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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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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
Yup. So how many analytic functions vanish on the given set (and not elsewhere)? – Clive Newstead Dec 13 '12 at 16:23
Sir, i think i have got your point. Since zeros of analytic functions are isolated and the set does not contain any isolated zero ,hence we can conclude that the number of analytic functions vanish on the given set (and not elsewhere) is ZERO. Am i right? – user52976 Dec 13 '12 at 16:28
That's right :) Except it's wrong to say it doesn't contain any isolated zeros: it contains infinitely many! But the point is that it does contain a zero which isn't isolated. – Clive Newstead Dec 13 '12 at 16:29
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

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