This intersection has at most 2 components

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

-
I take it that $U$ and $V$ are subsets of $\Bbb R$? Because it’s not true without some such restriction. – Brian M. Scott Oct 18 '12 at 15:20
@BrianM.Scott I think yes, U and V are subsets of $\mathbb R$, in fact I'm studying this classification theorem for 1-manifolds: igt.uni-stuttgart.de/eiserm/lehre/2010/Topologie/… and I'm stuck in a corolary. – user42912 Oct 18 '12 at 15:28
Do you really mean union? Because for intervals one can get cases where union has two components, and other cases where union has one component (overlap of intervals). If it's intersection as you have it, then you can't get two components anyway. – coffeemath Oct 18 '12 at 15:30
Let the two intervals be $(a,b),(c,d)$ (since sets homeomorphic to $(0,1)$ or $(0,1]$ must be connected, and a connected subset of the reals is an interval). To save cases I'm using round brackets everywhere, but one can keep in mind that one or two of the round brackets could be replaced by square ones, in various orders. I of course am also assuming $a<b$ and $c<d$ since otherwise the intervals are empty. The "point" intervals $[c,c]$ are excluded as the initial intervals by your homeomorphic assumption, however one such might appear as an intersection, for example with $(-1,0]\cap[0,1)$. – coffeemath Oct 18 '12 at 15:30
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 $-\infty<c<+\infty$. You'll find that in all cases the intersection has only one component, or else is empty ($0$ components). – coffeemath Oct 18 '12 at 15:30

1 Answer

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?

-
I thought (since the "attempt outlined in the question mentioned looking at the real line), that the space was the set of reals. As B. Scott's remark shows, intersection can have 2 components on the 1-manifold of the unit circle. And of course for higher dimensional manifolds there can be any number of components. – coffeemath Oct 18 '12 at 16:21
yes, I proved the proposition 1, but differently. I proved saying that since ψ and ϕ are homeomorphisms, then ϕ(W) and ψ(W) are connected. The connected sets in the real line are the open intervals, thus ϕ(W) and ψ(W) has to be open intervals. – user42912 Oct 18 '12 at 18:18
I tried to do the same strategy to prove the corollary, but it didn't work. – user42912 Oct 18 '12 at 18:21
@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. – Brian M. Scott Oct 18 '12 at 19:44