Let $U$ be a domain and let $f\neq 0$ be an analytic map on $U$. Prove that the zero set $Z(f)$ of $f$ is most countable.
since $Z(f)$ is closed set.assuming $Z(f)$ to be uncountable,then it will have a limit point,hence that limit point will belong to $Z(f)$. which mean $f$ is zero map on $U$. which is contradiction .hence done.
but how to prove that an uncountable set in $\mathbb{C}$ has limit point?? any help
Let $A$ be your uncountable set and assume it is closed (otherwise take the closure). Consider the sets $A_n = B[0,n] \cap A$ (where $B[0,n]$ is the closed ball around 0 with radius $n$). Then one of the $A_n$, say $A_j$, must contain more than finitely many points (otherwise you have that $A$, as the union of countably many finite sets, is countable). Pick a sequence $(z_k)_k$ in $A_j$. Since $A_j$ is bounded, $(z_k)$ has an accumulation point which lies, by the closedness of $A_j$, again in $A_j$ and hence in $A$.