I want to verify if my reasoning is appropriate for the following question.
Let $U \subset \mathbb{C}$ be an open connected set and $f: U \to \mathbb{C}$ be a non constant analytic function. Consider the following two sets:
$\begin{align} &X= \{z \in U: f(z)=0\}\\ &Y= \{z \in U: f \text{ vanishes on an open neighbourhood of $z$ in $U$} \} \end{align}$
Then which of the following statements are true?
- $X$ is closed in $U$.
- $Y$ is closed in $U$.
- $X$ has empty interior.
- $Y$ is open in $U$.
My reasoning: All the options are correct as:
- Since $X= f^{-1}\{0\}$ and being inverse image of closed set it is closed in $\mathbb{C}$ as $f$ is continuous, now option (1) follows.
- For (2) and (4), if we let $z_0 \in Y$ then by the given condition we have $f \equiv 0$ on $U$ by the Identity theorem which gives us a contradiction and so $Y = \phi$. Thus it is open and closed in U.
- For (3), if there is an interior point then there is a nbd of the point inside $X$ and then by Identity thoerem we have $f\equiv 0$ on $U$ which is a contradiction, thus $X$ has empty interior.