# Problem related with analytic function

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.

-

Yes, you're on the right lines: is $0$ isolated in the set $\{ 0 \} \cup \left\{ \dfrac{1}{4n+1}\, :\, n \in \mathbb{N} \right\}$?
Given $\varepsilon > 0$, can you find an $n$ such that $\dfrac{1}{4n+1} < \varepsilon$?
No,sir.I do not think that here in the set $0$ is an isolated point. Here, we can choose $n$ to be arbitrarily large so that any nbd. of ${0}$ will contain points from the set ${1/(4n+1): n \in \mathbb N}$. Am i right? – user52976 Dec 13 '12 at 16:18
No... you were right in saying that there are no analytic functions vanishing on $S$. But this is because $S$ contains at least one non-isolated zero. – Clive Newstead Dec 13 '12 at 16:43