# Question about Noetherian topological spaces.

Theorem $$:$$

A Noetherian topological space $$X$$ has only a finitely many irreducible components. No component is contained in the union of the others.

I have tried to prove the first part of the theorem by the hint given by our instructor but I got stuck at last. Here's how I proceed to prove it.

Let $$M$$ be the set of all closed subsets of $$X$$ which cannot be written as a finite union of irreducible subsets of $$X.$$

Claim $$:$$ $$M = \varnothing.$$

If possible let $$M \neq \varnothing.$$ Then by the Noetherian property of $$X$$ we can say that $$M$$ has a minimal element. Call it $$Y.$$ Since $$Y \in M$$ it is not irreducible. So $$\exists$$ non-empty closed subsets $$Y_1,Y_2$$ of $$Y$$ with $$Y_i \neq Y$$ for $$i=1,2$$ such that $$Y=Y_1 \cup Y_2.$$ Since $$Y$$ is closed in $$X,$$ $$Y_i$$'s are closed in $$X$$ too. So by the minimality of $$Y$$ in $$M$$ we should have $$Y_i \notin M$$ for $$i=1,2.$$ So each $$Y_i$$ can be expressed as a finite union of irreducible subsets of $$X$$ and hence $$Y$$ can be expressed as a finite union of irreducible subsets of $$X,$$ a contradiction to the fact that $$Y \in M.$$ So $$M=\varnothing,$$ which proves our claim.

This implies that every closed subset of $$X$$ can be written as a finite union of irreducible subsets of $$X.$$ In particular $$X$$ can be written as a finite union of irreducible subsets of $$X.$$ Since every irreducible subset is contained in some irreducible component it follows that $$X$$ can be written as a finite union of irreducible components of $$X.$$ Let $$X = \bigcup\limits_{i=1}^{n} X_i$$ where each $$X_i$$ is a irreducible component of $$X$$ and $$X_i \neq X_j$$ for $$i \neq j.$$

Now let us take $$Y$$ to be any irreducible component of $$X.$$ Then $$Y = \bigcup\limits_{i=1}^{n} (X_i \cap Y).$$

At this moment our instructor claimed that $$Y=X_i \cap Y$$ for some $$1 \leq i \leq n.$$ But why does it imply that unless $$X_i \cap Y$$ are all closed subsets of $$Y$$ for all $$i$$? Because irreducible subsets $$Z$$ of $$X$$ are those which cannot be written as a union of two proper closed subsets of $$Z.$$

Please help me in this regard. Thank you very much.

## 1 Answer

The irreducible components of a Noetherian topological space are always closed.

This follows from two general facts. First, if $$X$$ is any topological space, and $$Y \subset X$$, then $$Y$$ is irreducible if and only if its closure is irreducible. Second, if $$X$$ is a Noetherian, and $$X_0 \subset X$$, then $$X_0$$ is an irreducible component of $$X$$ if and only if $$X_0$$ is maximal among the irreducible subsets of $$X$$.

• Exactly. I missed that $Y$ is an irreducible component of $X$, not merely an irreducible subset of $X.$ Because irreducible components are necessarily closed. Thanks @D_S for your kind help. – Dbchatto67 Feb 25 at 7:25