I want to prove that $\mathbb R ^n $ without a finite number of points is simply connected for $n\geq 3$. Let $X$ be that finite set of points. My idea is to prove this by induction on cardinality of $X$.
Base case with $|X|=1$ follows from the fact that $S^n$ is simply connected and it is a deformation retract of $\mathbb R^{n+1}$ without a point.
For inductive case with $n\geq 2$, I want to prove that there are two distinct parallel affine planes $P $ and $Q $ such that:
1) neither $P$ nor $Q$ intersect $X$;
2) if we denote with $A_+,A_-$ the connected components of $\mathbb R^n \setminus P\cup X$ and with $B_+,B_-$ the connected components of $\mathbb R^n \setminus Q\cup X$, such that $A_+$ contains $Q$ and $B_+$ contains $P$, then $X$ intersects $A_-$ and $B_-$ in at least one point.
(Maybe it is less complicate to prove that closure of $A_+ \cap B_+ $ does not contain any point of $X$)
Then I can apply inductive hypotesis, and $A_+$ and $B_-$ are simply connected. Moreover, $A_+ \cup B_+ = \mathbb R^n \setminus X$ and $A_+ \cap B_+$ is arc-connected, so I can apply Van Kampen.
How can I prove the existence of such two planes?
Best Answer
If you're happy with deformation retraction arguments, you can do this more quickly than that. Put disjoint balls around each point, and join them with thin paths in some order. I claim that $\mathbb{R}^n$ minus these points deformation retracts to the boundaries of these balls together with these thin paths; that is, $\mathbb{R}^n$ minus $k$ points is homotopy equivalent to a wedge sum of $k$ copies of $S^{n-1}$.