Zeros of function $f: \mathbb{C}^{n}\to\mathbb{C}$ I'm having trouble understanding how this condition $(*)$ comes about. I understand the proof, but do not clearly understand how the lemma follows from the proof. Any clarification or additional intuition provided would be greatly appreciated. Here is the lemma and its proof:
LEMMA:
Let $f_{1},\dots,f_{n}$ be the components of $F$, and let $k$ be the subfield of $\mathbb{C}$ generated by the coefficients of the polynomials $f_{i}$. There is a point $\xi$ in $\mathbb{C}^{n}$, fixed from now on, with the following property:

$(*)$ If $f: \mathbb{C}^{n} \rightarrow \mathbb{C}$ is a polynomial with coefficients in $k$, and $f(\xi)=0$, then $f(z) = 0$ for every $z$ in $\mathbb{C}^{n}$.

PROOF:
 Since $k$ is countable, there are only countably many polynomials with coefficients in $k$. The union of their zero-sets (ignoring the zero-polynomial) is thus a countable union of closed sets without interior, hence cannot cover the complete metric space $\mathbb{C}^{n}$. The lemma follows.

The proof shows that there exists a point $\xi\in \mathbb{C}^n$ that does not lie in the zero set of any nonzero polynomial $f\colon \mathbb{C}^n\to \mathbb{C}$ with coefficients in the (countable) field $k$. Thus if $f(\xi) = 0$ for some polynomial $f$ with coefficients in $k$, it follows that $f$ must be the zero polynomial, and hence $f\equiv 0$ on $\mathbb{C}^n$.
You did not specifically say where your confusion was, but hopefully this helps out!
