# If d.c.c. holds for irreducible closed sets is the topological space Noetherian?

We know that a topological space is said to be Noetherian if it satisfy d.c.c. on closed sets, i.e., every descending chain of closed subsets terminates. If in a topological space d.c.c. holds for irreducible closed sets is the topological space Noetherian ? ($Y \subset X$ is said to be irreducible if whenever $Y=Y_1 \cup Y_2$ where $Y_1$ and $Y_2$ are closed in $Y$ either $Y_1=Y$ or $Y_2=Y$)

I am not sure if it is true or not. Help me. Thanks.

No, this is not true. Indeed, most spaces have rather uninteresting irreducible closed sets, so that the descending chain condition on them is rather trivial. In particular, if $X$ is Hausdorff, the only irreducible subsets of $X$ are singletons (if $A\subseteq X$ and $x,y\in A$ are distinct, then they have disjoint neighborhoods $U$ and $V$, and then $A\setminus U$ and $A\setminus V$ are closed in $A$ and show $A$ is not irreducible). So any Hausdorff space satisfies the descending chain condition for irreducible closed subsets. However, most Hausdorff spaces are not Noetherian (for instance, the intervals $[0,1/n]$ are a descending sequence of closed subsets of $\mathbb{R}$).