Let $X_1,\ldots,X_k$ be lineary independent local sections of a vector bunlde $V$, $rank(V)>k$, defined on some open set $U$. Can I construct a local section $X_{k+1}$ on a (maybe smaller) open subset of $U$, such that $(X_1,\ldots,X_{k+1})$ are lineary independent on the smaller subset?
edit: I know that answer, however I would be interested in a more "elementary" argument.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Let $N$ denote the rank of $E$ and $n$ the dimension of the base. By working in a trivializing neighborhood for $E$, and identifying local sections of $E$ with $\Reals^{N}$-valued functions without using separate notation, the question boils down to:
The answer is clearly "yes". Consider the constant function $X_{k+1} = \Basis_{k+1}$. The orthogonal projection of $X_{k+1}$ into the space spanned by $(X_{j})_{j=1}^{k}$, and its squared magnitude, are given by $$ X_{k+1}' := \sum_{i=1}^{k} \langle X_{k+1}, X_{i}\rangle\, X_{i},\qquad \|X_{k+1}'\|^{2} = \sum_{i=1}^{k} \langle X_{k+1}, X_{i}\rangle^{2}. $$
By elementary linear algebra, $(X_{i})_{i=1}^{k+1}$ is linearly independent if and only if $X_{k+1} \not\in \operatorname{Span}(X_{i})_{i=1}^{k}$, if and only if $X_{k+1} \neq X_{k+1}'$. But the squared magnitude of $X_{k+1}'$ is $0$ at the origin, and is continuous as a real-valued function on $U$, so there exists an open set $U'$ containing the origin such that $\|X_{k+1}'\|^{2} < 1 = \|X_{k+1}\|^{2}$ on $U'$.