TITLE: Why do we have to deal with constructible sets?
QUESTION [10 upvotes]: I'm a beginner in algebraic geometry. Recently, I learned about constructible sets and Chevalley's theorem. In a Noetherian space, constructible sets are some kind of finite union of open and closed sets, and Chevalley's theorem states that if $f : X \to Y$ is a morphism of finite type between Noetherian schemes, then the image of any constructible set $C \subset X$ is also constructible.
Is there any good application of the theorem? I can't understand why we have to consider constructible sets. It resembles Borel sets in measure theory (but the situation is completely different since topology is far from Hausdorff), so maybe we can think of constructible sets as some sort of topologically simple sets. But such intuition doesn't give any result.

REPLY [8 votes]: The first time I discovered the use of constructible sets was through linear algebraic groups. 
A linear algebraic group $G$ is an affine variety such that the composition map $m : G\times G \rightarrow G$ and inversion map $i : G\rightarrow G$ are regular maps. 
Suppose we have a map of linear algebraic groups $\phi : G \rightarrow H$ (i.e. a regular map which is a group homomorphism). Since $\phi$ is continuous, the kernel will be a closed subgroup of $G$. What about its image? By Chevalley's theorem, $\phi(G)$ will be constructible and obviously it is a subgroup of $H$. Now it's not too hard to observe that a constructible subgroup of a linear algebraic group is closed (see Humphreys book below). So the image of $\phi$ is in fact a closed subgroup of $H$, i.e. a linear algebraic group!
Reference: Humphreys' book on linear algebraic groups, section 7.4.