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I was thinking about the following problem:

Let $S=\{0\}\cup \{ \frac{1}{4n+1}:n=1,2,3,4,\dots\}$ Then what is the total number of analytic function which vanish only on $S$?

I was trying to use the fact that zeros of analytic functions are isolated. But I could not progress further. Am I going in the right direction? Please help. Thanks in advance for your time.

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Yes, you're on the right lines: is $0$ isolated in the set $\{ 0 \} \cup \left\{ \dfrac{1}{4n+1}\, :\, n \in \mathbb{N} \right\}$?

Below is a hint: hover your mouse over the grey box to see it.

Given $\varepsilon > 0$, can you find an $n$ such that $\dfrac{1}{4n+1} < \varepsilon$?

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    Sorry sir.After posting my previous comment, I realized that i went wrong in my argument. So, i deleted my earlier comment which may mislead me( or even others who will go through it in future).I have got it,now. Thanks a lot for your time.2012-12-13