Lemma 5.12.15 (Alexander subbase theorem). Let $X$ be a topological space. Let $\mathcal{B}$ be a subbase for $X$. If every covering of $X$ by elements of $\mathcal{B}$ has a finite refinement, then $X$ is quasi-compact.

**Proof.**
Assume there is an open covering of $X$ which does not have a finite refinement. Using Zorn's lemma we can choose a maximal open covering $X = \bigcup _{i \in I} U_ i$ which does not have a finite refinement (details omitted). In other words, if $U \subset X$ is any open which does not occur as one of the $U_ i$, then the covering $X = U \cup \bigcup _{i \in I} U_ i$ does have a finite refinement. Let $I' \subset I$ be the set of indices such that $U_ i \in \mathcal{B}$. Then $\bigcup _{i \in I'} U_ i \not= X$, since otherwise we would get a finite refinement covering $X$ by our assumption on $\mathcal{B}$. Pick $x \in X$, $x \not\in \bigcup _{i \in I'} U_ i$. Pick $i \in I$ with $x \in U_ i$. Pick $V_1, \ldots , V_ n \in \mathcal{B}$ such that $x \in V_1 \cap \ldots \cap V_ n \subset U_ i$. This is possible as $\mathcal{B}$ is a subbasis for $X$. Note that $V_ j$ does not occur as a $U_ i$. By maximality of the chosen covering we see that for each $j$ there exist $i_{j, 1}, \ldots , i_{j, n_ j} \in I$ such that $X = V_ j \cup U_{i_{j, 1}} \cup \ldots \cup U_{i_{j, n_ j}}$. Since $V_1 \cap \ldots \cap V_ n \subset U_ i$ we conclude that $X = U_ i \cup \bigcup U_{i_{j, l}}$ 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Your email address will not be published.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.

## Comments (0)

There are also: