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Assume that $I$ is a compact interval in $\mathbb R$. Does the following statement (I hope true) follows from compactness or from connectedness of $I$?

For arbitrary family of open in $\mathbb R$ subsets $U_s$, $s \in S$, if $I\subset \bigcup_{s \in S} U_i$ then there are $n \in \mathbb N$ and $s_1, \ldots, s_n\in S$ such that $I\subset U_{s_1} \cup... \cup U_{s_n}$, and $U_{s_i} \cap U_{s_{i+1}}\neq \emptyset$ for $i=1,...,n-1$.

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This follows from compactness. Per definition: If $A$ is a compact set, then there exists for every open cover $(U_i)_{i\in I}$ (with I an arbitrary set) of $A$ a finite subset $J\subseteq\ I$ such that $U_{i_1},\cdots U_{i_n}$ covers whole $A$. The second condition follows from the fact that the sets $U_i$ are open.

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No, the second condition does not follow from compactness of $I$ and openness of the $U_i$, you need connectedness there. Otherwise you run into examples like $[1,2] \cup [4,5] \subset (0,3) \cup (3,6)$ and $(0,3) \cap (3,6) = \emptyset$. –  Lukas Geyer Oct 29 '12 at 14:19
    
Also, note that the second condition requires that you order the sets just right. I think you need an argument to show that this can always be done. –  Nate Eldredge Oct 29 '12 at 14:22
    
Oh yes true, sry i thought only over intervals form the form $[a,b]$ with a<b because then i follows from the fact that the covers are open. Thank you for correction :) –  Model of the Theory Oct 29 '12 at 14:23
    
Yeah, you can order the intervals in ascending order w.r.t. their left endpoints, and throw out intervals that are contained in other intervals, then the conditions will be satisfied. It is a little messy to write up, though, so I'll let you do it. :) –  Lukas Geyer Oct 29 '12 at 16:53
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