A topological space $V$ is said to be Noetherian if $V$ satisfies the accending chain condition on open subsets: If $U_{1} \subseteq U_{2} \subseteq ...$ is an increasing chain of open subsets of $V$, then there is an $n$ with $U_{n} = U_{n+r}$ for each $r\geq 0$. Show that the following statements are equivalent:
(a) The space $V$ is Noetherian space.
(b) Any nonempty collection $\lbrace U_{\alpha}\rbrace$ of open subsets of $V$ has a maximal element; that is, there is a $U \in \lbrace U_{\alpha}\rbrace$ not properly contained in any other element of $\lbrace U_{\alpha}\rbrace$.
(c) The space $V$ satisfies the descending chain condition on closed sets: If $C_{1} \supseteq C_{2} \supseteq ...$ is a decreasing chain of closed subsets of $V$, then there is an $n$ with $C_{n} = C_{n+r}$ for each $r \geq1$.
(a) $\Rightarrow$ (b). Simply apply Zorn's Lemma to any chain ordered by inclusion. My question is: how can I ensure that it is possible to compare the elements by inclusion?
(b) $\Rightarrow$ (c). I don't have any idea.
(c) $\Rightarrow$ (a). $A \subset B \Longrightarrow B^{c} \subset A^{c}$ and if $A$ is open, $A^{c}$ is closed. This is enough?
I would like some hints, and corrections.
The ascending open chain and descending closed chain conditions are clearly equivalent:
If we assume the open one, let $F_1 \supseteq F_2 \supseteq F_3 \ldots$ be a descending chain of closed sets in $X$. Then ,as $A \subseteq B$ iff $X \setminus A \supseteq X\setminus B$ for all subsets of $X$, we have that $U_n := X\setminus F_n$ is an ascending chain of open sets of $X$ and so there is some $N$ with $U_n = U_N$ for all $ n \ge N$. So this also means that $F_n = X\setminus U_n = X\setminus U_N = F_N$ for all $n \ge N$.
To go from closed to open, we also take complements. This is a common theme in topology: any definition purely in terms of open sets often has an equivalent formulation in terms of closed sets, purely by applying complements. E.g. for compactness we get the formulation that a family of closed sets with the finite intersection property has non-empty intersection, as a sort of dual to the every open cover has a finite subcover formulation.
The maximal element argument is a bit subtler; it needs a bit more detail, IMHO:
Suppose $X$ is Noetherian, and let $\mathcal{U}$ be any non-empty family of open subsets of $X$.
Then $(\mathcal{U}, \subseteq)$ is a poset (it’s just a subset of the partial order of the power set by inclusion); this is always a partial order (we don’t need it to be a chain). $\mathcal{U}$ has a maximal element iff this poset has a maximal element. The existence of a maximal element is guaranteed if we can check Zorn’s condition in this poset: every chain has an upperbound. But this is trivial: by assumption any chain terminates at some finite stage, and this element is then the required upperbound. So the lemma of Zorn guarantees the existence of a maximal element and we’re done.
If $X$ obeys the maximality condition, it is Noetherian: if $U_n$ forms a chain of open subsets, this is a family that must have some $U_N$ that is maximal w.r.t inclusion. So for $n \ge N$ we have $U_N \subseteq U_n$ by being a chain, and this inclusion cannot be proper by maximality of $U_N$ so $U_n = U_N$.