how to show that a open set of $S^3$ is simply connected? Let $B$ be the union of the compactification point and $(\mathbb R^3-X)$ in the one-point compactification of $\mathbb R^3$. (Here $X$ is a closed ball in $\mathbb R^3$.)
Then I think $B$ is somehow simply connected..but not able to prove it.
Can anyone help me with this?

We may take $X$ to be $\{x\in\mathbb R^3:\|x\|\le 1\}$.
The mapping $f:X^\circ\to B$ given by
$$
f(x) = \begin{cases} x/\|x\|^2 & \text{if }\|x\|>0, \\[6pt]
\text{the compactification point} & \text{if }x=0,  \end{cases}
$$
is a homeomorphism from $X^\circ$ to $B$ (where $X^\circ$ is the interior of $X$).  Therefore if $X^\circ$ is simply connected, then so is $B$.
