The Stacks project

Lemma 5.9.3. Let $f : X \to Y$ be a continuous map of topological spaces.

  1. If $X$ is Noetherian, then $f(X)$ is Noetherian.

  2. If $X$ is locally Noetherian and $f$ open, then $f(X)$ is locally Noetherian.

Proof. In case (1), suppose that $Z_1 \supset Z_2 \supset Z_3 \supset \ldots $ is a descending chain of closed subsets of $f(X)$ (as usual with the induced topology as a subset of $Y$). Then $f^{-1}(Z_1) \supset f^{-1}(Z_2) \supset f^{-1}(Z_3) \supset \ldots $ is a descending chain of closed subsets of $X$. Hence this chain stabilizes. Since $f(f^{-1}(Z_ i)) = Z_ i$ we conclude that $Z_1 \supset Z_2 \supset Z_3 \supset \ldots $ stabilizes also. In case (2), let $y \in f(X)$. Choose $x \in X$ with $f(x) = y$. By assumption there exists a neighbourhood $E \subset X$ of $x$ which is Noetherian. Then $f(E) \subset f(X)$ is a neighbourhood which is Noetherian by part (1). $\square$


Comments (2)

Comment #1362 by on

The indexing for the chain is a bit silly, on both occasions.

There are also:

  • 8 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Your email address will not be published.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 04Z8. Beware of the difference between the letter 'O' and the digit '0'.