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Suppose there is a finite linearly independent system of vectors $\{x_i\}$ in a linear normed space (which itself may be infinitely dimensional). Is is true that for a small "change" of each vector the resulting system stays linearly independent? Formally, I mean $\exists\varepsilon \forall \{y_i\}:\|y_i-x_i\|<\varepsilon\forall i\Rightarrow \{y_i\}$ is linearly independent.

I think that's true, but don't know a correct way to prove it. I had an idea to set the $\varepsilon_1$ to half of the distance between $y_1$ and the linear hull of others - then it's obvious that moving $y_1$ by this distance doesn't keeps linear independence. Then repeat this $n-1$ times for other vectors. However, this doesn't prove the original statements, as here there is no single $\varepsilon$ for all $i$s.

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Yes, it's true.

Suppose $V$ is a vector space with norm $\|\cdot\|$. Let $S_{n-1}$ denote the unit $n-1$ sphere in $\mathbb{R}^n$: $$ S_{n-1} = \{a\in \mathbb{R}^n : \sum_{i=1}^n a_i^2 = 1 \}. $$

Suppose $x_1,\ldots,x_n\in V$ are linearly independent. Consider the mapping $f:S_{n-1}\to\mathbb{R}$ given by $$ f(a) = \left\| \sum_{i=1}^n a_i x_i\right\|. $$ Since $S_{n-1}$ is compact, $\|\cdot\|$ is continuous, and $x_1,\ldots,x_n$ are linearly independent, there is an $M>0$ such that $f(a) > M$ for all $a\in S_{n-1}$.

Choose $\epsilon>0$ such that $M - \epsilon \sqrt{n}>0$.

Now consider any $y_1,\ldots,y_n\in V$ with $\|y_i - x_i\|<\epsilon$. For any $a\in S_{n-1}$, \begin{eqnarray} \left\|\sum_i a_i y_i \right\| &=& \left\|\sum_i a_i x_i + a_i(y_i - x_i)\right\| \\ &\ge& \left\|\sum_i a_i x_i \right\| - \sum_i |a_i| \ \left\|y_i - x_i\right\| \\ &\ge& \left\|\sum_i a_i x_i \right\| - \epsilon \sum_i |a_i| \\ &\ge& M - \epsilon \sqrt{n} \\ &>& 0. \end{eqnarray} It follows $y_1,\ldots,y_n$ are linearly independent.

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