Proof. Let $I$ be a set and for $i \in I$ let $X_ i$ be a quasi-compact topological space. Set $X = \prod X_ i$. Let $\mathcal{B}$ be the set of subsets of $X$ of the form $U_ i \times \prod _{i' \in I, i' \not= i} X_{i'}$ where $U_ i \subset X_ i$ is open. By construction this family is a subbasis for the topology on $X$. By Lemma 5.12.15 it suffices to show that any covering $X = \bigcup _{j \in J} B_ j$ by elements $B_ j$ of $\mathcal{B}$ has a finite refinement. We can decompose $J = \coprod J_ i$ so that if $j \in J_ i$, then $B_ j = U_ j \times \prod _{i' \not= i} X_{i'}$ with $U_ j \subset X_ i$ open. If $X_ i = \bigcup _{j \in J_ i} U_ j$, then there is a finite refinement and we conclude that $X = \bigcup _{j \in J} B_ j$ has a finite refinement. If this is not the case, then for every $i$ we can choose an point $x_ i \in X_ i$ which is not in $\bigcup _{j \in J_ i} U_ j$. But then the point $x = (x_ i)_{i \in I}$ is an element of $X$ not contained in $\bigcup _{j \in J} B_ j$, a contradiction. $\square$

## 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Your email address will not be published.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 08ZU. Beware of the difference between the letter 'O' and the digit '0'.