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In one dimension there is a theorem that states a non-constant analytic function has only finitely many zeroes on any closed, bounded subset of it's domain. I am interested in how this result might be extended to higher dimensions but seem unable to find a discussion of the topic. In particular I am interested in :

  1. Is it true that the zero set of an analytic function in a closed, bounded subset of its domain can be written as the union of a finite number of connected sets?
  2. Given that the above is (roughly) correct is it true that any connected subset of the zero set can be described by an analytic curve? For example if $g:D\subset\mathbb{R}^2\rightarrow\mathbb{R}$ is analytic (where $D$ is closed and bounded) and if $S$ is a connected subset of the zero set of $g$ must there be an analytic curve $f:[0,1]\rightarrow\mathbb{R}^2$ such that $$S = \{(x,y)\in\mathbb{R}^2| f(t) = (x,y) \mbox{ for some } t\in[0,1]\}?$$
  3. How would one go about proving these statements? It's not clear to me how the single variable proof might be generalized.
  4. Where might I find a discussion of this topic? My usual sources seem to have failed me.

Thank you for taking the time to read my question I'm greatful for any insight you can offer.

Edit: Changed the statement of the 1d case to include the requirement that the subset of the domain be closed.

Edit2: Changed points 1 and 2 to be consistent with edit 1.

Edgecase
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  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Mar 15 '18 at 20:01
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    @JoséCarlosSantos I'm not sure what you mean! What more would you have liked me to include in my question? Shall I make a list of all the sources I found that didn't discuss this question? Shall I write an essay on my failed attempts to formulate and prove these statements for myself? I know what I want to know and I asked about it as clearly and concisely as I could, it saddens me that your first reaction was to question my integrity! – Edgecase Mar 15 '18 at 20:13
  • You may have a look in the book 'Holomorphic Functions of Several Variables' by Ludger Kaup and Burchard Kaup. In Chapter 0, §7, they discuss removable singularties. (One consequence is that zero's never isolated in dimensions $\geq 2$!) – p4sch Mar 15 '18 at 20:27
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    @Edgecase Your question is fine. It's beyond fine, it's very well stated, showing a lot of thought. – zhw. Mar 15 '18 at 21:53
  • "In one dimension there is a theorem that states a non-constant analytic function has only finitely many zeroes on any bounded subset of it's domain." That's not true. For example $\sin (1/x)$ is analytic in $(0,1)$ and has zeros at $1/(n\pi),n=1,2,\dots$ Thus this analytic function has infinitely many zeros in $(0,1),$ which is bounded. I think you may want to replace "bounded" with "compact". – zhw. Mar 15 '18 at 22:00
  • @zhw Indeed you are correct, the theorem I was thinking of reads "finitely many zeroes on any CLOSED bounded subset of the domain" closed and bounded together of course imply compact. – Edgecase Mar 15 '18 at 22:12
  • Given that, did you want to make changes to 1. or 2.? – zhw. Mar 15 '18 at 22:26
  • @p4sch I've managed to get a copy and look it over and I can see its related but I'm not entirely clear on how results from complex analysis should be applied to the purely real case. My understanding is that even when restricted to the real line complex analytic functions must satisfy stricter conditions then real analytic functions. For example the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ given by $$f(x,y) = xy$$ is manifestly analytic and has an isolated zero at $(x,y)=(0,0)$. – Edgecase Mar 16 '18 at 20:12
  • Every real valued real analytic function on $\mathbb R$ extends to be complex analytic on an open subset of $\mathbb C$ containing $\mathbb R.$ – zhw. Mar 16 '18 at 22:00
  • An easier way to write the condition for 2: Simply say $S= f([0,1]).$ – zhw. Mar 16 '18 at 23:31

1 Answers1

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The answer to 2. is no. Take the function $g(x,y) = xy$ which is a nice real analytic function on $\mathbb R^2.$ The global zero set for $g$ is the union of the $x$ and $y$ axes. So consider $D$ to be the closed unit disc. The zero set of $g$ relative to $D$ is the "cross" $C=[-1,1]\times \{0\} \cup \{0\} \times [-1,1].$ Suppose $f:[0,1] \to C$ is analytic and surjective. Write $f(t) = (x(t),y(t)).$ Then $y(t)$ is $0$ on $f^{-1}([-1,1]\times \{0\}).$ That's an uncountable number of zeros for $y(t).$ Since $y$ is analytic, $y\equiv 0.$ Thus $f$ cannot be surjective, contradiction.

zhw.
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