Chapter 1, Exercise 1.7: Basic properties of noetherian spaces

(a) 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.

(b) A noetherian topological space is quasi-compact, i.e., every open cover has a finite subcover.

(c) Any subset of a noetherian topological space is noetherian in its induced topology.

(d) A noetherian space which is also Hausdorff must be a finite set with the discrete topology.

(a) We will verify that (i) is equivalent to (ii), that (iii) is equivalent to (iv), and that (i) is equivalent to (iii).

For the first equivalence, suppose $X$ is noetherian and let $\{X_\alpha\}$ be a family of closed sets that is nonempty. Pick a member $X_0\in\{X_\alpha\}$: if $X_0$ is minimal we are done. Otherwise, there is another set $X_1\in\{X_\alpha\}$ such that $X_1\subset X_0$. This process is iterated inductively. If no $X_i$ is minimal, then we construct a chain $\dots\subset X_2\subset X_1\subset X_0$ which does not stabilize, a contradiction. Conversely, if $X$ satisfies (ii) and $X_1\supset X_2\supset\dots$ is any descending chain, then the set $\{X_i\}$ has a minimal element and consequently the chain stabilizes.

The proof of the second equivalence is exactly the same, switching around the $\subset$ signs.

For the third equivalence, if $X$ is noetherian and $X_0\subset X_1\subset\dots$ is an ascending open chain in $X$, then $X_0^c\supset X_1^c\supset\dots$ is a descending closed chain which must stabilize, and since the complements stabilize so do the original sets. The other direction is identical, and the exercise is thus completed.

(b) Let $\{U_\alpha\}$ be any open cover for $X$ and consider the set $\Sigma$ of finite unions of the $U_\alpha$. Since $X$ is noetherian, $\Sigma$ has a maximal element $U=U_{\alpha_1}\cup\dots\cup U_{\alpha_r}$ by (a). For each $\alpha$, $U_\alpha\subseteq U$, and thus the elements of $U$ form a finite subcover.

(c) Equivalently by (a), we verify the ascending chain condition for open sets. Let $Y\subset X$ be arbitrary, and let $Y\cap U_0\subset U_1\subset\dots$ be an ascending chain of open sets in $Y$ where each $U_i$ is open in $X$. Then $U_0\subset U_1\subset\dots$ is an ascending chain in $X$ which stabilizes by assumption. Thus the intersection of this chain with $Y$ also stabilizes.

(d) Note that a noetherian space endowed with the discrete topology must be a finite set, since if $X$ was an infinite space with the discrete topology then we could trivially find non-stabilizing ascending chains of open subsets since every subset is open in $X$. Thus we need only show that a noetherian Hausdorff space has the discrete topology.

It is enough to show that each $Y\subset X$ is closed, or that $Y^c$ is open. For $y\in Y$, by Hausdorffness we can construct an open set $B_y$ such that the set of all $B_y$ as $y\in Y$ is an open cover for $Y$, and such that $B_y$ is disjoint from a neighborhood $C_x$ of any $x\in Y^c$. By (b) and (c), pick a finite subcover $\{B_{y_i}\}$ of $Y$. The $B_{y_i}$ correspond to subsets $C_{x_i}$ of $Y^c$. The set $C=\bigcup C_{x_i}$ is disjoint from $Y$, and contains all points of $Y^c$ by construction. Thus $Y^c=C$ is open.