Prove that the set ${\{} \frac{1}{x-c}{\}}_{\displaystyle\ c \in \mathbb{R}\setminus[0,1]}$ is linearly independent. Question: Let $V$ be the vector space of all real valued functions defined on the unit interval $[0,1]$.
Show that the set $\displaystyle\ \bigg{\{} \frac{1}{x-c}\bigg{\}}_{\ c \in \mathbb{R}\setminus[0,1]}$ is linearly independent.
Attempt:  Assume towards a contradiction that the set is linearly dependent. So, $\exists$ a finite subset $\displaystyle\bigg{\{}\frac{1}{x-c_i}\bigg{\}}_{i=1}^n$ which is linearly dependent.
Therefore, for some $(d_1,d_2,...,d_n)\neq (0,0,..,0)$ we must have
$\displaystyle f(x)=\sum_{i=1}^n\frac{d_i}{x-c_i}=\frac{d_1(x-c_2)...(x-c_n)+d_2(x-c_1)(x-c_3)...(x-c_n)+...+d_n(x-c_1)...(x-c_{n-1})}{(x-c_1)...(x-c_n)}=0$ for all $x\in[0,1]$.
Now, the denominator is $\neq 0$ $\forall x\in [0,1]$.
So, $f(x)=0$ only when $d_1(x-c_2)...(x-c_n)+d_2(x-c_1)(x-c_3)...(x-c_n)+...+d_n(x-c_1)...(x-c_{n-1})=g(x)=0$. However, $g(x)$ is a polynomial of degree $\leq n-1$ and $g(x)=0$ for every $x \in[0,1]$, which implies that number of zeros of $g(x)$ is $>deg(g(x))$, hence $g(x)$ must be identically equal to $0 \implies (d_1,d_2,...,d_n)= (0,0,..,0) $.
Therefore, our assumption that the given set is linearly dependent is not tenable, i.e. the set is linearly independent.
Is this correct?
 A: This argument is incomplete at the very last step; it's not clear that $g(x) = 0$ implies that the $d_i = 0$. You've assumed that the polynomials in the numerator are linearly independent which is essentially the result you're trying to prove. A priori there could be some cancellation between them, e.g. already the leading coefficient $\sum d_i$ could be zero so you're not guaranteed that the polynomial has degree $n-1$.
The argument can be completed as follows. If the numerator vanishes on $[0, 1]$ then it must in fact vanish identically; that is, all of its coefficients as a polynomial must be zero, so it vanishes on all of $\mathbb{R}$. Now you can plug in each $c_i$ in turn which will tell you that $d_i = 0$. This amounts to a simple version of using analytic continuation to extend $\sum \frac{d_i}{x - c_i}$ to a meromorphic function and then computing its residues at each of its poles.
(Also, as AlexL says in the comments, it's completely unnecessary to frame this as a proof by contradiction. You are proving, directly, that the functions are linearly independent by verifying, directly, the definition of linear independence: that if a linear combination of them is zero then all of the coefficients must be zero.)
Other arguments are also possible; for example you can examine the growth rate of the Taylor series at $x = 0$, or take an analytic continuation and then take limits as $x \to c_i$.
As a challenge, the following more general result is true: the family of functions $\{ 1, \frac{1}{(x - c)^n}, \frac{1}{(x^2 + bx + c)^m}, \frac{x}{(x^2 + bx + c)^m} \}$ is linearly independent, where $\Delta = b^2 - 4c < 0$ and the exponents are $\ge 1$. (This is a special case of a description of a basis of the field of rational functions.) Now the cleanest way to proceed really is to work with $\mathbb{C}$ even though this is purely a statement about functions on $\mathbb{R}$.
A: A finite set of differentiable functions is linearly independent when its Wronskian is non-zero.
In this case the determinant leads to a vandermonde matrix \begin{align}&W(\frac{1}{x-c_1},\ldots,\frac{1}{x-c_n})
\\
&=\det\begin{pmatrix}(x-c_1)^{-1}&\cdots&(x-c_n)^{-1}\\
-(x-c_1)^{-2}&\cdots&-(x-c_n)^{-2}\\
\vdots\\
(-1)^{n-1}(n-1)!(x-c_1)^{-n}&\cdots&(-1)^{n-1}(n-1)!(x-c_n)^{-n}\end{pmatrix}\\
&=\frac{(-1)^a\prod_{k=1}^{n-1}k!}{(x-c_1)^n\cdots(x-c_n)^n}\det\begin{pmatrix} (x-c_1)^{n-1}&\cdots&(x-c_n)^{n-1}\\
\vdots\\
1&\cdots&1\end{pmatrix}\\
&=\frac{(-1)^{a'}\prod_{k=1}^{n-1}k!\prod_{i<j}(c_i-c_j)}{\prod_i(x-c_i)^n}
\end{align}
The fraction is zero iff $c_i=c_j$ for some $i\ne j$.
