Decomposition of Noetherian space into irreducible subsets I am trying to relate two (maybe not) different decompositions of a noetherian topological space into irreducible subsets, given in Ravi Vakil's notes on algebraic geometry.

Exercise 4.6.N : Let $X$ be a topological space, then any point is contained in a an irreducible component.

It follows that any space $X$ is the union of its (closed) irreducible components, but there is not statement of uniqueness.

Proposition 4.16.14 : Let $X$ be a noetherian topological space, and $Z \subseteq X$ a closed (non-empty) subset. Then there is a
  unique decomposition $Z = Z_1 \cup \cdots \cup Z_n$ where the $Z_i$'s
  are irreducible closed subsets, none containing another.

So in particular $X$ is a finite union of irreducible closed sets, but here it seems like they may not be its components.
The exercise is just a small exercise, while the proposition seems to be more important. So my question is : Why is the decomposition in the proposition more important than the one in the exercise ?
Is it because of the uniqueness statement ? But I think the decomposition of $X$ into its irreducible closed components also satisfies this uniqueness condition.
Is it because it applies to any closed subset $Z$ ? But $Z$ also has the decomposition of the exercise. However, the subsets are irreducible in $Z$ and not in $X$, so this is the point ?
Thank you for your help.
 A: I think the most important thing about the proposition is the finiteness part: any Noetherian topological space can be written as a finite union of irreducible closed subsets.  
This is not true of, say, $\mathbb{R}$ with the usual topology. In fact, the only irreducible closed subsets of $\mathbb{R}$ are the singletons, so while it is true that $\mathbb{R}$ is a union of its irreducible closed sets (points), it is an uncountable union of these sets. No information about the topology of $\mathbb{R}$ is gained by knowing this decomposition.
If you want to think about things algebraically, you probably know that in algebraic geometry most Noetherian spaces you deal with are spaces that are (at least locally) given by the spectrum of a Noetherian ring. The Noetherian condition for rings is important in commutative algebra because it is a finiteness condition, saying that the ring can't be too crazy. It then shouldn't be surprising that the usefulness of Noetherianity on the geometry side of things is that it imposes a finiteness condition on the spaces you deal with. 
The uniqueness part of the proposition is also important, of course, but once you know that a Noetherian space can be written as a finite union of irreducible closed subsets, it doesn't take any deep reasoning to derive the uniqueness. 
