Definition. A topological space $X$ is called noetherian if it satisfies the descending chain condition for closed subsets: for any sequence $Y_{1} \supseteq Y_{2} \supseteq \ldots$ of closed subsets, there is an integer $r$ such that $Y_{r}=Y_{r+1}=\ldots$.
Show that the following conditions are equivalent for a topological space $X$ :
(i) $X$ is noetherian;
(ii) every nonempty family of closed subsets has a minimal element;
(iii) $X$ satisfies the ascending chain condition for open subsets;
(iv) every nonempty family of open subsets has a maximal element.
proof : i can show $(i) \iff (iii)$ (because complement of open set is close and if $A \subset B \implies B^c \subset A^c $ ) and $(ii) \implies (i)$ :
Suppose $(ii)$ holds and let $\{F_n\}_{n \in \mathbb N}$ be a sequence of decreasing closed subsets. By hypothesis, there is $k \in \mathbb N$ such that $F_k \subset F_n$ for all $n$. But since the sequence is decreasing, we have $F_n \subset F_k$ for all $n \geq k$. From here it follows directly $X$ is noetherian.
I think we can use Zorn's Lemma for $(iii) \implies (iv)$.
Best Answer
Your proof that (ii) implies (i) isn’t quite right: the hypothesis guarantees only that $\{F_n:n\in\Bbb N\}$ has a minimal element, not that it has a minimum element. Thus, there is a $k\in\Bbb N$ such that no $F_n$ is a proper subset of $F_k$. But we know that $F_n\subseteq F_k$ for $n\ge k$, so we must have $F_n=F_k$ for $n\ge k$.
To show that (i) implies (ii), suppose that $\mathscr{F}$ is a family of closed subsets of $X$ with no minimal element. Then for each $F\in\mathscr{F}$ there is an $F'\in\mathscr{F}$ such that $F'\subsetneqq F$, and you can recursively construct a strictly decreasing infinite sequence of closed sets.
You can then prove the equivalence of (ii) and (iv) the same way that you proved the equivalence of (i) and (iii).