Prove that $\mathbb R ^n $ without a finite number of points is simply connected for $n\geq 3$

Solution 1:

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 $S^{n-1}$.

Solution 2:

Here is a longwinded way involving tedious path surgery.

First we need to show that $Y=\mathbb{R}^n\setminus X$ is path connected. Pick $x,y \in Y$.Choose $d \neq 0$ such that $d \bot (x-y)$. Pick $\lambda \in \mathbb{R}$, $t \in [0,1]$ and let $p_\lambda(t) = x + t(y-x) + (1-|2t-1|) \lambda d$. (It is easy to see that $p_\lambda$ is the polygonal path $(x,{x+y \over 2}+ \lambda d, y)$.)

Since $x-y,d$ are linearly independent, it is easy to see $p_{\lambda_1}=p_{\lambda_2}$ iff $\lambda_1 = \lambda_2$ . In fact, if $t_1, t_2 \in (0,1)$, then $p_{\lambda_1}(t_1) = p_{\lambda_2}(t_2)$ iff $\lambda_1 = \lambda_2$ and $t_1 = t_2$.

Each $p_\lambda$ is a path between $x,y$ and at most $|X|$ of these can intersect $X$. Since there are an uncountable number of such paths, there is at least one path joining $x,y$ hence the set is path connected and so is connected.

Now suppose $x_0 \in Y$ and $\gamma:[0,1] \to Y$ is a closed path based at $x_0$. Since $Y$ is open, we see that $\gamma$ is homotopic to a polygonal closed path in $Y$ also based at $x_0$. Hence we may take $\gamma$ to be polygonal, that is, straight lines joining a finite number of points $x_0=\gamma_0,...,\gamma_m = x_0$.

Now consider the finite collection of points $A=\{\gamma_k\} \cup X$. Pick a hyperplane $H$ passing through $x_0$ such that the orthogonal projections onto $H$ of the points in $A$ are distinct. (There are only a finite number of orientations of the hyperplane such that two points in $A$ project to the same point.) Let $\Pi$ be the orthogonal projection operator (projects onto the subspace parallel to $H$), and let $h$ be the normal of the hyperplane.

Since $Y$ is open, we see that $B(x_0,\epsilon) \subset Y$ for some $\epsilon>0$. Let $H_\eta = H+ \{ \eta h\}$. By choosing $\eta$ sufficiently small, we can shift the hyperplane such that it intersects $B(x_0,\epsilon)$ but passes through none of the points $X$.

Let $\phi$ be the (affine) orthogonal projection onto $H_\eta$.

Let $B= \{ y | \Pi y = \Pi x \text{ for some } x \in X \}$ (a finite collection of lines perpendicular to $H_\eta$). By construction, none of the points $\gamma_k$ lie in $B$, but it is possible that some segment $[\gamma_i, \gamma_{i+1}]$ intersects $B$.

Suppose a segment $[\gamma_i, \gamma_{i+1}]$ intersects $B$. Pick a direction $d$ that is perpendicular to $\gamma_{i+1}-\gamma_{i}$ and $h$ (this is where $n\ge 3$ comes in). As above, define $p_\lambda(t) = \gamma_i + t(\gamma_{i+1}-\gamma_{i}) + (1-|2t-1|) \lambda d$, and let $N= \{ \lambda | p_\lambda([\gamma_i, \gamma_{i+1}]) \cap B \neq \emptyset \}$. Note that $N$ is finite, hence there is some $\delta>0$ such that $p_\lambda([\gamma_i, \gamma_{i+1}])$ does not intersect $B$ for $\lambda \in (0,\delta]$. Hence $p_\lambda([\gamma_i, \gamma_{i+1}])$ does not intersect $X$ for $\lambda \in [0,\delta]$. Hence we can continuously modify the path $\gamma$ by adding the point ${\gamma_i +\gamma_{i+1} \over 2}+ \delta d$ while remaining in $Y$. Repeat this process for all segments that intersect $B$. Hence the original path is homotopic in $Y$ to a curve that does not intersect $B$.

The points $x_0, \phi(x_0)$ are in $B(x_0,\epsilon)$ and since the ball is convex, we can see that the modified curve is homotopic in $Y$ to the same curve with the points $x_0, \phi(x_0)$ prepended (that is, the points on the path are $x_0, \phi(x_0), x_0=\gamma_0, ...$). In a similar manner, add the points $\phi(x_0), x_0$ to the end of the path.

The modified path looks like $x_0, \phi(x_0), \gamma_1, ...,\gamma_n, \phi(x_0), x_0$ (the $\gamma_i$ are the modified points).

Now consider the map $\theta_t(x) =(1-t)x+ t \phi(x)$ apply the map to the portion of the curve $\phi(x_0), \gamma_1, ...,\gamma_n, \phi(x_0)$. Hence the modified path is homotopic in $Y$ to the path $x_0, \phi(x_0), \phi(\gamma_1), ...,\phi(\gamma_n), \phi(x_0), x_0$, and since the points $\phi(x_0), \phi(\gamma_1), ...,\phi(\gamma_n), \phi(x_0)$ lie in the convex set $H_\eta \subset Y$ the curve is homotopic to the curve $x_0, \phi(x_0), x_0$, and since the ball is convex, this curve is homotopic in $Y$ to the constant curve $t \mapsto x_0$.

Solution 3:

Actually, the first comment on my question helped me to find a simple solution.

The proof is by induction on $|X|$

If $|X|=1$, then WLOG we can suppose $X=\{0\}$: then $\mathbb R^n \setminus \{0\} $ deforms into $S^{n-1}$ that is simply connected for $n\geq 3$.

If the thesis is true for $|X|<k$, let us prove that the thesis is true for $|X|=k$. WLOG, we can suppose that $X=\{p_1 ,..., p_k\}$, with $(p_i)_1\leq (p_{i+1})_1 $ (if $x\in \mathbb R^n$, then by definition $(x)_1$ is the first coordinate of $x$ with respect to canonical basis). We can suppose also that there exists $i$ such that $(p_i)_1<(p_{i+1})_1$. There is no loss of generality because $k\geq 2$ and so there are at least two points of $X$ that have distinct coordinates. Define then $\delta = \frac {(p_{i+1})_1 - (p_{i})_1}{3}$, and name $A $ the open set of points such that first coordinate is $>(p_{i+1})_1 - 2\delta$ and $B$ the open set of points such that first coordinate is $<(p_i)_1 +2\delta$. Then $A$ and $B$ are homeomorphic to $\mathbb R ^n $ and $A$ and $B$ both intersect $X$ in less than $k$ points. Then, by inductive hypotesis, $A\setminus X$ and $B\setminus X$ are both simply connected; moreover, $A\cap B = A\setminus X \cap B\setminus X =$ a convex set, and $A\cup B \setminus X = \mathbb R ^n \setminus X$. Then we can apply Van Kampen.

Solution 4:

Definition of simply connected only requires trivial fundamental group together with path-connectedness. The argument for both cases is intuitively the same, just utilize an extra dimension to go around the holes.

If it so happens that the straight line between two points passes through a hole, cut the line before and after the hole and append a half-circle around it. Inductively this gives a path between any two points in the desired space.

For the triviality of the loop classes, consider locally around one of the holes. Fit a plane to the largest segment of the loop possible and if the hole is coplanar, homotope in any "normal" direction (relative to the plane) to go around. Induct on the number of holes and that should be it. This is not very rigorous, but it can probably be parameterized if one tries hard enough.