TITLE: Unit sphere in $\mathbb{R}^\infty$ is contractible?
QUESTION [23 upvotes]: Let $\mathcal{T}_{\infty}= \left\{ U \subset \mathbb{R}^{\infty}: \ U \cap \mathbb{R}^n \in \mathcal{T}_n, \text{ for } n=1,2,... \right\} $.
Of course $\mathcal{T}_{\infty}$ is topology in $\mathbb{R}^{\infty}$. How to prove that $S^{\infty} = \{ v \in \mathbb{R}^{\infty} : \ \|v\|=1 \}$ is contractible? 
:)
Can we find homeomorphism without fixed point from $D^{\infty} = \{ v \in \mathbb{R}^{\infty} : \ \|v\| \le 1 \}$ onto $D^{\infty}$? I was trying to find such homeomorphism, but I failed...

REPLY [11 votes]: We may notice that $\mathbb{S}^{\infty}= \bigcup\limits_{n \geq 1} \mathbb{S}^n$ where $\mathbb{S}^n$ is included into $\mathbb{S}^{n+1}$ thanks to $$(x_1, \dots, x_n) \mapsto (x_1, \dots, x_n,0).$$ In particular, $\mathbb{S}^n$ is a subcomplex of $\mathbb{S}^{n+1}$ (this construction is also described in Hatcher's book). 
Because attaching a $m$-cell does not change the $n$-th homotopy group when $m>n$, we deduce that $$\pi_n( \mathbb{S}^{\infty})= \pi_n(\mathbb{S}^{n+1})= 0, \ \forall n \geq 1.$$
According to Whitehead theorem, a weakly contractible CW complex is contractible. Therefore, $\mathbb{S}^{\infty}$ is contractible.
Of course, the proof given by Daniel Rust is much more elementary, but I find interesting to see how adding cells kills successively the homotopy groups in order to make $\mathbb{S}^{\infty}$ contractible.