Definition 1 :Suppose $(X,\mathcal{E})$ is a $CW$ complex. A subspace $S\subseteq X$ is said to be a $\textbf{subcomplex}$ of $(X,\mathcal{E})$ if there exists $\mathcal{E}'\subseteq \mathcal{E}$ such that
1- $S=\bigcup_{e'\in \mathcal{E}'\subseteq \mathcal{E}}e'$
2- If for $e\in \mathcal{E}$ we have $e\subseteq S$ then $\overline{e}\subseteq S$.
Definition 2:
A $\textbf{CW complex}$ is a cell complex $(X,\mathcal{E})$ satisfying:
$(C)$: For each $e\in \mathcal{E}$, $\overline{e}$ is contained in a union of finitely many cells in $\mathcal{E}$
$(W)$: The topology of $X$ is coherent with the family of closed subspaces $\{ \overline{e}$ $:$ $e\in \mathcal{E}$ $\}$
Theorem:
Suppose $X$ is a $CW$ complex and $Y$ is a subcomplex of $X$. Then $Y$ with the inherited cell structure from $X$ is a $CW$ complex:
Proof: Since a subspace of hausdorff space is hausdorff, $Y$ is hausdorff. Since $Y$ is a subcomplex of $X$, there exists an $\mathcal{E}'\subseteq \mathcal{E}$ such that $Y$ is the union of all the members in $\mathcal{E}'$ and $Y$ contains the closure of every cell of $X$ which it contains. We aim to show that $(Y,\mathcal{E}')$ is a $CW$ complex. Let us first show $(C)$ holds: Let $e \in \mathcal{E}'$. Then, $cl_{Y}(e)= \overline{e}\cap Y \subseteq \bigcup_{e'\in \mathcal{E}}\bigcup_{i=1}^ne_i\cap e'$ , where $(e_i)_{i=1}^n \subseteq \mathcal{E}$. Since $\mathcal{E}$ partitions $X$, we deduce $(C)$. Now to show $(W)$, we must show that the subspace topology on $Y$ is coherent with the family $\{ cl_{Y}(e)$ $:$ $e\in \mathcal{E}'\}$. First note that if $e\subseteq Y$ is a cell in $\mathcal{E}$ then $cl_{Y}(e)=\overline{e}\cap Y = \overline{e}$. That is, the subspace closure coincides with the closure. It therefore suffices to show that $S\subseteq Y$ is closed in $Y$ if and only if $S\cap \overline{e}$ is closed in $\overline{e}$ for every $e\in \mathcal{E}'$. The forward direction is obvious. For the backwards direction, suppose $S\cap \overline{e}$ is closed in $\overline{e}$ for every $e\in \mathcal{E}'$. We aim to show that $S$ is closed in $Y$. Note that if $\mathcal{E}= \mathcal{E}'$ then $X=Y$ and so we are done. Suppose $\mathcal{E}'$ is properly contained in $\mathcal{E}$. Thus choose some $e\in \mathcal{E}$ for which $e$ is not in $\mathcal{E}'$. Thus, $e$ is not contained in $Y$. Then $\overline{e}\backslash e$ is contained a union of finitely many cells in $X$…… How do I complete the proof from here?
Best Answer
Ok, so I'm showing that if $S\subseteq Y$ is such that $S\cap \overline{e}$ is closed for every $e\in\mathcal{E}'$, then $S\subseteq Y$ is closed. Since $Y$ is closed, it suffices to show $S\cap\overline{e}$ is closed for $e\in\mathcal{E}\setminus\mathcal{E}'$. For such an $e$, by (C), there exist finitely many cells $e_i$ such that $\overline{e}\subseteq\bigcup e_i$. Then: $$S\cap \overline{e}=(\bigcup e_i)\cap S\cap \overline{e} $$ We can select only those cells $e_i'$ which are in $\mathcal{E}'$, as the other ones cannot contribute to the intersection with $S\subseteq Y$. But then by 2-: $$S\cap \overline{e}=(\bigcup \overline{e_i}')\cap S\cap \overline{e} $$ By our assumption, $S\cap \bigcup\overline{e_i}'=\bigcup (S\cap\overline{e_i}')$ is closed, so $S\cap\overline{e}$ is also closed.