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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.

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    $\begingroup$ Hint: An uncountable set in $\mathbb C$ must have a cluster point. $\endgroup$ – user99914 Jun 22 '15 at 6:04
  • $\begingroup$ @John I had a feeling that that might be the case. Can you direct me to the theorem that guarantees that? $\endgroup$ – Miz Jun 22 '15 at 6:11
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    $\begingroup$ Never mind. I got how an uncountable set has a limit point. Because if it does not you can put each of them inside their own individual open ball and identify a rational point with each. $\endgroup$ – Miz Jun 22 '15 at 6:20
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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).

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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.

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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 $\Bbb C$.

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    $\begingroup$ There is no reason to believe there is a sequence of zeros of the form $a + \frac1n$ just because there are uncountably many zeros. $\endgroup$ – Erick Wong Nov 13 '15 at 8:07

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