Question

Here I mean the version with all but finitely many components zero.

Answer 1

This is the swindle, isn't it?

There's an elegant way to phrase this with lots of sines and cosines, but working it all out is too much like hard work. Here's the quick and dirty way.

Let $T: S^\infty \to S^\infty$ be the "shift everything down by 1" map.

Then for any point, $x$ and $T(x)$ are not colinear, so the line between them does not go through the origin. We can therefore define a homotopy from the identity on $S^\infty$ to $T$ by taking the homotopy $t x + (1 - t)T(x)$ and renormalising so that it is always on the sphere (incidentally, although you are working in $\ell^0$, by talking about a sphere you implicitly have a norm).

Then we simply contract the image of $T$, which is a codimension 1 sphere, to a point not on it, say $(1,0,0,0,0,...)$. Again, we can use 'orrible sines and cosines, but renormalising the direct path will do.

(Incidentally, there's nothing special about which space you are taking the sphere in. So long as your space is stable in the sense that $X \oplus \mathbb{R} \cong X$ then this works)

Added a bit later: Incidentally, if you want to work in a space that doesn't support a norm (such as an infinite product of copies of $\mathbb{R}$) you can still define the sphere as the quotient of $X$ without the origin by the action of $\mathbb{R}^+$. The argument above still works in this case.

Added even later: Revisiting this in the light of the duplicate: http://mathoverflow.net/questions/38763/is-lp-mathbbr-minus-the-zero-function-contractible, the key property on $T$ is that it be continuous, injective, have no eigenvectors, and be not surjective. These conditions imply the following:

1. injective ⟹ the end-point of the homotopy is not the origin
2. no eigenvalues ⟹ the homotopy does not pass through the origin en route
3. not surjective ⟹ there is a point not in the image to which the image can be contracted
4. continuous ⟹ the homotopy is jointly continuous

Finally, there's no difference between the sphere and the space minus a point (indeed, without a norm the "space minus a point" is easier to deal with). Indeed, the homotopy described here actually works on the "space minus a point" and is just renormalised to work on the sphere.

Answer 2

Another nice solution to a similar question is at http://katlas.math.toronto.edu/drorbn/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible:

Let H=L^2([0,1]) and define S^\infty = {x \in H : ||x||=1}.

Claim. S^\infty is contractible.

Proof. For any t \in [0,1] and any f \in H define f_t(x)= f for 0<x<t and f_t(x)=1 for t<x<1. Observe that t --> f_t/||f_t|| is continuous and gives the desired retraction to the point f=1.

Answer 3

3 proofs on Wikipedia - basically the same arguments as above. The Hilbert space part is superfluous.

Answer 4

Kind of late to the party, but the (weak) contractibility follows from pi_i(S^\infty) = 0 for i>0.

Answer 5

Are follow up questions allowed? What does "version with all but finitely many components zero" mean? Is this different from taking the limit of n-spheres?

Answer 6

Here are my thoughts on the matter. However, this is not too much more than what is done above. I think...

$\quad$ We seek to show that a homotopy from the identity map of $S^{\infty}$ ($id_{S^{\infty}}$) to a constant map can be constructed and thus it must be null-homotopic, $i.e.$, contractible. Let $T: S^\infty \to S^\infty$ be the "shift everything 'down' by 1" map given by $(x_1, x_2, x_3,...) \mapsto (0, x_1, x_2,...)$. Then for any point, $x$ and its image $T(x)$, the line between them does not go through the origin.

$\quad$ We can therefore define a homotopy from the identity on $S^\infty$ to T by taking a homotopy and renormalizing, so that it is always on the sphere, as follows. Let $f_t: \mathbb{R}^{\infty} \setminus 0 \rightarrow \mathbb{R}^{\infty} \setminus 0$ be given by
$$f_t(x_1,x_2,...) = (1-t)(x_1, x_2,...) + tT(x_1, x_2,...).$$ (Note that the vector $f_0 = id_{\mathbb{R}^{\infty}}$ and that $f_t$ takes nonzero vectors to nonzero vectors $\forall t \in \left[ 0, 1\right]$.) Then we can renormalize it to ensure everything is still on $S^{\infty}$, $i.e.$, $$\frac{f_t}{\left| f_t\right|} = F(x,t) : S^{\infty} \times I \rightarrow S^{\infty}.$$ Thus we now have that $id_{S^{\infty}} \simeq T$. Or, in other words, $F$ gives a homotopy from the identity map of $S^{\infty}$ to the map $(x_1, x_2, x_3,...) \mapsto (0, x_1, x_2,...)$.

$\quad$ Then we simply contract the image of $T$ , which is a codimension 1 sphere, to a point not on it, say $(1,0,0,0,0,...) = N$ (north pole). So let $g_{t} : \mathbb{R}^{\infty} \setminus 0 \rightarrow \mathbb{R}^{\infty} \setminus 0$ be given by $$g_t(x_1, x_2, ...) = (1-t)(0, x_1, x_2,....) + t(1, 0, 0, ...).$$ Now observe that $g_0 = f_1$, $f_0 = id_{S^{\infty}}$, and $g_1 = N$ (a constant). Again we can renormalize to guarantee everything is still on $S^{\infty}$, $i.e.$, $$\frac{g_t}{\left| g_t\right|} = G(x,t) : S^{\infty} \times I \rightarrow S^{\infty}.$$ Furthermore, we have that $\frac{g_0}{\left| g_0\right|} = \frac{f_1}{\left| f_1\right|}$. Therefore, it follows that $T \simeq N$ ($G$ gives a homotopy from
$T$ to the north pole) and since the composition of homotopies is again a homotopy we have that $id_{S^{\infty}} \simeq N$. Hence the desired result follows and we conclude that $S^{\infty}$ is contractible.