Is $U \cap \mathbb{R^3} \cong \mathbb{S^2}\times \mathbb{R^1} $ If $U \subseteq\mathbb{S^3}$ is open and nonempty, Is $U \cap \mathbb{R^3} $ homeomorphic to $\mathbb{S^2}\times \mathbb{R^1}$? how to imagine this? 
 A: Let $U=\mathbb{S}^3$ and consider
$$U\cap\mathbb{R}^3 = \{(x,y,v,w)\in\mathbb{R}^4|x^2+y^2+v^2+w^2 = 1\}\cap\{(x,y,v,w)\in\mathbb{R}^4|w=0\}$$
$$ = \{(x,y,v,w)\in\mathbb{R}^4|w=0\mbox{ and }x^2+y^2+v^2=1\}\cong \mathbb{S}^2.$$
$\mathbb{S}^2$ does indeed contain some copies of $\mathbb{S}^1\times\mathbb{R}$ up to homeomorphism, but certainly not all open non-empty subsets are homeomorphic to $\mathbb{S}^1\times\mathbb{R}$.
I see you changed the sapce to be obtained to $\mathbb{S}^2\times\mathbb{R}$. From the above discussion it should be clear that this can never be accomplished.
If $U\subseteq\mathbb{S}^3$, then $U\cap\mathbb{R}^3\subseteq \mathbb{S}^3\cap\mathbb{R}^3\cong\mathbb{S}^2$. As $\mathbb{S}^2\times\mathbb{R}\not\subseteq\mathbb{S}^2$ no $U$ gives $U\cap\mathbb{R}^3\cong\mathbb{S}^2\times\mathbb{R}$.
A: They may have wanted you to think of the 3-sphere as $\mathbb{R}^3$ plus a point. Then if $U$ is the sphere minus any other point, $U\cap\mathbb{R}^3$ is homeomorphic to $\mathbb{S}^2\times \mathbb{R}$.
