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?