Equivalent conditions for a topological space to be Noetherian

Question

Problem

Show that the following statements are equivalent:

(a) $X$ is a noetherian topological space

(b) Every non-empty family of closed subsets of $X$ has a minimal element.

(c) If $$U_1 \subset U_2 \subset ... \subset U_i \subset U_{i+1} \subset ...$$ is an increasing sequence of open subsets of $X$, then there is $i_0 \in \mathbb N$ such that $U_i=U_{i_0}$ for all $i \geq i_0$.

(d) Every non-empty family of open subsets of $X$ has a maximal element.

Attempt at a solution

It is very easy to see that $(a) \iff (c)$ and $(b) \iff (d)$. I've also proved $(b) \implies (a)$ and, using that implication, that $(d) \implies (c)$. This is the proof:

Suppose $(b)$ 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.

Now suppose it holds $(d)$ and take an arbitrary increasing sequence $\{U_n\}_{n \in \mathbb N}$ of open subsets. Let $U_j$ be the maximal element, if now we consider the decreasing sequence of closed subsets $\{F_n\}_{n \in \mathbb N}$ such that $F_n={U_n}^c$, then, since $(d) \implies (b) \implies (a)$, then there exists $i_0$ such that $F_i=F_{i_0}$ for all $i \geq i_0$, but then $U_i={F_i}^c={F_{i_0}}^c=U_{i_0}$ for all $i \geq i_0$.

I would appreciate some help to show $(a) \implies (b)$, $(c) \implies (d)$.

Answer 1

Hint for (c) $\implies$ (d): apply Zorn's Lemma to a given collection of open subsets ordered under inclusion. This gives that (a) $\implies$ (b) as well, since you've already shown (a) $\iff$ (c) and (b) $\iff$ (d).

A comment on your proof of (b) $\implies$ (a): take care with the definitions of maximality and minimality. You write

Suppose $(b)$ 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 $n$.

This is true, but only because $\{F_n\}_{n \in \mathbb N}$ is a totally ordered set under inclusion. Given a collection $\mathcal S$ of subsets ordered by containment, $A \in \mathcal S$ is minimal if for all $B \in \mathcal S$,

$$A \subset B \implies A=B. \,\,\, (*)$$

This is generally not the same as saying that $A \subset B$ for every $B \in \mathcal S$, since it might happen that $A \not\subset B$ and $B \not\subset A$, in which case $(*)$ still holds.