I am reading about CW complexes. My book uses the following definitions.
Definition: A cell complex $X$ is a Hausdorff space which is the union of disjoint subspaces $e_{\alpha} (\alpha \in \mathcal{A})$ called cells satisfying:
a) To each cell we associate an integer $n \geq 0$ called its dimension. If $e_{\alpha}$ has dimension $n$ we often use the notation $e_{\alpha}^n$ for this cell. We write $X^n$ for the union of all cells $e_{\alpha}^k$ with $k \leq n$. $X^n$ is called the $n$-skeleton.
b) If $e_{\alpha}^n$ is an $n$-cell, there is a "characteristic map" $\chi_{\alpha}: (B^n, S^{n-1}) \to (X, X^{n-1}) $ such that $\chi_{\alpha} |_{B^n – S^{n-1}}$ is a homeomorphism from $B^n – S^{n-1}$ onto $e_{\alpha}^n$.
Definition: If $A \subset X$ is a subset of $X$, we define $K(A)$ to be the intersection of all subcomplexes containing $A$. If $A \subset B$, $K(A) \subset K(B)$. Hence if $p \in e, K(p) = K(e) =K(\overline{e})$. Thus $K(A)$ is a subcomplex.
Definition:
C: $X$ is said to be closure finite if for each cell $e_{\alpha}^n$, $K(e_{\alpha}^n)$ is a finite subcomplex.
W: $X$ is said to have the weak topology if for each subset $F \subset X$, $F$ is closed iff $F \cap \overline{e}_{\alpha}^n$ is compact for each cell $e_{\alpha}^n$.
A cell complex satisfying (C) and (W) is called a CW complex.
Question:
If you look at definition in Hatcher's book, where he defines cell complexes inductively, the first step is to start with a discrete set $X^0$. Therefore, in this case, the fact that $X^0$ has discrete topology is just because of the definition. However, in my case, I don't see why this is necessarily true. However, the result is used in proving certain results later, and I wonder how to show that it is true by my definition.
I was thinking that it has to do with the weak topology; my approach would be to show that every subset is closed, for which it suffices to show that intersection of that subset with every cell is closed. I'm not sure if that leads anywhere; I don't know how to proceed.
Best Answer
The characteristic map is a continuous map from a compact space to a Hausdorff space, hence it is closed (a classic result in point-set topology), meaning that it maps closed sets to closed sets, so that $e_{\alpha}^{0}$ is closed, so that $\overline{e_{\alpha}^{0}} = e_{\alpha}^{0}$. Then for any subset $F \subset X$, $F \cap \overline{e_{\alpha}^{0}} = F \cap e_{\alpha}^{0} = e_{\alpha}^{0}$, which is a single point, which is always compact, so $F$ is always closed, completing the proof.