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Are the zeros of a non-constant real analytic function $f$ from a finite dim, real vector space $V$ to the real numbers $\mathbb{R}$ which takes values in $[0,1]$ always a countable set?

Update:

Is it possible for there to be an analytic function which is not constant in any of its variables (i.e. dimensions of $V$) which has uncountably many zeros?

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    $\begingroup$ Can someone elaborate what we mean by a real analytic function in the case that $V \not \subseteq \mathbb R$? $\endgroup$ – Stefan Mesken Sep 3 '15 at 15:37
  • $\begingroup$ @Stefan Of course if the domain is "a vector space" this is not so clear. But if the domain is, say, $\Bbb R^n$ it just means that $f$ is given by a power series in a neighborhood of each point. $\endgroup$ – David C. Ullrich Sep 3 '15 at 15:42
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    $\begingroup$ The generic case (in some sense) is that the zeros of a differentiable function $f \colon U \to \mathbb{R}$, where $U$ is an open subset of $\mathbb{R}^n$, form an $n-1$-dimensional set. $\endgroup$ – Daniel Fischer Sep 3 '15 at 15:54
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    $\begingroup$ Yes, it's true for $V=\Bbb R$. (And yes, it seems clear to me that that's why one might think it was true in general.) $\endgroup$ – David C. Ullrich Sep 3 '15 at 16:09
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    $\begingroup$ @Benjamin Yes it's true for $V=\mathbb R$. My point is that it isn't true for any higher dimension, which is the purpose of the generalization. Is there a motivation for adding restrictions like $[0,1]$ and non-constant in each variable? Normally one adds such restrictions to separate counterexamples from an actual example, which is why I asked if you had one. If the only examples are dimension 1, then the obvious separation is the number of dimensions in $V$. But if you have an example in dim $2$ that would add greatly to the question. $\endgroup$ – Erick Wong Sep 5 '15 at 15:08
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No. For instance, define $f:\Bbb R^2\to[0,1]$ by $f(x,y) = x^2/(1+x^4)$.

For the revised question, "not constant in any of its variables"", change that to $f(x,y)=(x+y)^2/(1+(x+y)^4)$.


Just so you get your money's worth: Yes, it's true for $V=\Bbb R$. Two ways to see this: (i) in that case $f$ is the restriction to $\Bbb R$ of a function holomorphic in some neighborhood of $\Bbb R$ in the plane.

(ii) Say $x_n\to x$ and $f(x_n)=0$ for every $n$. It follows that $f^{(k)}(x)=0$ for every $k$...

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    $\begingroup$ Not sure why you didn't go for $f(x,y) = x$ instead :D . Since the answer has been deleted, I cannot comment on Benjamanins comment, but $f(x,y) = x-y$ shows that such a function does not have to be constant in one variable. $\endgroup$ – Stefan Mesken Sep 3 '15 at 15:48
  • $\begingroup$ @Stefan Because he asked about $f:V\to[0,1]$. Details... $\endgroup$ – David C. Ullrich Sep 3 '15 at 15:50
  • $\begingroup$ Oh, I missed that part of the question. Thanks. $\endgroup$ – Stefan Mesken Sep 3 '15 at 15:51
  • $\begingroup$ In this case, $f(x,y) = \frac{x^2}{2+2x^4} - \frac{y^2}{2+2y^4}$ answers Benjamins question. $\endgroup$ – Stefan Mesken Sep 3 '15 at 15:52
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Consider $f(x, y) = \sin^2(x)$.

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