- Show that the following are equivalent for a topological space \(X\):
\(X\) is Noetherian;
every nonempty family of closed subsets has a minimal element;
\(X\) satisfies the ascending chain condition on open subsets;
every nonempty family of open subsets has a maximal element.
A Noetherian topological space is quasicompact.
Any subset of a Noetherian topological space is Noetherian in its induced topology.
A Noetherian space which is also Hausdorff is a finite set with the discrete topology.
Solution
(a) \(1 \implies 2\): given a family \(\ts{Y_j}\) of closed sets, choose \(Y_1\); if it is not minimal there is \(Y_2 \subsetneq Y_1\), and so on. The resulting descending chain stabilises, and where it stabilises is a minimal element. \(1 \iff 3\) and \(2 \iff 4\) are complementation. \(4 \implies 3\): a chain of opens is a family, so it has a maximal element, which is where it stabilises.
(b) Let \(\mcu\) be an open cover and set \(V_n = \Union_{i \leq n} U_i\). Then \(V_1 \subseteq V_2 \subseteq \cdots\) is an ascending chain of opens, so it stabilises at some \(V_n\), and \(V_n = \Union_i U_i = X\) is a finite subcover.
(c) Let \(U_1 \subseteq U_2 \subseteq \cdots\) be opens in \(Y\), so \(U_i = V_i \intersect Y\) with \(V_i\) open in \(X\). The family \(\ts{V_i}\) has a maximal element \(V_n\), and then \(U_n = V_n \intersect Y\) is maximal in the chain.
(d) Decompose \(X = \Union_{i \leq n} C_i\) into irreducible components. Each \(C_i\) is Hausdorff, and an irreducible Hausdorff space is a point: if \(x \neq y\) in \(C_i\) had disjoint neighbourhoods \(U \ni x\), \(V \ni y\), then \(U\) would be a nonempty open that is not dense, contradicting irreducibility. So \(X\) is a finite union of closed points, hence finite and discrete.
Source Collections
Dependencies
D-9DIKBIrreducible and Noetherian spacesuses