Decomposition into irreducibles of a Noetherian topological space 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 $i,j$?

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.
