Why is $S^n \cong \bar{B^n}$? In class when my professor first introduced complexes and cells, he gave some examples of closed $n-$cells, (where he defines X as a closed $n$-cell if $X\cong \bar{B}^n$) and one thing he said was 


$S^n\cong \bar{B^n}$ is a closed $n$-cell


I'm really struggling to see why $S^n\cong \bar{B}^n$. I should note that I am $\textbf{very}$ new at topology and thinking about these types of geometric figures, so please forgive me and bear with me. So,  $S^n$ is basically the boundary of $B^{n+1}$ right? And $B^n$ is filled in where $S^n$ is just a boundary. So it's really hard for me to picture how this works.
Further, he says 


$S^n\setminus \{p\}\cong R^n$ is an open n-cell


But that's essentially the same as saying $S^n\setminus \{p\}\cong B^n$ Because $B^n\cong R^n$, right? So what does removing one single point from the n-sphere do so drastically to make it go from a closed cell to an open cell
Thanks for any help!
 A: I'm guessing $B^n_r = \{\textbf{x}: \|\textbf{x}\| < r\}$ and so $\overline{B^n_r} = \{\textbf{x}: \|\textbf{x}\| \leq r\}$. Yes, this is a closed $n$-cell or more simply put, a closed $n$-dimension disk. However, $S^n \not \cong \overline{B^n_r}$ since the latter is a manifold with boundary. I think the notation the professor was going for, was one that Hempel uses in his 3-manifolds book i.e $\overline{B^n}(S^{n-1}) = \textbf{n}-\textrm{cell}$. Here Hempel means, if you fill in $\partial B^n = S^{n-1}$, then that gives a $\textbf{n}$-cell.
Also, if you take $S^n \setminus \{p\}$ then you can use stereographic projection to get a homeomorphism between the punctured sphere and $\mathbb{R}^n$. Again, for simplicity, you should just think of popping a hole in a sphere and expanding that puncture so that the result flatens to a plane. This is of course the visual, displaying $S^2 \setminus \{p\} \cong \mathbb{R}^2$. 
A: I very well may be misinterpreting the question, but I believe if you "pierce" $S^n$ at any point and flatten it into an $n$-dimensional object, it becomes $B^n$. You can visualize this using $S^2$ and $B^2$. 
