Entire function with uncountably many zeros Suppose that an entire function $f$ has uncountably many zeros. Is it true that $f=0$?
I have no idea how to proceed with this. Perhaps some theorem that I am not aware of. I have done an undergraduate course in complex analysis.
 A: Lets consider open ball $B_n$ at center zero and radius $n$, a natural number, then  $\mathbb{C}=\cup_{n\in N} B_n$. Now if none of $B_n$ contains uncountably many complex zeros of $f$, zeros of $f$ will become countable, a contradiction. So suppose $B_k$ for some natural number $k$, contains uncountably many complex zeros of $f$. Hence zeros of $f$ has a bounded uncountable subset in $B_k$. By Bolzano-Weirstrass theorem it has limit point. Now by Identity theorem in complex analysis $f$ is identically zero.  
A: As John already pointed out, an uncountable subset $A\subseteq\Bbb C$ has necessarely an accumulation point. Then the identity principle for holomorphic functions, says that a function which is $0$ on $A$, is necessarely $0$ on its whole domain (which is $\Bbb C$ if your function is entire).
A: If there are uncountably many zeroes of $f(z)$ and suppose $a$ (say) be their limit point then $f(a+\frac 1n)=0$ where $n\in \Bbb N$. By corollary of Identity theorem $f(z)$ is identically zero on $\Bbb C$.
