$C^0([a,b])$ is an infinite dimensional vector space I am proving that $C^0([a,b])$ is an infinite dimensional vector space. The fact that it is a vector space is clear. But I cannot understand how to prove that it has infinite dimension.
Let $F:=\{f_n:[a,b]\to\mathbb{R},x\mapsto x^n:n\in\mathbb{N}\}$. $\mid F\mid=+\infty$. But how can I prove that $F$ is made of linearly independent vectors?
 A: If some finite linear combination
$$\sum_{k=0}^n a_kx^k$$
is the function $0$, then the polynomial
$$\sum_{k=0}^n a_kx^k$$
has infinitely many roots, namely, the elements of $[a,b]$. That means that the polynomial is $0$, that is, $a_k=0$ for any natural $k$.
Hence, $F$ is lineary independent.
A: Consider arbitrary finite collection of vectors $(x^{n_i})_{i\in\mathbb{N}_m}$ in $F$. Assume we have
$$
\sum_{i=1}^m\alpha_i x^{n_i}=0
$$
This eqaulity tells us that we have a polynomial which is identically zero. Therefore all its coefficients are zero: $\alpha_i=0$ for all $i\in\mathbb{N}_m$. This means that $(x^{n_i})_{i\in\mathbb{N}_m}$ is linearly independent in $F$. Since we considered an arbitrary finite collection of vectors in $F$ and they turns out to be linearly independent, then the whole system $F$ is linearly idependent too.
A: Take a finite subset of elements $f_{n_1},\ldots,f_{n_k}$ of $F$ and WLOG assume that
$$n_1<\cdots<n_k$$
and let $a_{n_1},\ldots,a_{n_k}\in \Bbb R$ such that:
$$a_{n_1}f_{n_1}+\cdots+a_{n_k}f_{n_k}=0$$
so by differentiating the above equality $n_k$ times we find that $a_{n_k}=0$ and then by differentiating  $n_{k-1}$ times we find that $a_{n_{k-1}}=0$ and so on we prove that all the coefficients are zero and then the elements are linearly independent. 
