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Let $\Omega\subset (0,1)$ be a nowhere dense set which has no lower and upper bound in $(0,1)$ and for which $\Omega^{d}\cap(0,1)=\Omega$ ($\Omega^{d}$ denotes here the set of all limit points of $\Omega$; let also $\Omega^{\pm d}$ denote the set of all two-sided limit points of $\Omega$). Is it true that then the set $(0,1)\backslash \Omega^{\pm d}$ is the set which is the union of the disjoint closed intervals $[a,b]$, where $a$ and $b$ are left-sided and right-sided limit points of $\Omega$, respectively? I think it's true but have no idea how to proceed with the proof. Thank you for any replies.

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I'm not sure it matters that $\Omega$ is nowhere dense. It seems more important that (a) every point in $\Omega$ is some kind of limit point, and (b) $(0, 1) \setminus \Omega$ is open, so it is a union of disjoint open intervals. What can you say about the end points of those intervals? –  Niels Diepeveen Jun 1 '12 at 1:24
    
I would say the end points of these intervals are just left-sided and right-sided limit points of $\Omega$. Am I right? If it were true, it would be the end of the proof. But how to show it? –  John Jun 1 '12 at 15:41
    
You are right. If for some $x \in \Omega$ there is an interval $(x, y)$ that does not intersect $\Omega$ then by definition $x$ is not a right-sided or two-sided limit point, so by (a) it must be a left-sided limit point. –  Niels Diepeveen Jun 2 '12 at 1:10
    
Now I see your point. Thank you for your tips. They were very helpful. –  John Jun 2 '12 at 18:35

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