Set $f(x)=\sum a_n x^n$ , $f$ is well defined on a nonempty set $A$ . If $f(x) = 0$ for all $x \in A$ , can we prove that $a_n=0$ for all $n$? Set $f(x)=\sum a_n x^n$ ,$x \in C$ . $A$ is a subset of $C$ , $f$ is well defined on $A$ and only on $A$ ,$A-0$ is nonempty. If $f(x) = 0$ for all $x \in A$ , can we prove that $a_n=0$ for all $n$ ?  
My attempt:
If $B$ is an open set in $A$ , then $f$ has derivatives of all orders in $B$ . Since $f=0$ in $B$ , $f^{(n)}(x)=0$ in $B$ , this implies that $a_n = 0$ .
I'm not sure whether this proof is right , and I think there might have some other direct and rigorous proof . 
 A: I don't think the set necessarily has to have any topology on it, so we can create counterexamples where the idea of an "open" set doesn't make any sense. For example, if $A=\mathbb{F}_2$, we can choose $a_0=0,a_1=a_2=1$ to get
$$x+x^2,$$
a polynomial which is identically $0$ on $\mathbb{F}_2$ but still has nonzero coefficients. 
A: Assume $f(x)=\sum a_nx^n$ converges exactly on $A\subset\mathbb{R}$ with $A\setminus\{0\}\neq\emptyset$. Let $r\in A\setminus\{0\}$. Because every infinite series has a radius of convergence, such that everything strictly within that radius makes the series converge, while everything strictly outside the radius makes the series diverge, we find that $\{x\in\mathbb{R}:|x|<|r|\}\subset A$. Because $|r|>0$ we find an interval where the series of $f$ converges.
For the final step, use the fact that an analytic function is entirely defined by its local behaviour. If an analytic function is identically $0$ on some non-discrete set (for example an interval) then all power series coefficients are $0$.
A: For simplicity, I will assume $A$ is connected otherwise repeat the proof to each component. Observe first that if $f(z)$ is analytic on $A$, then it has an open neighborhood where it's analytic on so $A$ has infinitely many elements. I will even weaken the assumptions given and only use that we need a limit point inside $A$ (this is known as the identity theorem).
So by Heine-Borel, we know there exists a limit point $z^*$ inside $A$ such that $f(z_k) = 0$ for all $z_k \rightarrow z^{*}$. Then by analyticity \begin{equation} f(z) = a_k(z-z^*)^k + \sum_{n=0}^{\infty} a_{k+1+n}(z-z^*)^{k+1+n} \end{equation} where $k$ is chosen as the minimal integer such that $a_k \neq 0$ (otherwise we'll be done). Then \begin{equation} \frac{f(z)}{(z-z^*)^k} = a_k + \sum_{n=0}^{\infty} a_{n+1+k}(z-z^*)^{n+1} \end{equation} But observe as for all $z_j$ the left hand side is 0, we see by continuity at $z=z^*$, the left hand side is 0. But at $z^*$ the sum is 0, so it implies $a_k$ must be zero; therefore, we have a contradiction. In particular, this shows that every point where the power series of $z^*$ converges is identically 0.
In particular we have shown that $U = \{z: f(z) = 0\}$ is open. But also since $f$ is continuous, so the pre-image of a singleton must be closed, which means $U$ is closed, open and non-empty. Then as $A$ is connected (or repeat the proof to each connected component), we see $U = A$. So $f$ is identically $0$ over A. 
