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Let $X$ be a topological space with a finite basis $B$, and let $S$ be an open cover of $X$. Then every $U\in S$ is a finite union of members of $B$.

Let $S'\subseteq S$ be a subcover such that members of $S'$ are disjoint. Then $S'$ is finite. Hence X is compact.

How can I show that $S'$ exists? And why is $S'$ finite?

I guess it'd also work if I can show that $S\subseteq B$.

Sid Caroline
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    $S'$ doesn't necessarily exist. For instance, if $X = {1, 2, 3}$ and the basis is $B = {\emptyset, {2}, {1, 2}, {2, 3}, {1, 2, 3}}$ (in fact, that's the entire topology) and $S = {{1, 2}, {2, 3}}$, then there is no disjoint subcover. – Arthur Sep 11 '17 at 10:15

2 Answers2

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By the definition every open subset of $X$ can be written as a union of some subcollection from $B$. Now $B$ is finite so there is finitely many subcollections of $B$ and therefore $X$ has finitely many open subsets. In other words the topology $\tau$ on $X$ is a finite set.

So if we take any collection $S$ of open subsets (it doesn't even have to be a cover, any collection) then it already is finite! Simply because $S$ is a subset of $\tau$ which is finite. So just put $S'=S$ and you're done.

Side note: you typically don't assume that members of the (sub)cover are disjoint. This is not required for compactness. And indeed in most cases (e.g. $X$ connected) this can't even happen.

freakish
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    This answer is shorter and easier to follow than mine. But if I'm allowed to be a bit haughty, my answer is still better because it uses the finiteness of the basis directly instead of pointing out the intermediate result that the topology is finite (I don't really mean that, but I'm a bit miffed that I didn't think of it, so I have to say something). – Arthur Sep 11 '17 at 10:31
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For each point $x$ in $X$, pick an open neighbourhood $U_x\in S$ of $x$. By definition of basis, for each $x$ and corresponding $U_x$, there is an open basis neigbourhood $V_x\subseteq U_x$ of $x$ (i.e. $V_x\in B$).

There are only finitely many elements in the basis. Therefore, only finitely many distinct basis elements could have been chosen. Pick points $x_1,x_2,\ldots,x_n \in X$ such that $V_{x_1}, V_{x_2}, \ldots, V_{x_n}$ are all the chosen basis elements. Now $S' = \{U_{x_1}, U_{x_2}, \ldots,U_{x_n}\}$ is a finite subset of $S$.

Take an arbitrary $x\in X$. Then there must be an $x_i$ such that $V_x = V_{x_i}$. Since $x\in V_x = V_{x_i} \subseteq U_{x_i}$, $x$ is covered by $S'$. This shows that $S'$ covers all of $X$, and we are done.

Arthur
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