In David Mumford's book Algebraic Geometry I, Complex Projective Varieties the proof of (3.25) Specialization Priciple on page 53 contains an argument I not understand.

General assumptions: all our varierties are over $\mathbb{C}$. The statement is:

(3.25) Specialization Priciple Let $Z \subset \mathbb{P}^n \times \mathbb{P}^m $ be $r$-dimensionally subvariety and $X =p_1(Z) \subset \mathbb{P}^n$ for $p_1: \mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^n $. Suppose $\operatorname{dim} X=r $ and let

$$\phi= \text{ res } p_1: Z \to X $$

is almost everywhere (= on an open set) finite to one. [... ]

 

Then the map

$$ F: X_1 \to \mathbb{N}, x \mapsto \# \phi^{-1}(x)$$

where $X_1 := \{x \in X \ \vert \ X \text{ smooth at } x \text{ and } \phi^{-1}(x) \text{ finite } \}$ is lower semi-continuous in the Zariski topology on $X_1$.

Lower semi-continuous means that for every $n \in \mathbb{N}$ the set $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is closed.

The first part of the proof shows that $ F: X_1 \to \mathbb{N}$ is lower semi-continuous in the classical topology, recall a smooth complex variety can be canonically endowed with classical analytical topology considering it as a complex manifold.

Mumfords remarks that we can observe that for every $n \in \mathbb{N}$ the set $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is constructable, ie a union of finite number of locally closed sets in Zariski topology.

Then Mumford claims that this implies that $F$ is lower semi-continuous in the Zariski topology.

That's the point I not understand. Does anybody see why the impliciation $F$ lower semi-continuous with resp classical topology and that $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is constructable imply $F$ lower semi-continuous with resp Zariski topology?

In David Mumford's book Algebraic Geometry I, Complex Projective Varieties the proof of (3.25) Specialization principle on page 53 contains an argument I not understand.

General assumptions: all our varieties are over $\mathbb{C}$. The statement is:

(3.25) Specialization principle. Let $Z \subset \mathbb{P}^n \times \mathbb{P}^m $ be $r$-dimensionally subvariety and $X =p_1(Z) \subset \mathbb{P}^n$ for $p_1: \mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^n $. Suppose $\operatorname{dim} X=r $ and let

$$\phi= \text{ res } p_1: Z \to X $$

is almost everywhere (= on an open set) finite to one. [... ]

Then the map

$$ F: X_1 \to \mathbb{N}, x \mapsto \# \phi^{-1}(x)$$

where $X_1 := \{x \in X \ \vert \ X \text{ smooth at } x \text{ and } \phi^{-1}(x) \text{ finite } \}$ is lower semi-continuous in the Zariski topology on $X_1$.

Lower semi-continuous means that for every $n \in \mathbb{N}$ the set $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is closed.

The first part of the proof shows that $ F: X_1 \to \mathbb{N}$ is lower semi-continuous in the classical topology, recall a smooth complex variety can be canonically endowed with classical analytical topology considering it as a complex manifold.

Mumford remarks that we can observe that for every $n \in \mathbb{N}$ the set $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is constructible, ie a union of finite number of locally closed sets in Zariski topology.

Then Mumford claims that this implies that $F$ is lower semi-continuous in the Zariski topology.

That's the point I not understand. Does anybody see why the implication $F$ lower semi-continuous with resp classical topology and that $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is constructible imply $F$ lower semi-continuous with resp Zariski topology?