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Hatcher, Algebraic Topology, Exercise 0.16 reads:

Show that $S^\infty$ is contractible.

Let's look at the definition Hatcher gives for $S^\infty$:

There are natural inclusions $S^0 \subset S^1 \subset \cdots \subset S^n$, but these subspheres are not subcomplexes of $S^n$ in its usual cell structure with just two cells. However, we can give $S^n$ a different cell structure in which each of the subspheres $S^k$ is a subcomplex, by regarding each $S^k$ as being obtained inductively from the equatorial $S^{k−1}$ by attaching two k cells, the components of $S^k−S^{k-1}$. The infinite-dimensional sphere $S^\infty = \bigcup_k S^k$ then becomes a cell complex as well.

I know how to show that $S^\infty$ is contractible in various ways, but they all include bringing in a lot of stuff that hasn't been mentioned up to this point in Hatcher -- for instance, you basically have to know that this cell structure is the same as the one induced on the unit sphere in $\mathbb{R}^\infty$ (a space which isn't even mentioned in Hatcher so far as I can tell), and then construct a fairly tricky series of homotopies.

There's absolutely nothing like this in Chapter 0 of Hatcher -- every construction and proof in the book up to that point is done via fairly straightforward operations on the cells. The obvious implication is that there should be some such solution to this exercise as well, but I don't see one.

The only (possible) hint is that the previous exercise asks one to enumerate the subcomplexes of $S^\infty$ with the given cell complex structure. Unless I'm mistaken, these are the spheres and hemispheres of each dimension, together with the empty complex and $S^\infty$ itself. But I don't see how that helps.

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Geometrically speaking, since $S^k$ is obtained inductively from the equatorial $S^{k−1}$ by attaching two $k$ cells, you can squeeze $S^{k−1}$ into one point in one of two $k$ cells. And then continue squeezing.

Just apply same strategy used in proposition 0.16 (which is the infinite concatenation of homotopies using $t$-interval $\left(\frac{1}{2^{n+1}},\frac{1}{2^{n}}\right)$) to make it a deformation retraction.

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