The Stacks project

Example 5.8.10. Let $Y$ be an infinite set, furnished with the topology whose closed sets are $Y$ and the finite subsets of $Y$. Then $Y$ is Kolmogorov but not quasi-sober. However, the family of its singletons (which are its irreducible components) is a covering by discrete and hence sober spaces.

Comments (0)

There are also:

  • 8 comment(s) on Section 5.8: Irreducible components

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Your email address will not be published.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 0B33. Beware of the difference between the letter 'O' and the digit '0'.