# General topology question-simply connected, connected boundary

Suppose $\Omega\subset\mathbb{R}^3$ be a bounded connected open set. Then $\Omega_C\subset\Omega$ is an open set such that $\overline{\Omega_C}\subset\Omega$ and such that $\Omega_I:=\Omega\setminus\Omega_C$ is connected.

Is it true that if $\Omega_I$ is simply connected then

• $\Omega_C$ is simply connected and
• $\partial\Omega$ has just one connected component?

## 1 Answer

The first part ($\Omega_C$ simply connected) is not necesarily true. In fact $\Omega_C$ may not even be connected. Consider for instance $\Omega=(0,5)^3, \Omega_C=(1,2)^3\cup (3,4)^3$. Then $\Omega_I$ is indeed simply connected (it is path connected, and for any two paths with the same endpoints in $\Omega_I$, we can find an homotopy between those paths that "goes around" $\Omega_C$).

The second part ($\partial \Omega$ is connected) is false as well. Consider for instance $\Omega=B(0,5)\setminus\{0\}$ and $\Omega_C=B(1,0.5)$. Again, $\Omega_I$ is simply connected. However, $\partial \Omega=\{0\}\cup \partial B(0,5)$ has two connected components.

In general, to decide whether some sets have connected boundary, this result is quite useful.