about noetherian topological space

Question

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 $(iii) \implies (iv)$.

Answer 1

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).

Answer 2

Take a pick on how to show the equivalence, they're all quite straightforward:

(i) implies (ii): Suppose $X$ is Noetherian and let $\mathcal{F}$ be a non-empty family of closed sets that has no minimal element (so we strive for a contradiction now). Let $F_0 \in \mathcal{F}$ (it's non-empty!). By assumption it's not minimal in $\mathcal{F}$, so there exists $F_1 \in \mathcal{F}$ such that $F_1 \subsetneq F_0$. If we have constructed (for some $m \ge 1$) $F_m \in \mathcal{F}$ such that $F_m \subsetneq F_{m-1}$, then by the non-minimality again we can pick $F_{m+1} \in \mathcal{F}$ such that $F_{m +1} \subsetneq F_m$, so we can construct by recursion a sequence $(F_n)_{n\ge 0}$ from $\mathcal{F}$ that is strictly decreasing by inclusion, and this directly contradicts $X$ being Noetherian. There is no need for full Zorn here (just DC, dependent choice, if we care about how much of AC we use).

(i) is equivalent to (iii) because we can take complements and turn an increasing sequence of open sets into a decreasing one of closed sets and vice versa.

(iii) implies (iv) by the same recursion argument but replacing non-minimal by non-maximal etc.

(iv) implies (iii) is easy: suppose that $(G_n)_n$ is an increasing by inclusion family of open sets. By (iv) $\{G_n\mid n \in \Bbb N\}$ has a maximal element $G_m$. But then if $n \ge m$, $G_m \subseteq G_n$ by increasingness and equality must hold by maximality of $G_m$. So the sequence stabilises at $m$ and (iii) is shown.

(ii) implies (i): exactly like (iv) implies (iii) with minimal instead of maximal.

(ii) and (iv) are equivalent by the same complements argument (complements reverse inclusions so maximal and minimal elements are interchanged).