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.

  • $\begingroup$ Oh, I meant a path-connected neighborhood of $X$, which I denote $M$. Right, sorry. $\endgroup$ – max Nov 5 '13 at 10:09
  • $\begingroup$ This isn't true for $n=2$, $X$ the open interval $(-1,1)$ on the $x$-axis, and $M$ the open unit ball about the origin. Do you mean $X$ to be closed, maybe? $\endgroup$ – Jeremy Rickard Nov 5 '13 at 10:27
  • $\begingroup$ Originally, no. Now yes. I never actually put to much thought outside of a properly embedded manifold. $\endgroup$ – max Nov 5 '13 at 10:29

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.

  • $\begingroup$ Thank you! My homology theory is weak, so I was unable to constuct such an argument. $\endgroup$ – max Nov 5 '13 at 23:14
  • $\begingroup$ It's a good observation that we somehow need to use the simply connectedness of $R^n$. For instance it would be false if we replaced $R^n$ with an annulus. $\endgroup$ – Nate Eldredge Nov 6 '13 at 5:20

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