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Let $c\in\mathbb{C}^n$ and let $c_\delta \in B(c, \delta),$ the $\delta-$ ball around $c.$ Further let $V^I$ and $V_\delta^I$ denote the set of isolated solutions of the polynomial systems $F(\mathbf{x}) =c$ and $F(\mathbf{x}) =c_\delta$ respectively. Here $F:\mathbf{C}^n \rightarrow \mathbf{C}^n.$ Then is it true that for any $\epsilon >0,$ one can find a small enough $\delta > 0$ so that

1) for each $x^* \in V^I,$ $\exists$ $y^* \in V^I_\delta$ such that $||x^* - y^*||_2 < \epsilon$

2) for each $y^* \in V^I_\delta, \exists$ $x^* \in V^I$ such that $||x^* - y^*||_2 <\epsilon$.

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Answer to 1) should be always true, I think. However, for 2), you need some sort of properness condition. More precisely, if $F$ is not proper in a neighborhood of $F^{-1}(c)$, then there is a curve $\gamma(t)$ such that as $t$ goes to infinity, $\gamma(t)$ goes to infinity and $F(\gamma(t))$ goes to $c$. Since there can be only finitely many isolated solutions of $F(x) = c$,your condition (2) will be violated. – auniket Mar 27 '12 at 4:52
When do you say a map $F$ is proper? – Suresh Mar 27 '12 at 5:16
$F$ is proper in a neighborhood of $F^{-1}(c)$ means that there is an open set $V$ containing $c$ such that $F$ restricted to $F^{-1}(V)$ is proper (i.e. for every compact subset $Z$ of $V$, $F^{-1}(Z)$ is also proper). E.g. $F := \mathbb{C}^2 \to \mathbb{C}^2$ be defined by $u = x^2y - x + y$ and $v = xy$ (where $(u,v)$ are the coordinates in the 'target'). Then $F$ is not proper at $F^{-1}(c)$ for every $c$ on the line $v = 1$. – auniket Mar 27 '12 at 5:32
Did u mean $F^{-1}(Z)$ is compact instead of proper? – Suresh Mar 27 '12 at 5:45
Yeah! Careless mistake ... – auniket Mar 27 '12 at 5:49

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