Identity theorem for $\mathbb{R}^n$ This question is follow up to an interesting question I found here. 
The results of this question states the following:

If $U$ is a domain, and $f,g$ are two real-analytic functions defined
  on $U$, and if $V\subset U$ is a nonempty open set with $f\lvert_V
 \equiv g\lvert_V$, then $f \equiv g$. If the domain is one-dimensional
  (an interval in $\mathbb{R}$), then it suffices that $f\lvert_M \equiv
 g\lvert_M$ for some $M\subset U$ that has an accumulation point in
  $U$.

I have a few question about this theorem:


*

*I was looking for a reference for this. However, in the previously pointed out reference A Primer of Real Analytic Functions by Krantz and Parks, I was not able to locate this theorem.  I would appriciate if someone could point me to proper reference of this theorem. 

*My main question, is about the second part of the theorem.  Specifically, I am interested in why there is such a difference going from $\mathbb{R}$ to $\mathbb{R}^n$. That is in one-dimension we can assume that $M$ is just a set with an accumulation point, but in $\mathbb{R}^n$ we have to assume that $M$ is an open set.  I would really like see a counter example that demonstraes that assuming that $M$ is a set with accumulation points is not sufficient in $\mathbb{R}^n$.

*I would really like for you to speculate or suggest an extra assumptions on functions $f$ and $g$ such that it suffices to consider $M$ to be only a set with an accumulation point. 


For 2 and 3: the functions $0$ and $x$ are a counterexample if all you want is an accumulation, and since a general analytic function will have an $n-1$ dimensional $0$ set, the best you can do is what Mark van Leeuwen suggests here.
