Exercise 0.15

Solution. Recall that any $S^n$ is constructed as a CW complex consisting of $S^{n-1}$, $e_1^n$, and $e_2^n$, where we identify $\partial e_i^n\simeq S^{n-1}$. Thus, $S^n={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^n(e_1^k\cup e_2^k)$.

Now suppose $\varnothing \ne A\subseteq S^{\infty }$ is a subcomplex. Then, $A$ contains some $k$-cell $e_i^k$ for $i=1$ or $2$. Then, since for $A$ to be closed we must have $S^{k-1}=\partial e_i^k\subseteq A$ as well, we see that $A$ must contain every $\ell$-cell for $\ell <k$. Thus, any subcomplex $A$ is in the collection:

for supposing $A\ne \varnothing$, if there is a maximal $n$ such that $e_i^n\subseteq A$ for $i=1$ or $2$, then $A=S^n$ or $e_i^n\cup S^{n-1}$, and if not, then for any $n$, $A$ must contain some $e_i^{n^{\prime }}$ where $n^{\prime }>n$, and so $A$ contains all $e_i^n$ by the argument above, i.e., $A=S^{\infty }$. □