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I'm struggling with the proof that says a Noetherian topological space $X$ is the finite union of closed irreducible subsets. In particular with this part:

First observe that every nonempty set of closed subsets of $X$ has a minimal element, since otherwise it would contain an infinite strictly descending chain.

I get that a chain $Y_1\supsetneq Y_2\supsetneq\ldots$ should terminate by the Noetherness of $X$. But why is it not possible to have a nonempty set of closed subsets, in which $Y_i\nsupseteq Y_j$ for all $i,j$?

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Suppose that we have a family of closed sets $\mathcal{F}$ without a minimal element. So pick $F_0 \in \mathcal{F}$. Then $F_1 \in \mathcal{F}$ exists such that $F_1 \subsetneq F_0$, as otherwise $F_0$ would have been minimal. As $F_1$ is not minimal either, some $F_2 \in \mathcal{F}$ exists with $F_2 \subsetneq F_1$ as well, and by recursion we have such a sequence $F_n$ in $\mathcal{F}$ with $F_{n+1} \subsetneq F_n$ for all $n$. But this sequence contradicts being Noetherian. So such a family cannot exists.

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  • $\begingroup$ But here you pick $F_1\subsetneq F_0$. My question is on wether we can always do so. Or equivalently: "Why can't we have $F_i\nsubseteq F_j$ for all $i,j$? $\endgroup$ – gebruiker May 30 '15 at 15:32
  • $\begingroup$ Like I said, otherwise $F_0$ would already be minimal. And minimal means that there is no proper subset, so not minimal means the opposite. $\endgroup$ – Henno Brandsma May 30 '15 at 15:34
  • $\begingroup$ Oh I see now. In my case the minimal element wouldn't be unique, but still minimal. Thank you for the help. $\endgroup$ – gebruiker May 30 '15 at 15:37
  • $\begingroup$ Graag gedaan hoor ! $\endgroup$ – Henno Brandsma May 30 '15 at 16:18

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