If $f: X \rightarrow \mathbb{P}^n$ is a regular map and $U \subset X$ is constructible, then $f(U)$ is constructible.

Question

I'm reading "Algebraic Geometry: A First Course" by Joe Harris, and I find myself stuck on one part of his proof for the following theorem. I copied his proof below, and bolded the part that I'm stuck on.

Let $X \subset \mathbb{P}^m$ be a quasi-projective variety, $f: X \rightarrow \mathbb{P}^n$ a regular map, and $U \subset X$ any constructible set, then $f(U)$ is a constructible subset of $\mathbb{P}^n.$

Proof: The key step in the proof is to establish an a priori weaker claim: that the image $f(U)$ contains a nonempty open subset $V \subset \overline{f(U)}$ of the closure of $f(U).$ Given this, we set $U_1 = U \cap (X - f^{-1}(V))$ and observe that the theorem for $U$ follows from the theorem for $U_1.$ We then apply the claim to $U_1$ and define a closed subset $U_2 \subset U_1$ and so on; since the Zariski topology is Noetherian, a chain of strictly decreasing closed subsets of a constructible set is finite, and the result eventually follows.

I'm just not seeing what's going on right there. Why would proving the theorem for $U_1$ imply that the theorem is true for $U?$ There are a few equivalent definitions for constructible, so it's probably just that I don't understand all of them well enough.

EDIT:

A constructible set $Z \subset \mathbb{P}^n$ may be defined to be a finite disjoint untion of locally closed subsets $U_i \subset \mathbb{P}^n,$ that is, a set expressible as $$Z = X_1 - (X_2 - (X_3 - \dots - X_n))\dots)$$ for $X_n \subset \dots \subset X_3 \subset X_2 \subset X_1$ a nested sequence of closed subsets of $\mathbb{P}^n.$ Equivalently, we may define the class of constructible subsets of $\mathbb{P}^n$ to be the smallest class including open subsets and closed under the operations of finite intersection and complementation.

Answer 1

Hint: the biggest thing to do here is to figure out what the relationship between $f(U)$ and $f(U_1)$ is. The rest of this will be spoiler-tagged so you can go through it at your own pace but still have access to a full solution.

Warm up: Is $U_1$ constructible? Why?

Yes, it is - it's the intersection of the closed subset $X\setminus f^{-1}(V)$ with the constructible set $U$.

To begin: Write down a statement relating $f(U)$ and $f(U_1)$.

We have that $U_1=U\cap (X\setminus f^{-1}(V))$ for $V$ open. So $f(U_1)$ is $f(U)$ but without $V$. In symbols, this means $f(U_1)=f(U)\cap (Y\setminus V)$. What does that say about the constructibility of $f(U_1)$ in relation to $f(U)$?

Next:

If $f(U)$ is constructible, then $f(U)\cap (Y\setminus V)$ is too, but this is exactly $f(U_1)$. Conversely, if $f(U_1)$ is constructible, then $f(U)$ is the (disjoint) union of $V$ and $f(U_1)$, so $f(U)$ is constructible too. So $f(U_1)$ is constructible iff $f(U)$ is, assuming we can find $V$ as claimed.