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Let $U$ and $V$ two sets homeomorphic either to the open interval (0,1) or the half-open interval [0, 1), then their intersection has at most two components.

My attempt was show that the open connect sets in the real line are the open intervals ones. The problem is the intersection of open subsets has at most one component. Anyone can help me in this part?

Thanks

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    Now consider the possible orderings of the four points $a,b,c,d$ on the number line. (as usual an endpoint can be $+\infty$ or $-\infty$ on the intervals, and we use the usual convention for finite $c$ that -\infty. You'll find that in all cases the intersection has only one component, or else is empty ($0$ components).2012-10-18

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Seeing the paper changes things completely. $U$ and $V$ are not (necessarily) subsets of $\Bbb R$; they’re subsets of a manifold $X$ that are homeomorphic (via maps $\varphi$ and $\psi$, respectively) to $(0,1)$ or $[0,1)$. And yes, $U\cap V$ can have two components: let $X$ be the unit circle in the plane, let $U$ be the set of points of $X$ whose polar coordinates are $\langle 1,\theta\rangle$ for $-\frac{\pi}4<\theta<\frac{5\pi}4$, and let $V$ be the set of points of $X$ whose polar coordinates are $\langle 1,\theta\rangle$ for $\frac{3\pi}4<\theta<\frac{9\pi}4$. Then the intersection of $U$ and $V$ consists of two disjoint open arcs of $X$.

Have you tried to prove Proposition 1 following the hint given?

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    @user42912: The connected sets in $\Bbb R$ are the intervals; they don’t have to be open. And showing that $\varphi[W]$ and $\psi[W]$ are open intervals isn’t enough anyway: you have to show that they are **outer** intervals.2012-10-18