The Stacks project

Lemma 5.9.4. Let $X$ be a topological space. Let $X_ i \subset X$, $i = 1, \ldots , n$ be a finite collection of subsets. If each $X_ i$ is Noetherian (with the induced topology), then $\bigcup _{i = 1, \ldots , n} X_ i$ is Noetherian (with the induced topology).

Proof. Let $\{ F_ m\} _{m \in \mathbf{N}}$ a decreasing sequence of closed subsets of $X' = \bigcup _{i = 1, \ldots , n} X_ i$ with the induced topology. Then we can find a decreasing sequence $\{ G_ m\} _{m \in \mathbf{N}}$ of closed subsets of $X$ verifying $F_ m = G_ m \cap X'$ for all $m$ (small detail omitted). As $X_ i$ is noetherian and $\{ G_ m \cap X_ i\} _{m \in \mathbf{N}}$ a decreasing sequence of closed subsets of $X_ i$, there exists $m_ i \in \mathbf{N}$ such that for all $m \geq m_ i$ we have $G_ m \cap X_ i = G_{m_ i} \cap X_ i$. Let $m_0 = \max _{i = 1, \ldots , n} m_ i$. Then clearly

\[ F_ m = G_ m \cap X' = G_ m \cap (X_1 \cup \ldots \cup X_ n) = (G_ m \cap X_1) \cup \ldots (G_ m \cap X_ n) \]

stabilizes for $m \geq m_0$ and the proof is complete. $\square$

Comments (0)

There are also:

  • 6 comment(s) on Section 5.9: Noetherian topological spaces

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0053. Beware of the difference between the letter 'O' and the digit '0'.