My question is about a part of a given proof for the identity theorem of analytic functions. This part shows that two analytic functions are identical in a neighbourhood of an accumulation point of the set in which they agree (see below).
(Theorem:) Let $f_1, f_2$ be two analytic functions in a connected open set $\Omega\subset\mathbb{C}$ and suppose that $E \equiv \{z\in\Omega\mid f_1(z) = f_2(z)\}$ has an accumulation point in $\Omega$. Then $\forall z \in \Omega:f_1(z) = f_2(z)$
(Proof:) Let $z_0$ be an accumulation point of $E$ and take $B(z_0, R)$ be some neighbourhood of $z_0$. By analyticity of $f_1, f_2$, $f_1(z) = \sum_{n=0}^\infty a_n(z – z_0)^n, f_2(z) = \sum_{n=0}^\infty b_n(z – z_0)^n$ as the series expansion of the said functions. Suppose that $\exists k\in\mathbb{N}:a_k\neq b_k$ and let $k'\in\mathbb{N}$ be the smallest such index. Then $f_1(z) – f_2(z) = (a_{k'} – b_{k'})(z – z_0)^{k'}(1 + g(z, z_0)) \neq 0$ when $z\neq z_0$ and $|g(z, z_0)| < 1$. It follows that $f_1(z)\neq f_2(z)$ for all $z$ close enough to $z_0$, a contradiction. Thus $f_1$ and $f_2$ must agree on a neighbourhood of $z_0$.
I do not understand the contradiction. Is it due to taking any point $z' \in E\cap B(z_0, R)$? But even in that case why couldn't $f_1$ and $f_2$ differ on $B(z_0, R)\setminus E$?
Best Answer
The point of the argument is that the function $f_1-f_2$ is analytic and nonzero on $B(z_0,r)\setminus\{z_0\}$ for some small enough $r>0$. But because $z_0$ is an accumulation point of $E$, there exists a sequence $\{z_n\}$ with $f_1(z_n)-f_2(z_n)=0$ for all $n$; that is, for $n$ big enough you have $z_n\in B(z_0,r)\setminus\{z_0\}$, and that's the contradiction because $f_1-f_2$ is nonzero on the punctured ball.
Finally, because $f_1-f_2$ is zero on a disk and the disk is inside the connected open set $\Omega$, we get that $f_1-f_2=0$ on al of $\Omega$.