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Let $S = \{0\}\cup \{\frac{1}{4n+7} : n =1,2\ldots\}$. How to find the number of analytic functions which vanish only on $S$?

Options are

a: $\infty$

b: $0$

c: $1$

d: $2$

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    Hint: the zeroes of an analytic function can't be arbitrarily close together.2012-05-13
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    @Pedro Then what would be conclusion sir?2012-05-13
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    @srijan Are there pairs of points in $S$ which are arbitrarily close together?2012-05-13
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    @Pedro Sir i dont think that there are such points which are arbitrarily close to each other . we can always find $\epsilon>0$ such that distance between any two points will be greater than from chosen $\epsilon$.2012-05-13
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    @pedro Can you write the answer with little explanation please? I would be very much thankful to you.2012-05-13
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    Actually the zeros of an analytic function can be arbitrarily close together, e.g. $\sin(z^2)$. But the zero set an analytic function whose domain is a connected open subset of the plane cannot have an accumulation point in the domain, unless the function is identically zero.2012-05-13
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    @jonas Than you sir for your response. What about this problem sir?2012-05-13
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    @srijan: The last sentence of my comment is directly applicable to this problem.2012-05-13
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    @jonas Ok sir perhaps i have to think little more.2012-05-13
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    @srijan: Have you seen the "[identity theorem](http://en.wikipedia.org/wiki/Identity_theorem)"? (If you scroll down to "an improvement" you will find stuff about accumulation points.) That article refers to $2$ analytic functions $f$ and $g$, but the version Pedro and I are alluding to is the special case where $g=0$. So suppose that you have an analytic function $f$ that is zero on $S$. Can you find a way to apply the identity theorem to make a conclusion about $f$? By the way, are your analytic functions assumed to be defined on the entire plane?2012-05-13
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    @jonas Thanks for your help sir. Ya our analytic functions assumed to be defined on the entire plane.2012-05-13

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