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A Zariski space is a topological space with the property that every descending chain $F_1\supset F_2\supset F_3\dots$ of closed sets is eventually constant. Show that every Zariski space can be expressed as a finite union $X=Y_1\cup Y_2\cup\dots$, where the $Y_i$ are closed and irreducible and $Y_i\not\subset Y_j$ for $i\neq j$.

What if the Zariski space is composed of an infinite number of disjoint closed sets? Even though the space may satisfy the property that the chain $F_1\supset F_2\supset F_3\dots$ of closed sets is eventually constant, we may still not be able to express the space as the finite union of closed sets, as infinite union of closed sets is not closed.

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    $\begingroup$ But if the union is the entire space, then the entire space is both closed and open in a trivial sense. $\endgroup$ – user99680 Jun 17 '14 at 6:39
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    $\begingroup$ Then the whole space is irreducible (as not the union of closed proper subsets). $\endgroup$ – martini Jun 17 '14 at 6:39
  • $\begingroup$ @user99680- Isn't the entire space anyway clopen in all topologies? How does this lead to a contradiction? $\endgroup$ – freebird Jun 17 '14 at 6:52
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This arguments is given shortly after we begin to study the algebraic geometry by Hartshorne, yet to restate the argument may be helpful for someone doesn't have access to the resource and questions for this.

Let $\mathcal{U}$ be a set of closed subsets of $X$ that cannot be written as a finite union of irreducible closed subsets.

Suppose $\mathcal{U}$ is non-empty, then it contains at least one minimal element $Y$ because of the descending chain condition of closed sets (i.e. for a descending chain of closed sets $\{F_j\}$, there is $r\geq 1$ s.t. $F_s=F_{s+1}$ for all $s\geq r$).

Since $Y$ is not irreducible, we can write $Y=Y_1\cup Y_2$ for some closed proper subsets $Y_1,Y_2\subseteq X$, where $Y_i$ can be written as a finite union of irreducible closed subsets of $X$ by the minimality of $Y$, thus $Y$ is, which is contradiction (of course, this expression is not unique at all, but reducibility implies that $Y$ at least contains closed proper subset, say $Y_1$. Putting $Y_2 = Y\cap\overline{Y-Y_1}$ gives a decomposition by two closed subsets for instance).

We now showed that Zariski space can be expressed as a finite union of irreducible closed subsets.

Now to show the uniqueness, suppose $X=\cup_{i=1}^r Y_i=\cup_{j=1}^s Y'_j$, where $Y_i, Y'_j$ are irreducible closed subsets such that each summand not to be contained to another for different indices. Since we can write

$$ Y_i = Y_i\cap X = \cup_{j=1}^s Y_i\cap Y'_j, $$

$Y_i=Y_i\cap Y'_j$ for some $j$ because of its irreducibility, and that $Y_i\subseteq Y'_j$. By reversing the arguments, we have $Y_k\supseteq Y'_j$ for some $k\leq s$. By definition and because $Y_i\subseteq Y_k$, $k=i$, which proceeds to the conclusion.

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This naturally begs the question if there are any spaces with the property you mentioned, that are also Zariski-spaces. But the general proof idea seems to work naturally for any kind of Zariski-space. Hint: If our space itself is irreducible, then we have such a finite expression at hand. If not, how could we proceed?

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