What is the closure of space of polynomials in a dense subspace along with a marked point equal to? EDIT
Let $\mathbb{C}^{m*}$ be the space of non zero polynomials of degree at most 
$d$ in two variables. So an  element of this space is essentially 
$$ f:=f_{00} + f_{10} x + f_{01} y + \ldots f_{0d} y^d  $$
where not all the coefficients are zero. 
Hence, this space can be identified with $\mathbb{C}^{m*}$, where $m = \frac{d(d+3)}{2}+1$. 
Let $\psi: \mathbb{C}^{m*} \times \mathbb{C}^2 \rightarrow \mathbb{C} $ 
be the evaluation map, ie 
$$ \psi(f, x,y) = f(x,y) = f_{00} + f_{10} x + f_{01} y + \ldots f_{0d} y^d.$$
I have two questions: 
1) Given $(f ,x_0,y_0) \in \mathbb{C}^{m*} \times \mathbb{C}^2$ such that 
$\psi(f, x_0,y_0) =0 $ and a polynomial $f_t$ that is  sufficiently "close" to $f$,
does there exist a point $x_t, y_t$ close to $x_0, y_0$ such that 
$$ \psi(f_t, x_t, y_t) =0 ~~?$$ 
Note that this would be true if $(x_0, y_0)$ was a smooth point of $f$. 
But I am not making this assumption. 
2) Let $ \mathcal{S}$ be a dense subspace of $\mathbb{C}^{m*}$. Define the 
space $\mathcal{F}$ as
$$ \mathcal{F} := \{ (f, x,y) \in \mathcal{S} \times \mathbb{C}^2: \psi(f, x,y) =0 \} $$
Is it true that 
$$ \overline{\mathcal{F}} = \{ (f, x,y) \in \mathbb{C}^{m*} \times \mathbb{C}^2: \psi(f, x,y) =0 \} $$ 
where 
$\overline{\mathcal{F}}$ denotes the closure of $\mathcal{F}$ inside 
$\mathbb{C}^{m*} \times \mathbb{C}^2$? 
The basic idea being that if $ \psi(f, x,y) =0$ but $f \notin \mathcal{S} $ 
we can choose a sequence $ f_n \in \mathcal{S} $ converging to $f$. 
And by 1), there exists a sequence (possibly non unique) $(x_n, y_n) \in \mathbb{C}^2$
such that 
$$ \psi(f_n, x_n, y_n) =0 $$ 
This sequence $(f_n, x_n,y_n)$ should converge to 
$(f, x,y)$. 
 A: (1) Yes. For notational simplicity, assume $(x_0, y_0)=(0,0)$. After a linear change of coordinates, we may assume that $f(x,0)$ is not identically zero; say $f(x,0) = x^n g(x)$ with $g(0) \neq 0$. Choose a small disc $D$ around $0$ so $f(x,0)$ has no zeroes for $x \in D \setminus \{ 0 \}$. 
Choose $f_t$ close enough to $f$ that $f$ is nonvanishing on $\partial D$. Then $z(t) := \frac{1}{2 \pi i} \int_{\partial D} \frac{\partial f_t(x,0)/\partial x}{f_t(x,0)} dx$ is a continuous function of $t$. But $z(t)$ is the number of roots of $f_t(x,0)$ inside $D$, so it is integer valued and constant. So $f_t(x,0)$ has $n$ roots inside $D$ for all $t$ sufficiently close to $0$. 
A: If I understood the first question right, it sounds as follows: suppose that $\psi$ is a polynomial in $N>1$ variables over $\mathbb C$; can its set of zeroes in $\mathbb C^N$ have an isolated point?
The answer is 'no': connected components of a, say, affine algebraic set in Zariski topology and in the classical topology are the same, so this isolated point would be an irreducible component of the set of zeroes of $\psi$. However, each irredicuble component of the zero set of a non-zero polynomial in $\mathbb C^N$ must have dimension $N-1$.
