Proposition 9

For a topological space $X$ to be Noetherian, it is necessary and sufficient that every open set in $X$ be quasi-compact.

To show that the condition is necessary, it is sufficient, by virtue of Proposition 8, to prove that every Noetherian space $X$ is quasi-compact. Let $(U_i)_{i \in I}$ be an open covering of $X$; the set of finite unions of sets $U_i$ is non-empty and hence admits a maximal element $V = \bigcup_{i \in H} U_i$, where $H$ is a finite subset of $I$. By definition, $V \cup U_i = V$ for all $i \in I$ and hence $V = X$.

Conversely, suppose that every open set of $X$ is quasi-compact and let $(U_i)$ be an increasing sequence of open subsets of $X$. The union $V$ of the $U_i$ is open and hence quasi-compact; as $(U_i)$ is an open covering of $V$, there is a finite subfamily of $(U_i)$ which is a covering of $V$ and hence $V = U_n$ for some index $n$, which proves that the sequence $(U_i)$ is stationary.