CW decomposition of topo space

Question

we know that $S^n = e^0 \sqcup e^n.$ Then CW decomposition of $S^n$ the following $$S^n = \bigcup\limits_{i = 0}^nX^{(n)}, X^{(0)}\subseteq X^{(1)} \subseteq \ldots \subseteq X^{(n)},$$ where $X^{(0)} = e^0, X^{(n)} = e^n, X^{(1)},\ldots, X^{(n-1)} \mbox{ are empty}.$

My question is CW decomposition of $S^n$ above right or wrong??

I think that replace $X^{(n)} = e^n$ by $X^{(n)} = S^n$??

Can you explain for me? Thankyou very much!

Answer 1

If $X^{(1)}$ has $X^{(0)}\not=\emptyset$ as a subset, it can't be empty... In fact, the $(n-1)$-skeleton is the same as the $0$-skeleton in this case, because the $k$-skeleton is the union of all cells of dimension at most $k$. By the same statement, $X^{(n)}=S^n$, because it contains all the cells of $S^n$.

Answer 2

A really easy $CW$ decomposition of $S^n$ is given by using only an $n$-cell (say $e^n$) and a $0$-cell (say $e^0$). Consider the attaching map given by $f: \partial e^n \to e^0$ and define $X = e^n \bigsqcup_f e^o$.