Why is $\mathbb{R}^n-\{x\}$ homeomorphic to $S^{n-1}\times\mathbb{R}$? I am trying to understand the following intuitively (or with a bit of reasoning): 
For a point $x\in\mathbb{R}^n$, the complement $\mathbb{R}^n-\{x\}$ is homeomorphic to $S^{n-1}\times\mathbb{R}$.
I could understand by taking out a point, there is a hole in $\mathbb{R}^n$ and somehow homeomorphic to $S^{n-1}$. What I struggle to understand is the $\times\mathbb{R}$ part. What does the $\mathbb{R}$ do? What is the shape of $S^{n-1}\times\mathbb{R}$?
Any understandable explanation for this? Thanks.
 A: Take $\mathbb{R}^n-\{x\}$, and consider the sphere $S^{n-1}$ on $\mathbb{R}^n-\{x\}$. Push what is "inside" the sphere down, and what is "outside" the sphere up (of course, what is pushed to where is just a matter of arbitrary taste). When you do this adequately, you will get a cylinder $S^{n-1} \times \mathbb{R}$. 
Do this in the case $\mathbb{R}^2 - \{x\}$ to visualize better.
A: Recall that $S^n \setminus \{p\} \simeq \mathbb{R}^n$. Now if you take $S^n \setminus\{p,q\}$ this is homeomorphic to $\mathbb{R}^n \setminus \{\textrm{pt}\}$. Let $p,q$ be the north an south poles respectively. The image of the sphere with two punctures is now a band. Here $S^{n-1}$ is the equator and $\mathbb{R}$ is the curved segment extending from the top of the band to the bottom. The actual map is given by $x \mapsto \left(\frac{x}{\|x\|},\|x\|\right)$.
A: Any point in $\Bbb R^n-\{0\}$ can be specified uniquely by choosing a direction (i.e. a point on the unit sphere $S^{n-1}$) and a distance (i.e. a point in $\Bbb R^+$). So clearly, there is a natrual correspondence between $\Bbb R-\{0\}$ and $S^{n-1} \times \Bbb R^+$, and this correspondence turns out to be a homeomorphism. To finish the problem, you just need to find a homeomorphism from $\Bbb R^+$ to $\Bbb R$ (I suggest the logarithm, but there are many others).
