Does the zeros of an analytic function $f$ form a discrete set?

Asked Jun 28 '15 at 19:26

Active Jun 28 '15 at 19:28

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A function $f$ is called analytic if locally it is given by a convergent power series.

Let $ U \subset \mathbb R^n $ be an open set and $f : U \to \mathbb R$ be non zero analytic function. Does the zeros of $f$ form a discrete set?

I'm unable to find counterexample.Please help!

edited Jun 28 '15 at 19:28

asked Jun 28 '15 at 19:26

Dontknowanything

the zeros of analytic complex function are indeed a discrete set. Don't know about $\mathbb R^n$ though :) – Ant Jun 28 '15 at 19:30
@Ant ofcourse thats true! – Dontknowanything Jun 28 '15 at 19:34
This post on the identity theorem for real-analytic functions is what you're after, I believe. – msteve Jun 28 '15 at 19:37
1

You seem to be looking at "real-analytic" functions. On $\Bbb R ^2$ consider the function $f(x,y) = y.$ – zhw. Jun 28 '15 at 19:45
@zhw. Thank you.Does the result holds for n=1? – Dontknowanything Jun 28 '15 at 19:53
Sure, and whatever proof you know for holomorphic functions can be used for $n=1.$ – zhw. Jun 28 '15 at 19:54
Ohh I see.Thank you.Regards, – Dontknowanything Jun 28 '15 at 19:55

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