# Unit sphere in $\mathbb{R}^\infty$ is contractible?

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?

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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…

#### Solutions Collecting From Web of "Unit sphere in $\mathbb{R}^\infty$ is contractible?"

You’ll find a proof that the infinite dimensional sphere is contractible on page 88 of Allen Hatcher’s Algebraic Topology kindly hosted for free by him on his website.

The proof gives an explicit homotopy between the identity map and the constant map on the sphere $S^{\infty}$.

Let $f_t\colon\mathbb{R}^{\infty}\rightarrow \mathbb{R}^{\infty}$ be given by $f_t(x_1,x_2,\ldots)=(1-t)(x_1,x_2,\ldots)+t(0,x_1,x_2,\ldots)$. For all $t\in[0,1]$, this map sends nonzero points to nonzero points, so $f_t/|f_t|$ is a homotopy from the identity map on $S^{\infty}$ to the map $(x_1,x_2,\ldots)\mapsto (0,x_1,x_2,\ldots)$. We then define a homotopy from this map to the constant map at $(1,0,0,\ldots)$ by setting $g_t(x_1,x_2,\ldots)=(1-t)(0,x_1,x_2,\ldots)+t(1,0,0,\ldots)$. The homotopy is then given by $g_t/|g_t|$. The composition of these two homotopies then gives a homotopy from the identity map to the constant map, and so $S^{\infty}$ is contractible.

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.