Proposition 8

(i) Every subspace of a Noetherian space is Noetherian. (ii) Let $(A_i)_{i \in I}$ be a finite covering of a topological space $X$. If the subspaces $A_i$ of $X$ are Noetherian, $X$ is Noetherian.

(i) Let $X$ be a Noetherian space, $A$ a subspace of $X$ and $(F_i)$ a decreasing sequence of subsets of $A$ which are closed in $A$; then $F_i = \overline{F_n} \cap A$ and the closures $\overline{F_n}$ of the $F_i$ in $X$ form a decreasing sequence of closed subsets of $X$. As this sequence is stationary, so is the sequence $(F_i)$.

(ii) Let $(G_n)_{n \geq 0}$ be a decreasing sequence of closed subsets of $X$; by hypothesis, each of the sequences $(G_n \cap A_i)_{n \geq 0}$ is stationary. As $I$ is finite, there is an integer $n_0$ such that, for $n \geq n_0$, $G_n \cap A_i = G_{n_0} \cap A_i$, for all $i \in I$. But

and hence the sequence $(G_n)$ is stationary and $X$ is Noetherian.