Show that the complement of a finite set of points in \(\mathbb{R}^n\) is simply-connected if \(n \geq 3\).
Solution
Goal. Show \(\RR^n \sm \theset{p_1, \dots, p_k}\) is simply connected for \(n \ge 3\).
1. \(\RR^n \sm \theset{p_1, \dots, p_k}\) is path-connected for \(n \ge 2\).
any two points can be joined by a path avoiding the finitely many removed points (there is plenty of room in dimension \(\ge 2\)).
2. It suffices to show \(\pi_1(\RR^n \sm \theset{p_1, \dots, p_k}) = 0\).
simply connected means path-connected and trivial fundamental group.
3. \(\RR^n \sm \theset{p_1, \dots, p_k}\) deformation-retracts onto a wedge of \(k\) copies of \(S^{n-1}\).
place small disjoint \((n-1)\)-spheres around each removed point; the complement deformation-retracts onto the wedge of these spheres (for \(n \ge 2\)).
4. \(\pi_1\qty(\bigvee_{i=1}^k S^{n-1}) = 0\) for \(n - 1 \ge 2\).
\(\pi_1(S^{n-1}) = 0\) for \(n - 1 \ge 2\), and the fundamental group of a wedge is the free product of the fundamental groups, so \(\pi_1 = \ast_{i=1}^k \pi_1(S^{n-1}) = 0\).
5. Hence \(\pi_1(\RR^n \sm \theset{p_1, \dots, p_k}) = 0\).
3 and 4 (homotopy invariance of \(\pi_1\)).
6. Q.E.D.
1 and 5 show the complement is simply connected.