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I believe that the following statement is true:

Let $E$ be a connected open subset of $\mathbb{R}^2$. For any $n$ distinct points in $E$, there exists a connected and simply connected open set $G \subset E$ that contains those points.

How can I show this? My original idea was to try to show this by induction, by arguing that there is a simple curve from the $n-1$-th point to the $n$-th point that doesn't intersect the set $G_{n-1}$ more than once.

But then, I realized I don't know how to show that given 2 points in an open connected set, there is a simple curve connecting the two. I thought that given any curve connecting two points, one could construct a simple curve, but I wasn't sure of how to deal with the case when there are infinitely many self-intersections.

Note: The statement I'm trying to show is equivalent to showing that given $n$ distinct points, there is a simple curve that connects the $n$ distinct points. But unfortunately I don't know how to show this either.

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"But then, I realized I don't know how to show that given 2 points in an open connected set, there is a simple curve connecting the two." Any path-connected Hausdorff space is arc-connected, meaning that any two distinct points lie in a subspace which is homeomorphic to the unit interval. – LostInMath Aug 11 '11 at 13:05
@LostInMath - Well, I guess that solves for the problem for $n=2$ :) – Braindead Aug 11 '11 at 13:22
@LostInMath - Where can I find a proof of this statement? – Braindead Aug 11 '11 at 16:20
It's much simpler to show that an open subset of euclidean space is connected iff it is path connected. This is not very hard at all. – Grumpy Parsnip Aug 11 '11 at 17:46
I've included a proof of the stronger statement that any two points can be joined by a PL arc. – Grumpy Parsnip Aug 11 '11 at 17:59
up vote 5 down vote accepted

Here's a suggestion. Connect the first two points by an arc, as LostInMath mentions. Now draw an arc from the third point to the second point. This might hit the first arc, so actually look at the first place it does, and forget the rest of the arc. This gives you a Y-shaped tree (or an interval if the arc makes it all the way to the second point). Repeat this process to get a tree joining all of the $n$ points. A tree is simply connected so will work just as well as a line going through all the points. Now take a small neighborhood of the tree to get a simply connected open set. To see that this neighborhood is contractible requires some technique, but to make things simpler, I claim that the arcs that we chose to make the tree can be assumed to be piecewise linear. That is, they can be assumed to be a finite union of straight line segments. It is not hard to show (see below) that any connected open set in Euclidean space has the property that any two points can be joined by such PL arcs. Now showing that a neighborhood of a tree comprised of PL arcs is contractible is not difficult.

Here is a proof of the PL connectedness statement. First note that whether two points are connected by a PL line is an equivalence relation. So we can partition the open set into equivalence classes under this relation. Now I claim that an equivalence class is open. This is because for any point $x$, I can find a ball centered around $x$ contained in the big open set. I can join $x$ to any other point in the ball by a radial line segment. Hence all points in the ball are in the same equivalence class. So an equivalence class is open. So by connectivity there can be only one equivalence class.

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I think that the construction of the simply connected neighborhood is still a non-trivial problem and begs for the rigorous proof. – Ilya Aug 11 '11 at 16:56
@Gortaur: you can triangulate your space so that the tree is a subcomplex. Now take a regular neighborhood. – user641 Aug 11 '11 at 17:33
@Gortaur: I agree that it is a nontrivial statement. – Grumpy Parsnip Aug 11 '11 at 17:47
@Steve: This now obviously works since I've proven the tree is made of PL arcs. – Grumpy Parsnip Aug 11 '11 at 18:00

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