TITLE: Regarding linear independence on a normed-linear space given a condition
QUESTION [12 upvotes]: Let $(X,\|\cdot\|)$ be a normed linear space and $x_{1}, x_{2}, \ldots, x_{n}$ be $n$ linearly independent vectors in $X$. Show that there exists $\epsilon > 0$ such that if $y_{1}, y_{2}, \cdots, y_{n} \in X$ with $\|y_{i}\| < \epsilon$, $i = 1,2,\ldots,n$, then $x_{1} + y_{1}, x_{2} + y_{2},\ldots,x_{n} + y_{n}$ are also linearly independent vectors in $X$.

I've been thinking over this for a few days and I'm not really getting anywhere.
I've tried to start with the definition of linear independence for the $x$-vectors and then working towards the linear independence of $x+y$ given the condition on $y$, but I'm not getting anywhere.
May someone offer a hint or a solution?

REPLY [2 votes]: Linear independence is an open condition, which is most easily seen by observing that $x_1, \dotsc, x_n$ are independent iff $x_1\wedge \dotsb\wedge x_n \neq 0$. So for any neighborhood $U$ of $x_1\wedge \dotsb\wedge x_n$ in $\Lambda^n X$, there is a neighborhood $V$ in $X\times\dotsb\times X$ mapping into $U$ under the wedge map. If we choose $U$ away from the origin, then any $\epsilon$ keeping in the neighborhood $V$ will do the job.