Show that the complement of a finite set of points in $\Bbb R^n$ is simply-connected if $n \ge 3$.
I know this question has been answered here a few times, but I'm asking about a detail of the usage of Van Kampen's theorem in the proof.
If $\{x_1, \dots, x_k\}$ are the points removed, then for $k=1$ we have $\Bbb R^n – \{x_1\} \simeq S^{n-1}$ and $S^{n-1}$ is simply-connected for $n\ge 3$.
Now when $k \ge 2$ since there are finitely many points there exists a hyperplane $H$ such that $H$ splits up $\{x_1, \dots x_r\}$ and $\{x_{r+1}, \dots,x_k\}$. If $A$ is an open cube minus the first $r$ points and $B$ a cube minus the remaining points, then $\Bbb R^n – \{x_1, \dots, x_k\} = A \cup B$. Van Kampen now says that $$\pi_1(\Bbb R^n – \{x_1, \dots, x_k\}) \cong \pi_1(A) * \pi_1(B) /N$$ but since $\pi_1(A) = \pi_1(B)=0$ the fundamental group is trivial thus $\Bbb R^n – \{x_1, \dots, x_k\}$ is simply-connected.
I think that there are few "holes" in the proof. Van Kampen requires that $A \cap B$ is path-connected and non-empty so this handwavy argument about cubes is not probably going to fly. I think that I would need to have that the cube $A$ will go slightly higher than $H$ and $B$ slightly lower than $H$ in order to $A \cap B$ to be path-connected and non-empty?
I'm also wondering if I this would be helpful to do with using the one-point compactification $S^n$ of $\Bbb R^n$ and consider $S^n – \{x_1, \dots, x_k\}$?
Best Answer
The proof is more or less correct. However, I would not use cubes but half spaces.
Here is my suggestion. We proceed by induction.
The base case $k=1$ has been treated.
Now consider $k+1$ points $x_1,\ldots, x_{k+1}$. Since $x_1 \ne x_{k+1}$, we have $x_1^j \ne x_{k+1}^j$ for some coordinate $j$. Choose $r,s$ strictly between $x_1^j$ and $x_{k+1}^j$ such that $r < s$ and $x_i^j \notin (r,s)$ for all $i$. This is possible because there are only finitely many values $x_i^j$. Define $$A = \{(z^1,\ldots,z^n) \in \mathbb R^n \mid z^j > r\},$$ $$B = \{(z^1,\ldots,z^n) \in \mathbb R^n \mid z^j < s\} .$$
These are open half spaces (which are homeomorphic to $\mathbb R^n$). Each of $A,B$ contains at most $k$ of the points $x_i$ (note that $x_1, x_{k+1}$ lie in exactly one of $A, B$). By induction hypothesis we know that $A' = A \setminus \{x_1,\ldots, x_{k+1}\}$ and $B' = B \setminus \{x_1,\ldots, x_{k+1}\}$ are simply connected. We have $A \cup B = \mathbb R^n$, thus $A' \cup B' = \mathbb R^n \setminus \{x_1,\ldots, x_{k+1}\}$. Moreover $A' \cap B' = \{(z^1,\ldots,z^n) \in \mathbb R^n \mid z^j \in (r,s) \}$ which is path-connected . Now van Kampen applies.
If you know the result for $\mathbb R^n$, you can easily derive the analog for $S^n$. In fact, $S^n$ minus $k$ points is homeomorphic to $\mathbb R^n$ minus $k-1$ points.