A generalization of Jordan curve theorem to connected open sets in the plane

Question

Problem

(Fulton's Algebraic Topology: A First Course, Problem 5.23) Let $U\subseteq\mathbb R^2$ be any connected open set in the plane.

1. If $X\subseteq U$ is homeomorphic to $[0,1]$, then $U\setminus X$ is connected.
2. If $X\subseteq U$ is homeomorphic to $S^1$, then $U\setminus X$ has 2 connected components.

Thoughts

The proof of Jordan's curve theorem on Fulton's book is based on Mayer-Vietoris sequence for de Rham cohomology group, which relies heavily on the fact that the universal space is $\mathbb R^2$ (or homeomorphic to it). If $U$ is not simply connected, it seems that I cannot simply apply the original proof. My idea is to find a neighborhood of $X$ which is homeomorphic to $\mathbb R^2$. I have no clear idea how to do that.

Any idea? Thanks!

Answer 1

For $(1)$ take $V_1=\Bbb R^2\backslash X$ and $V_2=U$, by the Jordan Curve Theorem $V_1$ is connected.
On the other hand $V_1\cup V_2=\Bbb R^2$ hence $H^1(V_1\cup V_2)=0$.
As $V_1$ and $V_2$ are connected and $H^1(V_1\cup V_2)=0$ by using the exercise5.8 it follow that $V_1\cap V_2$ is connected, but $V_1\cap V_2=U\backslash X$.
For $(2)$ use same argument.