Connectedness of a certain subset of the plane

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?

Solution 1:

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 $D_{i}$.]

Solution 2:

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.)