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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.

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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 $X$.

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  • $\begingroup$ 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. $\endgroup$ – Dbchatto67 Feb 25 at 7:25

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