phi_\alpha\}$ be a partition of unity subordinate to this cover and denote by $A_i$ the collection of subsets of $A$ with $i+1$ elements. Given $a=\{\alpha_0,\dots,\alpha_i\}\in A_i$, denote by $W_a$ the set consisting of those $b\in B$ such that $\phi_\alpha(b)\lt\phi_{\alpha_0}(b),\dots,\phi_{\alpha_i}(b)$ for all $\alpha\neq\alpha_0,\dots,\alpha_i$. Then the collection of $n+1$ open subsets $U_i:=\cup_{a\in A_i} W_a$ covers $B$ and is such that your bundle restricted to each $U_i$ is trivial.