Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a topological space with a fixed basis. Can we check the compactness of $X$ just by showing that every open covering using basis elements has a finite subcover? I have a problem showing it for the following reason:

Let $\{E_{\alpha}\}_{\alpha \in J}$ be any open cover of $X$. Then each $E_{\alpha}$ is union of basis elements, and so is $X$. Thus we can write $X = \cup_{j=1}^{n}B_{j}$ for finitely many basis elements $B_{j}$. But I want to have a finite subcover of the collection $\{E_{\alpha}\}_{\alpha \in J}$, and only with this argument, it is not necessarily true that $B_{j}$ are members of this collection.

share|improve this question
More is true: we can even use covers by elements from a subbase (a collection such that all finite intersections from it form a base). This is called Alexander's subbase lemma, and is quite useful.See en.wikipedia.org/wiki/Subbase#Alexander_subbase_theorem e.g. –  Henno Brandsma Feb 3 '13 at 13:21

2 Answers 2

up vote 2 down vote accepted

Replace every $E_\alpha$ with the set $\{B\mid B\subseteq E_\alpha, B\text{ is a basic open set}\}$. When you have obtained a finite subcover of basis elements $B_1,\ldots,B_n$ then you can choose $E_1,\ldots, E_n$ such that $B_i\subseteq E_i$. Now we have: $$X=\bigcup_{i=1}^n B_i\subseteq\bigcup_{i=1}^n E_i=X$$

Note that this method does not even require the axiom of choice, we only make finitely many arbitrary choices!

share|improve this answer

Yes, you can. Let $\mathscr{B}$ be a base for the topology on $X$, and let $\mathscr{U}$ be an arbitrary open cover of $X$. For each $U\in\mathscr{U}$ there is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$. Let $\mathscr{V}=\bigcup_{U\in\mathscr{U}}\mathscr{B}_U$; then $\mathscr{V}$ is an open cover of $X$ by basic open sets, and $\mathscr{V}$ refines $\mathscr{U}$. Let $\{V_1,\dots,V_n\}$ be a finite subcover of $\mathscr{V}$. For $k=1,\dots,n$ there is a $U_k\in\mathscr{U}$ such that $V_k\subseteq U_k$. Now $\{U_1,\dots,U_n\}$ is a finite subcover of $\mathscr{U}$.

More generally, it’s sufficient to show that every open cover has a finite open refinement. If $\mathscr{U}$ is an open cover of $X$, and $\mathscr{R}$ is a refinement of $\mathscr{U}$, then for each $R\in\mathscr{R}$ we can choose a specific $U_R\in\mathscr{U}$ such that $R\subseteq U_R$, and the family $\{U_R:R\in\mathscr{R}\}$ is then a subcover of $\mathscr{U}$ whose cardinality is at most that of $\mathscr{R}$. (It might be less, since it’s possible that the map $R\mapsto U_R$ is many-to-one.)

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.