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 :
- 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?
- 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 $$S = \{(x,y)\in\mathbb{R}^2| f(t) = (x,y) \mbox{ for some } t\in[0,1]\}?$$
- How would one go about proving these statements? It's not clear to me how the single variable proof might be generalized.
- 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.
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.