For $n$ continuous linearly independent functions on an interval $I$, are there subintervals however small that preserve the independence? This is just something that I thought might be true, purely by intuition. However, when I tried proving it I got stuck and now I actually doubt whether it's right or not. This is my problem:
Let $f_1, f_2, \dots f_n$ be continuous functions on an interval $I$ and linearly independent on it. Then, is it true that $\forall \epsilon > 0,\hspace{0.1cm} \exists (x \in I, 0 < \eta \leq \epsilon)$ such that the functions are linerly independent on $(x-\eta, x+\eta)$?
I tried approaching the problem by contradiction:
Suppose that for some $\epsilon > 0,\hspace{0.1cm} \forall (x\in I, 0 < \eta \leq \epsilon)$, the functions are linearly dependent on $(x - \eta, x + \eta)$. So I need to prove that the functions are linearly dependent on $I$. It seems that dividing $I$ into a finite number of intervals of length less than $2\epsilon$ doesn't work. For example, the functions
$$ f_1(x) = \begin{cases} -x+2, & x \in (0,1) \\ 1, & x\in [1,2] \\ 2x-3, & x \in (2,2.5) \\ -2x+7, & x\in [2.5, 3) \\ 1, & x\in [3, 4) \\ x-3, & x \in (4,5) \end{cases} \\ f_2(x) = \begin{cases} 4x-2, & x \in (0,1) \\ 2, & x \in [1,2] \\ 5x-8, & x \in (2, 2.5) \\ -5x+17, & x \in [2.5, 3) \\ 2, & x \in [3,4] \\ -4x+18, & x \in (4,5) \end{cases} \\ f_3(x) = \begin{cases} 2x+2, & x \in (0,1) \\ 4, & x \in [1,2] \\ 6x-8, & x \in (2, 2.5) \\ -6x+22, & x \in [2.5, 3) \\ 4, & x \in (3,4) \\ -2x+12, & x \in [4,5) \end{cases} $$ are linearly dependent on $(0, 2), (1, 4), (3, 5)$, but not on $(0, 5)$. An example of dependency is $-8f_1 + 2f_2 + f_3 \equiv 0$ on $(1, 4)$ and $2f_1 + f_2 - f_3 \equiv 0$ on $(0, 2) \cup (3, 5)$. To prove the independence, note that $af_1 + bf_2 + cf_3 \equiv 0$ on $I \implies a + 2b + 4c = 0 \wedge -a + 4b + 2c = 0 \wedge 2a + 5b + 6c = 0$, but since $\left|\begin{matrix} 1 & 2 & 4 \\ -1 & 4 & 2 \\ 2 & 5 & 6 \end{matrix}\right| \ne 0$, the system only has $(0,0,0)$ as a solution, so the functions are linearly independent.
 A: Your claim isn't true for all such cases if $n > 1$.
If you want to test claims about continuous functions, one of the easiest non-constant examples are triangle functions, or "hat functions", for how their graph looks like.
Now if you partition your interval $I$ into $2n-1$ equal subintervals, and define the function $f_i$ as a hat function over the $(2i-1)$st subinterval (say having function value $1$ at the top of the hat), then those functions are lineraly independent over the whole I. That's because if you assume to have any linear combination where
$$\forall x \in I: \sum_{i=1}a_if_i(x)=0$$
then you can choose any index $i$ and choose $x_i$ from inside (not the boundary of) the $(2i-1)$st subinterval. By definition we have $\forall k \neq i: f_k(x_i)=0$, so the above sum, when evaluated at $x=x_i$ just becomes
$$a_if_i(x_i)=0.$$
But since we know that $f_i(x_i) > 0$ ($x_i$ is from the inside of the subinterval), that necessarily means $a_i=0$, and this can be done for any $i$.
OTOH, if you choose $\epsilon < \frac12\frac{|I|}{2n-1}$, ($|I|$ denotes the length of the interval), then any interval $(x-\eta, x+\eta)$ will have length less than $\frac{|I|}{2n-1}$, the length of the subintervals we initially divided $I$ into. Note that since the hat functions are defined over every other subinterval, any interval of length less then $\frac{|I|}{2n-1}$ can only overlap with at most one such interval (with a hat function over it).
Since $n>1$, there is at least one hat function (say $f_i$) whose function values are all zero inside the interval $(x-\eta, x+\eta)$, so the set of functions is not linearly independent over that interval, as
$$\forall x \in (x-\eta, x+eta): 0f_1(x)+0f_2(x)\ldots+0f_{i-1}(x) + 1f_i(x)+ 0f_{i+1}(x) + \ldots + 0f_n(x) = 0.$$
