\magnification1200

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\def\pf{\medskip\noindent{\sl Proof:}\ }
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\noindent{\bf Prob. 1 \& 3, p. 221.}
\item{1.} Let $X$ be a CW complex and $A\subset X$
a compact subset.
\itemitem{(a)}
Show that $A$ intersects only finitely many open cells of $X$.
Where do you use ``closure{--}finiteness''?

\pf
To help with this problem as well as part (b) to come, put a
partial ordering on the open cells of $X$ as follows:
$e_\alpha\prec e_\beta$ provided 
$\bar e_\beta\cap e_\alpha\neq\emptyset$ and 
$\dim e_\alpha < \dim e_\beta$.
Write $e_\alpha< e_\beta$ provided there exits a sequence
of cells $e_{\alpha_0}=e_\alpha$, $e_{\alpha_1}$, \dots
$e_{\alpha_r}=e_\beta$ such that $e_{\alpha_i}\prec e_{\alpha_{i+1}}$
for $0\leq i<r$.
Since the dimension of a cell is finite, any chain
$\cdots\prec e_{\alpha_r}\prec \cdots $ must have finite length.
Indeed, the maximal such length is the dimension plus 1.
Since $e_\alpha\prec e_\beta$ if and only if
$\dot e_\beta\cap e_\alpha\neq\emptyset$.
If we fix $e_\beta$, ``closure{--}finiteness'' implies that there are 
only finitely many open cells $e_\alpha$ with $e_\alpha\prec e_\beta$.
It follows that the set of open cells $e_\alpha$ with $e_\alpha<e_\beta$
is also finite.
This is a general result about partial orderings.
Let ${\cal I}(e_\beta,0)=\{e_\beta\}$.
Define ${\cal I}(e_\beta,r)$ inductively by
${\cal I}(e_\beta,r)=\{ e_\alpha\ \vert\ \exists\ \alpha\ {\rm such\ that\ }
e_\beta\prec e_\alpha\ {\rm and\ } e_\alpha\in{\cal I}(e_\beta,r-1)\}$ 
It follows by induction that the cardinality of 
${\cal I}(e_\beta,r)$ is finite for $0\leq r$.
For $r>\dim e_\beta$, ${\cal I}(e_\beta,r)=\emptyset$.
The set of open cells $e_\alpha$ with $e_\alpha<e_\beta$
is the union over $r$ of ${\cal I}(e_\beta,r)$ and this is finite.

Let $E$ denote any set of points such that $E\cap e_\alpha$
has finite cardinality for each open cell $\alpha$.
We claim that the subspace topology inherited by $E$ is the
discrete topology.
Let $F$ be any subset of $E$.
Note $F\cap \bar e_\beta\subset (F\cap e_\beta) \cup(
\displaystyle\mathop{\cup}_{e_\alpha<e_\beta} F\cap e_\alpha)$
and since the right{--}hand side is a finite union of finite sets,
the left{--}hand side is finite.
Since $X$ is Hausdorff, a finite set of points is closed so
$F\cap \bar e_\beta$ is closed in $bar e_\beta$.
Hence $F$ is a closed subset of $X$.
Since $F$ is an arbitrary subset of $E$, $E$ inherits the
discrete topology.

Finally, let $A\subset X$ be compact and let
$E=\{x_\alpha\}$ be a choice of one point for each open cell
$e_\alpha$ with $e_\alpha\cap A\neq\emptyset$.
Then $E$ inherits the discrete topology.
But all the discrete subsets of compact sets are finite.
Hence $E$ is finite.
\medskip
\itemitem{(b)} Show that $A$ is contained in a finite subcomplex.

\pf
Let $K\subset X$ be the union of all open cells $e_\alpha$
where $e_\alpha<e_\beta$ for some $e_\beta$ with
$A\cap e_\beta\neq\emptyset$.
By construction, $K$ is a union of finitely many open cells.
If $e_\alpha\subset K$ and $\dot e_\alpha\cap e_\beta\neq\emptyset$,
then $e_\beta\prec e_\alpha$, so $e_\beta\subset K$.
Hence $K$ is a subcomplex.
\bigskip
\item{3.}
For $X$ a CW complex, $X$ is compactly{--}generated.

\pf
We show
$i_X\colon X_C\to X$ is a closed map.
Let $A\subset X_C$ be closed.
Then $A\cap D$ is a closed subset of $D$ for all compact
$D\subset X$.
Since $\bar e_\alpha$ is the image of a compact set, it
is compact, so $A\cap \bar e_\alpha$ is a closed subset
of $\bar e_\alpha$ for all open cells $\alpha$.
But this means $A$ is closed in $X$.


\end