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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

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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:a2012-11-07
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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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$\newcommand{\ms}{\mathscr}\newcommand{\ext}{\operatorname{ext}}$Let $\ms{F}$ be a finite subset of $\ms U$, say $\ms F=\{\ext(a_1,b_1),\dots,\ext(a_n,b_n)\}$. Let $a=\max\{a_1,\dots,a_n\}$ and $b=\min\{b_1,\dots,b_n\}$; by hypothesis $a

$\qquad\qquad\qquad\qquad\qquad$if $x,y\in C$ and $x

This follows from the fact that $C$ is connected, though the details of the argument depend on just what definition of connectedness you’re using.