If $F:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is a real analytical mapping, it seems "natural" that the set of solutions to $F(x)=0$ should generally consist of isolated points only.
(Let us assume throughout that solutions do in fact exist. If helpful at all, one may also consider a case where all solutions are known to be in a compact subset of $\mathbb{R}^n$).
This is clearly the case for $n=1$ and $F\not\equiv 0$, as per the real analytic identity theorem.
Furthermore, for $f: \mathbb{R}^n \rightarrow \mathbb{R}$ the zero set consists of strata of dimension at most $n-1$, again given $f \not\equiv 0$ (Reference).
Therefore, when looking at a system of equations $F(x)=[f_1(x),f_2(x),...,f_n(x)]=0$, it seems natural to think that each additional $f_i$ reduces the dimension of the solution set by 1.
This of course in general, unless something goes wrong - clearly, the $f_i$ can't be equivalent, multiples of each other, or constructed from each other in a similar manner.
If we were dealing with linear functions, we'd clearly need linear independence. Is "functional independence" a appropriate generalization for the present purpose?
(Reference: Newns, 1967, Functional Dependence).
If yes, how does one prove it, if not, is there anything else to look for?.Is the general intuition on the $0$-dimensionality correct? To me, it seems like this should be the case, but I'm having a hard time finding discussions related to it.
Moreover, if this is generally true, how could one go about showing that it applies to a specific case?
Would it e.g. be enough to show that each for each $f_i$, there exists some $x_i$ such that $f_i(x_i)\neq 0$ but $f_{j\neq i}(x_i)=0$?
(Intuitively, that seems to show that each $f_i$ adds its own restriction to the solution space. But is this enough to establish that there's no region of dimension $\geq 1$ in $\mathbb{R}^n$ where the functions coincide?)Any references would be appreciated.
In case my intuition here is wrong, what are some counter-examples that are not obvious?