Let $P$ be certain property. Let $S \subset \mathbb{C}^n\times \mathbb{C}^m$ be a set of closed points such that for any point in $S$, it satisfies the property $P$. I know for any $x \in \mathbb{C}^n$, the set $(x, \mathbb{C}^m) \cap S$ is an open set (in Zariski topology, but may be empty) and for any $y \in \mathbb{C}^m$, the set $(\mathbb{C}^n,y)\cap S$ is an open set. Moreover, I know for any $s \in S$, there is an open set $U_s$ of **Euclidean topology** such that $s \in U_s \subset S$.

Can I conclude that when $S$ is nonempty, $S$ contains a nonempty open set (of Zariski topology).

This question comes from the intention to do Bertini type result for multipe linear systmes: Suppose for any variety with property $P$, then a general element of any basepoint free linear system also has property $P$. Let $X$ is a variety with property $P$, and $|L_i|, i=1,2$ be finite dimensional basepoint free linear systems, I what to say that $V = V_1 \cap V_2$ with $V_i \in |L_i|$ being the general elements also has property $P$. The difficulty is: when **fixed** $V_1$, one know that for general $V_2$, $V_1 \cap V_2$ has property $P$, but how to make both of $V_1, V_2$ being general?