Equivalence of definitions of $S^\infty$ Consider the following two definitions of the infinity-sphere $S^\infty$. Why do they define homeomorphic spaces?
$1)$ The set of points in $\mathbb R^\infty$ with distance $1$ from the origin.
$2)$ The CW complex with $2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells, and so on, with $2$ $n$-cells for each $n$ in general, such that the $n$-skeleton is $S^n$.
I can show (thanks to a helpful answer on a prior question I asked) that the CW-complex for $S^n$ is the same as the Euclidean space definition of $S^n$ for finite $n$, but I do not see how to show these limit spaces are homeomorphic. I have a vague idea that the proof might involve the idea of a colimit, but I know nothing about category theory. Any more elementary suggestions are also welcome.
 A: For $n<m$ we can isometrically embed $\mathbb R^n\to \mathbb R^m$ by mapping to the first coordinates / filling up with zeroes.
This way the limit of all $\mathbb R^n$ is $\mathbb R^\infty$, the set of all sequences $(x_n)_{n\in\mathbb N}$ with almost all terms $=0$, which still gets its topology from the scalar product $\langle x,y\rangle=\sum_{i=1}^\infty x_iy_i$ (that works because the sum is actually finite).
Consider the set $S^\infty:=\{x\in\mathbb R ^\infty\mid \lVert x\rVert = 1\}$.
For each $n\in\mathbb N$ and $\epsilon\in \{\pm1\}$ we have the subset 
$$C_{n,\epsilon}=\{x\in S^\infty\mid x_n\epsilon>0\land \forall k>n\colon x_k=0\}.$$
Then $C_{n,\epsilon}$ is just one hemisphere of $S^{n-1}\subset \mathbb R^n$ and the smaller ones make up the equator.
More precisely, the map 
$$(x_1,\ldots,x_{n-1})\mapsto (x_1,\ldots, x_{n-1},\epsilon \cdot\sqrt{1-x_1^2-\ldots-x_{n-1}^2}, 0,0, \ldots) $$
defined on the closed $(n-1)$-ball is a homeomorphism of its interior with $C_{n,\epsilon}$ and of its boundary with the union of all smaller $C_{n',\epsilon'}$ and at the same time $S^{n-1}\subset \mathbb R^{n-1}\subset \mathbb R^\infty$.
This shows the CW-structure you describe in "2)".
