Mistake in Hatcher A $1$-dimensional cell complex, $X$, is called a $\textit{graph}$.  A subset of $X$ is closed iff its intersection with the closure $\overline{e_\alpha}$ of each edge $e_{\alpha}$ ($1$-dimensional cell) is closed in $\overline{e_\alpha}$.  So every sequence in $X$ of points belonging to the interiors of disjoint edges is closed and doesn't converge.  This is correct but then Hatcher makes the following incorrect claim.  "This cell complex topology cannot be a metric topology if $X$ contains a vertex that is a common vertex of infinitely many edges; otherwise take a sequence of points in the disjoint interiors of these edges that gets closer and closer to the common vertex."  The common vertex is not in the closure of this sequence and hence there is no contradiction with the fact that, in a metric space, every point in the closure of a set is the limit of a sequence of points in the interior of the set.
 A: Hatcher's point is correct. Suppose $\alpha$ is a vertex common to infinitely many edges $E_i$ ($i\in\mathbb{N}$), and suppose $d$ were a metric on $X$ which induced the given topology. Let $x_i$ be a point on $E_i$ with $d(x_i, \alpha)<2^{-i}$.
On the one hand, we know by the definition of the topology on $X$ that $\{x_i:i\in\mathbb{N}\}$ is closed, and so $\alpha$ is not a limit point of $\{x_i:i\in\mathbb{N}\}$. On the other hand, for every $\epsilon>0$ there is some $i\in\mathbb{N}$ with $d(x_i, \alpha)<\epsilon$, so $\alpha$ is a limit point of $\{x_i:i\in\mathbb{N}\}$.
This gives us a contradiction, and so our original assumption - that $X$ is metrizable - was wrong.
A: Hatcher is correct. The contradiction is that in the CW topology no such sequence of points in the interior of distinct edges converges, but if we had a metric topology we could pick a point on the interior of the nth edge that was distance $1/n$ away from the common vertex. Since the distance between the nth point and the common vertex goes to 0, this converges which contradicts our earlier statement.
