Help with Linear Algebra proof that an infinite set of polynomials is independent Let $\mu_k(t) = t^k$, for $k = 0,1,2, \dots,$ and $t$ is real. I want to show that the infinite set $S = \{\mu_0, \mu_1 , dots \}$ is independent. To do this, set up
(1) $\sum_{k=0}^n c_kt^k= 0$, for all real $t$. 
Here's the part of the proof that I don't understand:
Putting $t=0$ in (1) we see that $c_0 = 0$. (Why? I don't see this, and it's my biggest stumbling block in the proof.) Now divide by $t$ in (1) and put $t=0$ again to find $c_1 = 0$. (So, I'm confused as to what is meant by $t$ in the proof. Is it all $t$? Also, why can we divide by $t$? Didn't we just say that $t=0$? Are there supposed to be subscripts?). Rinse and Repeat for the rest of the $c_k$.
I have one last question. The sum in (1) is finite. How does this prove that an infinite set is independent. Doesn't this show that only $n$ $\mu_i$'s are independent; rather than all $\mu_k$ for $k = 0, 1, \dots $ are independent.   
Update 1: Why $t$ are we concerning ourselves with the case when $t=0$?
 A: If $V$ is a vector space over a field $F$ then a subset $A\subset V$ is linearly independent if any finite subset $\{a_1,a_2,\ldots, a_n\}\subset A$, $\>n\in{\Bbb N}_{\geq1}$, is linearly independent. Note that, in algebraic terms, it wouldn't make sense to check under what circumstances an infinite sum $\sum_{k=1}^\infty \lambda_k a_k$ is $\>=0$, since such a sum would not be defined in the first place.
In your example $V$ is the space of all polynomials in an indeterminate $t$ with real coefficients. It is a matter of convention whether you consider a polynomial 
(I) as an infinite coefficient sequence $(c_0,c_1,c_2,\ldots)$
with only finitely many nonzero entries $c_k\in{\Bbb R}$, 
or (II) as a function
$$p:\ {\Bbb R}\to{\Bbb R},\qquad t\mapsto \sum_{k=0}^{n_p} c_k t^k\ .$$
Under both views the monomials $\mu_k$ form a linearly independent set: Under (I) we have
$$\sum_{k=0}^N\lambda _k\mu_k=(\lambda_0,\lambda_1,\ldots,\lambda_N,0,0,\ldots)=0$$
iff all $\lambda_k=0$. Under (II) we have to prove that $$p(t):=\sum_{k=0}^N\lambda_k\>t^k\equiv0\tag{1}$$
implies $\lambda_k=0$ for all $k$. A quick proof is
$$\lambda_k={1\over k!}\>p^{(k)}(0)=0\qquad(0\leq k\leq N)\ .$$ 
A more algebraic proof would set up an induction as follows: $(1)$ implies $\lambda_0=p(0)=0$ and therefore
$p(t)=t\>p_1(t)$ with
$$p_1(t)=\sum_{k=0}^{N-1}\lambda_{k+1}t^k\>\equiv0\ .$$
The induction hypothesis then guarantees $\lambda_1=\lambda_2=\ldots=\lambda_N=0$.
A: Let's give our polynomial a name: define $p(t)=\sum_{k=0}^nc_kt^k$. We are assuming $p(t)=0$ for all real $t$, and we want to show that $c_0=\ldots=c_n=0$. Now note that
$$p(t)=c_0+c_1t+c_2t^2+\ldots+c_nt^n$$
We know that the left hand side is zero for all real $t$. In particular, it must be true when $t=0$, so substituting $t=0$ into the right hand side we find
$$0=p(0)=c_0+c_1\cdot0+\ldots+c_n\cdot0^n=c_0$$
Hence our polynomial is now
$$0=p(t)=c_1t+c_2t^2+\ldots+c_nt^n=t(c_1+c_2t+\ldots+c_nt^{n-1})$$
So either $t=0$ or $c_1+c_2t+\ldots+c_nt^{n-1}=0$. This implies $c_1+c_2t+\ldots+c_nt^{n-1}=0$ for all $t\neq0$, but by taking the limit as $t\to0$, it must be true at $t=0$ too! This follows from the continuity of polynomials. Now we just repeat this procedure until we have shown every $c_i=0$. An induction proof would probably be more rigourous, but I think at this level "repeat until there are no $c_i$'s left" is sufficient.
But why does this show $S$ is independent? As you pointed out, $S$ is infinite whereas we only looked at the first $n$ terms. However addition is only really defined on finitely many terms; if you have some topology you can sometimes define infinite sums but in the general vector space case this is not possible. So, the only sensible definition of $S$ being independent is that every finite sum of $S$ is independent. If we have a finite subset $F\subset S$, then in particular there is a largest $n$ such that $\mu_n\in F$. Hence the above argument applied for that particular $n$ shows $F$ is independent, and thus by definition (and the fact $F$ was an arbitrary finite subset) we find that $S$ is independent.
