Proving a set is linearly independent in C[0,1] Prove that the set $P=\{f_n \in C[0,1]:f_n(t)=t^n,\ t \in [0,1],\ n \in \mathbb{N} \cup \{\ 0 \}\}$ is linearly independent subset of $C[0,1]$. What is the span $(P) $? What is the closure of span$(P)$ in $(C[0,1],d_\infty).$ \
I have difficulty to show that $P$ is linearly independent since for $t=0$, we have for any $n$, $$\alpha_1 (0)+\alpha_2 (0)+...+\alpha_n (0)=0$$
but we can have these $\alpha's \neq 0$??
Similarly for $t \in [0,1]$. So any hint or help would be appreciated it.  
 A: Clearly, the set $P$ is not finite. Therefore, to prove that $P$ is linearly independent, we need to prove that every finite subset of $P$ is linearly independent. Therefore, take any $m$ elements, say $\left\lbrace t^{n_1}, t^{n_2}, \cdots, t^{n_m} \right\rbrace$, where $t \in \left[ 0, 1 \right]$. If we say that there are scalars $\alpha_1, \alpha_2, \cdots, \alpha_m \in \mathbb{R}$ such that
$$\alpha_1 t^{n_1} + \alpha_2 t^{n_2} + \cdots + \alpha_n t^{n_m} = 0$$
we mean that this holds true for all $t \in \left[ 0, 1 \right]$. In particular, it holds true for all $t = \dfrac{1}{k}$, where $k = 1, 2, \cdots, m$. Thus, substituting these values, we get $m$ homogeneous equations in $m$ unknowns so that each of the $\alpha_i$s must be zero. Thus, we have proved the linear independence.
As for the next part, it is clear to observe that the span of $P$ is the set of all polynomials defined on $\left[ 0, 1 \right]$. Also, since the set of all polynomials is dense in $C \left[ 0, 1 \right]$, the closure of the span of $P$ is whole of $C \left[ 0, 1 \right]$.
