Why does space have the topology of a three sphere? Suppose that $U(x)$ is an element of the gauge group say $SU(2)$ and suppose  $U(x)=1$ as $|\vec{x}|\to\infty$. Then, why does space have the topology of $S^3$?
This is done in Srednicki page 571. Note that I'm not asking how to prove that $SU(2)\cong S^3$. What I'm asking is how to prove that when $U(x)=1$ as $|\vec{x}|\to\infty$ the space $\mathbb{R}^3$ is compactified to $S^3$ space.
 A: The question can be formulated more generally as:

Why is it that when we consider functions over $\mathbb{R}^n$ such that the limit as $|\vec{x}|\to\infty$ is the same in any direction then we can identify their domain with $S^n$?

Note that although the functions that we consider might be $\mathbb{R}^3\to SU(2)$, the statement is valid for any target space and any dimension $n$. The argument is the following:
First, notice that $\mathbb{R}^n$ is topologically the same as the open $n$-dimensional ball $B^n$. We can identify the functions in $\mathbb{R}^n$ (with existing limit in any direction as $|\vec{x}|\to\infty$) with the functions on the closed ball $\bar{B}^n$ by identifying $\mathbb{R}^n$ with $B^n$, and then assigning to every point in the boundary the limiting value of the function in that direction.
When a function has the same limit as $|\vec{x}|\to\infty$ in any direction, the corresponding function in $\bar{B}^n$ takes the same value in every point of the boundary. We can then identify all the points in the boundary, getting $S^n$ and a well-defined function over it. You can gain intuition about this identification by thinking about the case $n=2$. If we have the disk with all the boundary points glued together, we clearly get a $2$-sphere.
A: I think i got it,correct me if i am wrong.
We consider stereographic projection from the North pole $p$. Since Stereographic 
projection is a one-to-one correspondence between {$S^n−p$} and $R^n$ and since, 
$U(x)=1$ as $|\vec{x}|\to\infty$ we can regard $U(\infty)$ as the image of the 
point $p$ then, instead of having a map between $R^n$ and $SU(2)$ we can consider 
a map between $S^n$ and $SU(2)$ 
