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I'm familiar with this concept and it makes sense, but the proof for it is eluding me.

Let $p \in C$ and consider the set:

$\mathcal{U}=\{\operatorname{ext}(a,b)\mid p\in (a,b)\}$

Therefore no finite subset of $\mathcal{U}$ covers $C \setminus \{p\}$.

It makes sense that a finite number of exteriors will never cover the continuum $C$ ($C$ being nonempty, having no first or last point, ordered ($a), and connected) without $p$ since $p$ will be in exactly none of the subsets $\mathcal{U}' \subset \mathcal{U}$. I'm not sure if that's enough (or maybe even true) though. If anyone can point me in the right direction it will be very much appreciated. (Also, note that $\operatorname{ext}(a,b)$ is the same as $C \setminus [a,b]$. This clarification is simply based on notation.)

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    What do you mean by $\operatorname{ext}(a,b)$? $C\setminus(a,b)$?2012-11-07
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    I mean $C \setminus ((a,b) \cup {a} \cup {b})$. Sorry for not clarifying beforehand.2012-11-07
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    Same thing: $a$ and $b$ are elements of $C\setminus(a,b)$ already, since $(a,b)=\{x\in C:a.2012-11-07
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    Oh, but I meant to exclude them. That's why I put the parenthesis out. I mean to say that the exterior of $(a,b)$ won't include $a$ or $b$.2012-11-07
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    Sorry, I misread you previous comment. What you want, in more standard notation, is $C\setminus[a,b]$.2012-11-07
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    Oh, yes, hard brackets means the interval contains its endpoints. Sorry for the confusion.2012-11-07
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    No problem. I’ve posted a partial answer; if you need more, you’ll have to say exactly what your definition of *connected order* is, but I think that I’ve given a fairly good push in the right direction.2012-11-07
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    We shouldn't have to read the comments to understand the question. Please edit the body of the question to reflect the clarifications made.2012-11-07
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    I edited it. Thank you very much for the suggestion.2012-11-07

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