I apologize if this is too elementary for this site.


Given a closed subset, $X\subset \mathbb{R}^n$, given $X$, $X^C$ path-connected, show that any path-connected neighborhood of $X$, denoted $M$, has that $M-X$ is path-connected.


Both my professor and I are unable to solve it. It came up in the context of metric geometry (specifically, for continuous maps $S^{k}-\{(0, 1)\}\to \mathbb{R}^{k+1}$) , but it seemed to generalize.


If this is a standard result, where can I find it?


Thank you.



If a subset and its complement are path-connected, an neighborhood of the subset is path-connected
The answer seems to be yes.


Indeed, $M$ and $X^C$ are open subsets which cover $\mathbb{R}^n$ and whose intersection is $M \setminus X$. The Mayer-Vietoris sequence gives
$$H\_1(\mathbb{R}^n)\to H\_0(M\cap X^C)\to H\_0(M)\oplus H\_0(X^C) \to H\_0(\mathbb{R}^n)\to 0. $$
This sequence is isomorphic to 
$$0 \to H\_0(M\setminus X)\to \mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z} \to 0. $$
Therefore $H\_0(M\setminus X)$ is isomorphic to $\mathbb{Z}$, which means that $M\setminus X$ is path-connected.


Note that we did not use the fact that $X$ is path-connected, only that it is closed and that $M$ and $X^C$ are path-connected.


