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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.

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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 $\mathbb{R}$).

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  • $\begingroup$ Very good answer, many thanks. $\endgroup$ – user371231 Jan 31 '18 at 20:24

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