Let $U\subset \mathbb{C} $ be an open connected set and $f:U\rightarrow \mathbb{C}$ be a non-constant analytic function. Consider the following sets:-
$X=\{z\in U:f(z)=0\}$
$Y=\{z\in U:f $ vanishes on an open neighbourhood of z in $ U\} $
Then which of the following statements are true?
$(a)\;X$ is closed in $U$
$(b)\;Y$ is closed in $U$
$(c)\;X$ has empty interior
$(d)\;Y$ is open in $U$
My thoughts:-
For $(a)$:
$X=f^-\{0\}$ and $\{0\}$ is closed in $\mathbb{C}$ and inverse image of closed set under a continuous function is closed. So $X$ is closed in $U$.
For $(c)$:
Since zeroes of an analytic function are isolated, so it is true.
Now, for $(b)$ and $(d)$:
If $Y=\emptyset$, then it is both open and closed in $U$.
Else if $Y\neq \emptyset$, then $Y=U$ by uniqueness theorem which again is both closed and open in $U$. So $(b)$ and $(d)$ are true.
Please give a review on my answer and suggest possible corrections/refinements to it.