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Let $U$ be an open and connected subspace of the Euclidean plane $\mathbb{R}^2$ and $A\subseteq U$ a subspace which is homeomorphic to the closed unit interval. Is $U\setminus A$ necessarily connected?

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Is $A$ homeomorphic to the closed unit interval $[0,1]$? –  Arturo Magidin Aug 11 '11 at 16:56
Hmm. Here's an idea, but I'm not sure if it works. $A$ is compact, so it necessarily has positive distance to $U^c$; thus it should be possible to travel "along the boundary" of $U$ between any two points (but I'm not sure about this). So it suffices to show that any point can escape to the boundary while avoiding $A$, but this follows from the Jordan curve theorem. –  Qiaochu Yuan Aug 11 '11 at 17:00
@Lost: subspace (twice) in the sense of subset? –  Did Aug 11 '11 at 17:17
@Arturo: Yes, I meant the closed unit interval. I added the word "closed" to the question. –  LostInMath Aug 11 '11 at 17:31
This is really really close to the Jordan curve theorem. I'd guess that, like the Jordan curve theorem, it's true but hard to prove. I suspect that a Mayer-Vietores argument would work; something like the proof of the Jordan curve theorem one can find in Hatcher. –  MartianInvader Aug 11 '11 at 17:42
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Every subset $A$ of $\mathbb{R}^{2}$ homeomorphic to $[0,1]$ is "tame", that is, there is a self-homeomorphism of the plane $\varphi$, such that $\varphi(A)=[0,1]$ (citation needed :)). Then it follows that $A$ may be represented as an intersection $A=\cap_{i=1}^{\infty}D_{i}$ of a decreasing sequence of (closed) topological disks, so we have $D_{i}\subset U$ for some $i$ (sufficiently large). Now, to show that $U\backslash A$ is connected, let $x,y\in U\backslash A$ and $\gamma$ be a topological arc in $U$ connecting $x$ and $y$; then the set $L=(\gamma\backslash D_{i}^{0})\cup\partial D_{i}$ is connected. To prove this, let $L=L_{1}\cup L_{2}$ where $L_{1}$, $L_{2}$ are disjoint open subsets of $L$, then since $\partial D_{i}$ is connected, we may suppose that $\partial D_{i}\subset L_{1}$, but then each component $K$ of $\gamma\backslash D_{i}^{0}$ is also contained in $L_{1}$ ($K$ is intersecting $\partial D_{i}$ as$\ \gamma$ is connected!); thus $L=L_{1}$, so $L$ is connected. Now, since $L\subset U\backslash A$, it follows that any two points of $U\backslash A$ are contained in a connected set and therefore $U\backslash A$ is connected as well.

p.s. [Here $D_{i}^{0}$ is the interior of $D_{i}$.]

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Regarding the needed citation: mathoverflow.net/questions/57766/… –  Grumpy Parsnip Aug 11 '11 at 20:59
@symbo'leon: Thank you for the answer. Minor correction: choose suitable $D_i$ after picking $x,y\in U\setminus A$ to guarantee that $x,y\notin D_i^o$. The key fact for the proof seems to be that we are able to enclose $A$ inside a homeomorph of a disc contained in the open set $U$. I looked at the MO link provided by Jim Conant where they discuss tameness of arcs in the plane and began to wonder whether there is an elementary proof of this key fact or is it one of those results which is easy to believe but hard to prove... –  LostInMath Aug 11 '11 at 23:00
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Such an arc $A$ must necessarily be in the interior of $U$ and compact so we can find finitely many open $\epsilon$-balls, $U_1,U_2,\dotsc,U_n$ that cover the arc and whose closures are contained in $U$.

Recall that an open connected subset of Euclidean space is path connected. It should now be clear that $U \setminus \left(\cup U_i\right)$ is path connected (follow your original path until you hit the boundary if the $U_i$. Then follow it around clockwise or counterclockwise whichever you prefer until you hit the point where your path left them for the last time and follow the remaining bit of your original path). Hence it suffices to show the result for the open and simply connected set $\cup U_i$.

By the Riemann mapping theorem $\cup U_i$ is homeomorphic to $\mathbb{R}^2$. We finish off by applying Theorem 63.2 from Munkres's Topology:

Theorem 63.2 (A nonseparation theorem).    Let $D$ be an arc in $S^2$.   Then $D$ does not separate $S^2$.

(There's a technicality with the point at infinity that was added to compactify $\mathbb{R^2}$, but it shouldn't be hard to see how your arc connecting two points can stay some $\epsilon$ away from it.)

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You did not indicate how to choose the $U_i$s to ensure that $\cup_i U_i$ is simply connected, and it is not obvious to me. Could you please elaborate or give a hint? –  Jonas Meyer Aug 11 '11 at 17:55
If you want, you can approximate $A$ by a polygonal arc that close enough to $A$. Then cover that arc with finitely many balls. If you've chosen your polygonal arc close enough to $A$ then the balls will cover $A$ as well, I should think. –  kahen Aug 11 '11 at 18:02
@kahen: I'm not yet entirely convinced that your method of construction of a path joining any two points of $U\setminus\cup U_i$ works. But if we assume that it does, then couldn't we show the path-connectedness of $U\setminus A$ more easily by picking two points $a,b\in U\setminus A$, then the finitely many open balls $U_i$ with the additional condition that $a,b\notin\cup U_i$ and then constructing a path in $U\setminus\cup U_i$ joining $a$ and $b$ using your idea. –  LostInMath Aug 11 '11 at 20:25
Yeah, that does sound a bit too easy, LostInMath. I'm going to have to spend some time thinking about this... –  kahen Aug 11 '11 at 20:44
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