TITLE: Show that $\mathbb{R}^n\setminus \{0\}$ is simply connected for $n\geq 3$
QUESTION [8 upvotes]: Show that $\mathbb{R}^n\setminus \{0\}$ is simply connected for $n\geq 3$.

To my knowledge I have to show two things:

$\mathbb{R}^n\setminus \{0\}$ is path connected for $n\geq 3$.
Every closed curve in $\mathbb{R}^n\setminus \{0\}, n\geq 3$ is null-homotop.

The problem is that I do not know exactly how to show that.

REPLY [3 votes]: The easy part of the Seifert-van Kampen theorem and which does not require an algebraic context (e.g. groups or groupoids) is that if $X$ is the union of a family $\mathcal U$ of path connected and simply connected open sets $U_i$ such that any pairwise intersection $U_i \cap U_j$ is path connected, and all containing a "base point"  $a \in X$,  then $X$ is simply connected. The key part of the proof is conveyed by the following diagram

in which $\alpha$ is a loop  in $X$ at the base point $a$ subdivided so that each part $\alpha_i$ lies in a set say $U_i$ of $\mathcal U$. The start  point $x_i$ of $\alpha_i$ shown as $\circ$ need not be at the base point. However $x_i$ lies in  $U_{i-1} \cap U_i$ which is  path connected; so there is a path $\gamma_i$ in  $ U_{i-1} \cap U_i $  joining $x_i$ to the base point, and we set $\gamma_i$ to be constant for $i=0,n$. Now each path $\zeta_i= \gamma_i^{-1}\alpha_i\gamma_{i+1} $lies in $U_i$ which is simply connected, and so is homotopic  in $U_i$ rel end points to a constant path $\beta_i$. Hence $\alpha$ is null homotopic, using the "horizontal" composition of these homotopies. 
Note that even for this result we need a " horizontal composition of squares", and such an  intuition is the start of a $2$-dimensional theory. 
This type of argument was  generalised by J.F. Adams (in unpublished lecture notes) to prove any map $S^r \to S^n$ is null homotopic for $r <n$, see 7.6.1 of Topology and Groupoids.