Solution Set of a Polynomial System

Consider the polynomial System $F(x)-c=0,$ where $F:\mathbb{C}^n \rightarrow \mathbb{C}^n.$ Is it true that for almost all values of $c\in \mathbb{C}^n,$ the polynomial system will only have isolated solutions? Or in other words, for almost all values of $c,$ it will not a solution of positive dimension.

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This is true at least in some cases: Let $F = (f_1, \ldots, f_n)$. Let $V$ denote the zero set of a polynomial or set of polynomials. Then $V(F - c) = V(\{ f_1 - c_1, \ldots, f_n - c_n \}) = V(f_1 -c_1) \cap \cdots \cap V(f_n - c_n)$.
Using Bertini's Theorem, we can then prove that if $V_{n-1} := V(\{ f_1 - c_1, \ldots, f_{k-1} - c_{k-1} \})$ is a smooth variety of dimension $n-k+1$, there are either only finitely many values of $c_n$ such that $V(\{ f_1 - c_1, \ldots, f_{n} - c_{n} \})$ has dimension $> n-k$, or $f_n - c_n \in \langle f_1 - c_1, \ldots, f_n - c_n \rangle$ for all $c_n \in \mathbb C$. In the latter case, $V_{n-1} = \varnothing$, hence this case is irrelevant.
Thus, as long as $V_1, \ldots, V_n$ are all smooth varieties, the claim holds.
What if the constant term is allowed to vary only in $\mathbb{R}^n?$ In other words, is it true that there exists an open dense set $S$ of $\mathbb{R}^n$ such that for $c\in S,$ $F(x) - c = 0$ has only isolated solutions. Can the Bertini's theorem be extended to cover this situation? –  Suresh Apr 8 '12 at 4:03
If the result holds for $\mathbb C^n$, it will of course hold for $\mathbb R^n$. –  Johannes Kloos Apr 10 '12 at 18:48