Why a set is open in proof that manifold admit a finite cover where bundle is trivial This is the lemma 7.1 of Metric structures in differential geometry of Walschap.

Lemma 7.1 : Let $\xi$ denote a vector bundle over an $n$-dimensional manifold
$B$. Then $B$ can be covered by $n+1$ sets $U_o,\dots, U_n$ , where each restriction $\xi|_{U_i}$ is trivial.


proof : Choose an open cover of $B$ such that $\xi$ is trivial over each element.
It is a well-known theorem in topology that this (and in fact any) cover of an
$n$-dimensional manifold $B$ admits a refinement $\{ A_\alpha \}_{\alpha \in A}$ with the property that any point in $B$ belongs 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 all subsets of $A$ with $i + 1$ elements.
Given $a = \{ \alpha_0, \dots, \alpha_i \} \in A_i$, denote $W_a$ the set consisting of those $b\in B$ such that $\phi_\alpha(b) < \phi_{\alpha_0}(b), \dots, \phi_{\alpha_i}(b)$ for all $\alpha \neq \alpha_0, \dots, \alpha_i$.

To clarify, we may rewrite $W_a$ as
$$ W_a = \{ b \in B \mid \forall \alpha \in a, \forall \beta \in A \setminus a, \phi_\beta(b) < \phi_\alpha(b) \}. $$
Then the author conclude that $W_a$ is open for all $a$ but I don't understand why.
When I try to write the definition of $W_a$ with union and intersection of simpler opens I arrive to a infinite intersection which is not obviously open.
 A: The following proof is incomplete, containing an extra unwaranted assumption.   But maybe someone can fill in the gap.
For ease of notation, I'll use $\phi$ to denote the minimum of $\phi_{\alpha_1},... \phi_{\alpha_i}$.  So, $b\in W_{\alpha}$ iff $\phi_{\alpha}(b) < \phi(b)$ for all $\alpha\neq \alpha_1,...,\alpha_i$.
Suppose $b\in W_\alpha$.  We need to find an open neighborhood $U$ of $b$ with $U\subseteq W_\alpha$.  To that end, let $V_{b_1},....,V_{b_{n+1}}$ denote the $n+1$ elements of $\{V_\alpha\}_{\alpha \in A}$ which contain $b$.
Extra assumption:  The sets $V_1,...,V_{n+1}$ are distinct.
Set $V = V_{b_1}\cap ... \cap V_{b_{n+1}}$.  Then $V$ is an open set, and $b\in V$, so it's non-empty.  Note that every element of $V$ is contained in $n+1$ of the $V_\alpha$ (namely, $V_{b_1},..., V_{b_{n+1}}$, so these are all the sets in $\{V_{\alpha}\}$ which contain any point in $V$.
Since $b\in W_\alpha$, $\phi_{b_j}(b) <  \phi(b)$ for all $j = 1,...,n+1$.  For each $j=1,....,n+1$, let $U_j\subseteq V_{b_j}$ be the open set $\{c\in V_{b_j}: \phi_{b_j}(c) < \phi(c)\}$.
Consider the set $U = U_1\cap...\cap U_{n+1}\subseteq V$.  This is an intersection of finitely many open sets, so is open.  Also note that $b\in U_j$ for all $j$, so $U$ is a neighborhood of $b$.  Moreover, since $U\subseteq V$, any element $c\in U$ is in $V_{b_1},...,V_{b_{n+1}}$, but is in no other $V_\alpha$.
We claim that $U\subseteq W_\alpha$.  To see this, we must show for any $c\in U$ and for any $\alpha\notin \{\alpha_1,...,\alpha_i\}$, that $\phi_{\alpha}(c) < \phi(b)$.  We break into cases depending on whether $\alpha \in \{b_1,....,b_{n+1}\}$ or not.  If $\alpha$ is in this set, then by definition of $U_j$, we know that $0\leq \phi_{b_j}(c) < \phi(c)$.  If $\alpha$ is not in this set, then $c\notin V_\alpha$, so $\phi_{\alpha}(c) = 0 < \phi(c)$.  This completes the proof.
