# Finite sub cover for $(0,1)$

While learning topology one learns about compact set. The standard definition is:

• A set $X$ is said to be compact if open cover has a finite subcover.

Since $[0,1]$ is compact, if we take a open cover for this we should be able to get a finite subcover. I know, that $(0,1)$ is not compact, so there must exists some open cover for $(0,1)$ which doesn't admit any finite subcover. But how does one prove this fact?

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Generally, the nicest way to prove there exists some kind of something is just to give an example of such a something :-) – Peter LeFanu Lumsdaine Dec 1 '10 at 15:56

To prove that a set $X$ isn't compact is generally easy. You only need to give an example of a cover, that doesn't admit a finite subcover. To expand on Shai's answer.

Take

$\mathcal U := \{(0,n/(n+1)): n \in \mathbb N\}$

and observe that if you take any $U_1,U_2,\dots U_n \in \mathcal U$, then there exists some $k$ such that

$(0,k/(k+1))=\bigcup_{i=1}^n U_i.$

If this isn't obvious, note that for any two elements in $\mathcal U$ one is contained in the other. So if you take a finite number of them, then one of them will contain the rest. Furthermore since $(0,k/(k+1))$ does not contain $(0,1)$, $\mathcal U$ does not have a finite subcover, so $X$ is not compact.

I guess you should also show that $\mathcal U$ is a cover; I'll leave it as an exercise to the reader.

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Isn't this clear from $(0,1) = (0,1/2) \cup (0,2/3) \cup (0,3/4) \cup \cdots$?

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Let $A$ be a non-closed subset of a metric space, and let $x\in\overline A\setminus A$. Then $A$ is covered by open sets given by the complements of closed balls of positive radius centered at $x$, but not by any finite subcollection.

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Alternatively, use the fact that every compact subspace of a Hausdorff space is closed. Since (0,1) is not closed, it cannot be compact. (this is theorem 26.3 in Munkres)

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