The Stacks project

Lemma 5.12.6. Let $X$ be a quasi-compact topological space. If $\{ Z_\alpha \} _{\alpha \in A}$ is a collection of closed subsets such that the intersection of each finite subcollection is nonempty, then $\bigcap _{\alpha \in A} Z_\alpha $ is nonempty.

Proof. We suppose that $\bigcap _{\alpha \in A} Z_{\alpha } = \emptyset $. So we have $\bigcup _{\alpha \in A} (X\setminus Z_{\alpha })=X$ by complementation. As the subsets $Z_\alpha $ are closed, $\bigcup _{\alpha \in A} (X \setminus Z_{\alpha })$ is an open covering of the quasi-compact space $X$. Thus there exists a finite subset $J \subset A$ such that $X = \bigcup _{\alpha \in J} (X\setminus Z_{\alpha })$. The complementary is then empty, which means that $\bigcap _{\alpha \in J} Z_{\alpha } = \emptyset $. It proves there exists a finite subcollection of $\{ Z_{\alpha }\} _{\alpha \in J}$ verifying $\bigcap _{\alpha \in J} Z_{\alpha }=\emptyset $, which concludes by contraposition. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 5.12: Quasi-compact spaces and maps

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Your email address will not be published.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 005D. Beware of the difference between the letter 'O' and the digit '0'.