# Hatcher 0.15

Hatcher, Algebraic Topology, Chapter 0

15. Enumerate all the subcomplexes of $S^\infty$, with the cell structure on $S^\infty$ that has $S^n$ as its $n$-skeleton.

Proof. Recall that by definition, $S^\infty = \cup_{n=0}^\infty S^n$ (see page 7 in Chapter 0). Moreover, recall that for $n$ finite, $S^n$ can be given a cell structure in which each of the spheres $S^k$, $k\leq n$, is a subcomplex: This is done by way of attaching two $k$-cells to the $S^{k-1}$ considered as the equator as $S^k$ for all $k\leq n$. More precisely:

Begin with $S^0$ as two points $e^0_1=x_1$ and $e^0_2=x_2$; obtain $S^1$ by attaching two “arcs” (i.e., closed segments homeomorphic to the 1-dimensional “disk”) $e^1_1$ and $e^1_2$ to create a circle; next, obtain $S^2$ by attaching two 2-cells (i.e., closed regions homeomorphic to the 2-dimensional disk) $e^2_1$ and $e^2_2$. Continue inductively by attaching two $n$-cells $e^n_1$ and $e^n_2$, each homeomorphic to the $n$-disk $D^n$, to the resulting $S^{n-1}$. This yields a CW-complex structure on each $S^n$ which has two cells in each dimension $d=0,1,\ldots,n$ and which has $S^k$ as a subcomplex for each $k\leq n$. The construction for $S^k$, $k=0,1,2$, is shown in this solution to a previous exercise (see Figure 3 in that solution).

Therefore, we have that $S^n = \cup_{k=0}^n(e^k_1\cup e^k_2)$ and that $S^\infty$ is the union of all the $S^n$. Also, for any $n$, a subcomplex $X$ of $S^n$ is automatically closed (by definition of subcomplex) and therefore must contain the boundary $\partial e^k_i \cong S^{k-1}$ of any $k$-cell $e^k_i$ it contains, $i=1,2$. What does this mean? It means that any (closed) subcomplex $X$ of $S^\infty$ which contains a single $k$-cell must also include every single $j$-cell for $j. Thus, it follows that the only $k$-dimensional subcomplexes of $S^\infty$ are the hyper-hemispheres1 of the form $e^k_i\cup e^{k-1}_{1,2}\cup\cdots\cup e^1_{1,2}\cup e^0_{1,2}$ and so enumerating these for all $k=0,1,\ldots$ yields the final result.    $\square$

1. Thanks to Tarun Chitra for coming up with a good prose way to describe these objects geometrically. His work can be found here.