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...

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    $\begingroup$ Is $\mathcal{T}_n$ the Euclidean topology on $\mathbb{R}^n$? In what sense do you mean $U \cap \mathbb{R}^n$, since $\mathbb{R}^n$ is not a subset of $\mathbb{R}^\infty$? And most importantly, what is the norm $\|\cdot\|$ on $\mathbb{R}^\infty$? $\endgroup$ Commented Jan 19, 2013 at 21:12
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    $\begingroup$ $\mathbb{R}^{\infty}$ is a set of sequences $(a_1,a_2,...)$ where only finite number of coordinates $\neq 0$, and $\mathbb{R}^n$ is set of $(a_1,a_2,...,a_n,0,0,...)$. The norm is $\sqrt{\sum_{i=1}^{\infty}x_i^2}$. $\endgroup$
    – banas6
    Commented Jan 19, 2013 at 21:20
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    $\begingroup$ @Daniel: The right-shift map doesn’t meet the OP’s requirement that it map $D^\infty$ onto $D^\infty$. $\endgroup$ Commented Jan 19, 2013 at 21:32
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    $\begingroup$ Also: mathoverflow.net/questions/119362/… $\endgroup$
    – Asaf Karagila
    Commented Jan 19, 2013 at 22:25
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    $\begingroup$ Related: math.stackexchange.com/questions/459397 $\endgroup$
    – Watson
    Commented Dec 4, 2016 at 21:28

2 Answers 2


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.

  • $\begingroup$ Sorry I missed a bracket around the $1-t$, and thanks for pointing out the typo in Allen's name :). $\endgroup$
    – Dan Rust
    Commented Jan 19, 2013 at 21:23
  • $\begingroup$ Thank you Daniel, Brian and Nate :) I'm sorry for my requirements but I don't get for example why $f(t)/|f(t)|$ is a homotopy (of course why it's contiuous)? Can you give me a hint? Thanks :) $\endgroup$
    – banas6
    Commented Jan 19, 2013 at 23:05
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    $\begingroup$ Hopefully you agree that $f_t$ is continuous (this is common construction of a homotopy - for example the usual homotopy from the punctured plane to the circle). You should also be happy with the fact that $f_t(\alpha x)=\alpha f_t(x)$ holds for real numbers $\alpha$ and so $f_t$ preserves lines going though the origin (kind of like a rotation in the plane). Finally, you should also be happy that if $|x|=1$, then $|f_t(x)|=1$ and so $f_t$ is well defined if we restrict it to the sphere of radius 1. Restriction of a continuous map to a subspace is also continuous and so we're done. $\endgroup$
    – Dan Rust
    Commented Jan 19, 2013 at 23:25
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    $\begingroup$ Perhaps the easiest way to show this is to note that a basis of the topology of $\mathbb{R}^{\infty}$ is given by all balls $B_{r}(x)=\{y: ||x-y||<r\}$ for all $x\in\mathbb{R}^{\infty}$ and $r>0$ and then observe that $f_t$ sends all open balls to open balls for all $t$ (small exercise, but note that because $\mathbb{R}^{\infty}$ is a topological group, you only need to look at balls with center at the origin $(0,0,\ldots)$) and in an injective way. It follows that the inverse image of a basis element is a basis element and so $f_t$ must be continuous. $\endgroup$
    – Dan Rust
    Commented Jan 20, 2013 at 14:51
  • $\begingroup$ Hi Daniel, wonderful answer, thank you. But how can you make sure that $g_t \neq 0$? Thank you~ $\endgroup$ Commented Sep 10, 2013 at 16:58

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.

  • $\begingroup$ you might like to see the answers to this question $\endgroup$
    – Dan Rust
    Commented Aug 23, 2013 at 15:50
  • $\begingroup$ "Attaching $m$-cell to a CW complex $X$ does not change $\pi _n(X)$ with $m>n$". Is this conclusion comes from Cellular approximation theorem? Or is there any obvious way to show this fact? $\endgroup$
    – AG learner
    Commented Sep 14, 2015 at 16:35
  • $\begingroup$ It is Corollary 4.12 in Hatcher's book. He uses indeed the cellular approximation for pairs. $\endgroup$
    – Seirios
    Commented Sep 14, 2015 at 19:53
  • $\begingroup$ No cellular approximation theorem necessary: $\pi_n(S^\infty)=\pi_n(colim S^\infty)=colim \pi_n(S^\infty)=colim 0=0$. $\endgroup$ Commented Aug 28, 2016 at 2:25

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