Identity theorem Let $f$ be holomorphic functions on some domain $D\subset\mathbb{C}$, and let $S$ be the set of all zeros of $f$ which has a limit point in $D$. Then $f$ is identically zero in $D$.
Proof in short: Let $a$ be limit point of $S$ this means that there exist a sequence $a_n$ of zeros such that $a_n\rightarrow a$ as $n\rightarrow \infty$. Then $f(z)$ is identically zero on the disc $|z-a|<R$.
To complete the proof we have to show that $f(z)$ is identically zero on the whole domain $D$. For this purpose, we decomposed $D$ into two sets:
$E_1$=$z \in D$, where $z$ is a limit point of S
$E_2$=$z \in D$, where $z\not\in S$
Doubt 1 I am having difficulty to understand how the sets $E_1$ and $E_2$ are defined? Could someone explain me through example If possible?
$2$ How to show that set $E_1$ and $E_2$ are open? Book that I follow says that since $z\in E_1$ is a limit point of $S$, $f(z)$ is identically zero in a neighborhood of $z$. From here how can we deduce that each point of $E_1$ is an interior point of $E_1$
Thanks in advance. For any queries related to this question please feel free to put comments.
$E_1$ is the set of limit points of $S$. For instance, if $S$ is an open disk, then $E_1$ is a closed disk.
$E_2$ are all points in your domain which are not in $S$. Meaning - all points on which $f$ is non-zero.
$E_1$ is open by the argument you wrote - for any $z\in E_1$, there is a neighborhood $U$ of $z$ on which $f$ is identically $0$. In particular, all of the points in $U$ are also limit points of $S$ (which means that it is open, since you see that each point has a neighborhood contained in the set). Why is this true? Since $f$ is identically $0$ around each point in $U$, so you can get arbitrarily close to them via points on which $f$ is $0$.
Moreover, $E_2$ is open as the continuous preimage of the open set $\mathbb C - \{0\}$.
Hope this helps. Feel free to ask more.