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One calls a topological space noetherian if every decreasing chain of closed subsets becomes stationary.

In a lecture it was remarked that this is equivalent to every family $M$ of closed sets having minimal elements. However to me it looks like the direction

$X$ noetherian $\implies$ every family of closed sets $M$ has minimal element

requires use of Zorns Lemma: Every decreasing chain in $M$ must become stationary and thus be bounded, from Zorns Lemma it follows that a minimal element exists.

In the exercises we also showed that noetherian $\iff$ every open set is quasi-compact. Also here my proof of the $\implies$ direction used the Axiom of Choice.

My question: Are these statements actually dependent of axiom of choice or are my proofs just bad?

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  • $\begingroup$ I don't know if this is a duplicate, but it is certainly related: math.stackexchange.com/questions/1857112/… and there might be a few others. $\endgroup$ – Asaf Karagila Nov 2 '16 at 15:07
  • $\begingroup$ Without checking, I'd guess that given a counterexample of a Noetherian ring without choice, you can look at its spectrum to get a counterexample for Noetherian spaces without choice. $\endgroup$ – Asaf Karagila Nov 2 '16 at 15:18
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Without the axiom of choice we can take $X$ to be a strictly amorphous set and give it the discrete topology. Let $\mathscr{F}$ be the family of cofinite subsets of $X$. $\mathscr{F}$ is certainly a family of closed subsets of $X$, but it has no minimal element: given $F\in\mathscr{F}$, $F\setminus\{x\}\in\mathscr{F}$ for each $x\in F$.

On the other hand, suppose that $\{C_n:n\in\omega\}$ is a strictly decreasing chain of subsets of $X$. For $n\in\omega$ let $D_n=C_n\setminus C_{n+1}$; then

$$\{D_n:n\in\omega\}\cup\left\{\bigcap_{n\in\omega}C_n\right\}$$

is a partition of $X$ into finite sets, so $N=\{n\in\omega:|D_n|=1\}$ is infinite. Let $N=\{n_k:k\in\omega\}$ be the natural increasing enumeration of $N$; then the function from $\omega$ to $X$ that takes $k$ to the unique element of $D_{n_k}$ is an injection, which is impossible, since $X$ is Dedekind finite. Thus, every decreasing chain of closed subsets of $X$ stabilizes.

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  • $\begingroup$ Wow I have never heard of strictly amorphous sets, that is crazy. $\endgroup$ – s.harp Nov 5 '16 at 11:57
  • $\begingroup$ @s.harp: They are certainly pretty counter-intuitive! $\endgroup$ – Brian M. Scott Nov 5 '16 at 15:30

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