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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.

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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
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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!

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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.)

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