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This is from John M Lee's Topological manifold and the CW construction theorem (p139).

Theorem: Suppose $X_0 \subseteq X_1 \subseteq \cdots $ is a sequence of topological space satisfying the following conditions

(i) $X_0$ is a nonempty discrete space.

(ii)For each $n \ge 1$, $X_n$ is obtained from $X_{n-1}$ by attaching a (possibly empty) collection of $n$ cells.

Then $X= \bigcup_n X_n$ has a unique topology coherent with the family $\{X_n\}$ and a unique cell decomposition making it into a CW complex whose $n$-skeleton is $X_n$ for each $n$.

Proof: Give $X$ the final topology given by inclusion maps $i_n:X_n \rightarrow X$ ... The $n$-cells of any such decompositions are the components of $X_n \setminus X_{n-1}$...

I do not understand why the itacilized line is true. Each $n$ cell must be connected but why are they precisely the components? We do not even know what open sets partition them.

Note: The definition the book take for CW complex is that the space X has cell decomposition which satisfies closure finities and weakest topology with respect to the collection of closed cells.

Definition: A closed (open) $n$ cell is a topological space homeomorphic to $\bar{\mathbb{B}}^n$( $\mathbb{B}^n$).
If $X$ is a nonempty topological space, a cell decomposition of $X$ is a partition $\mathcal{E}$ of $X$ into subspaces that are open cells of various dimneions such that : for each cell $e \in \mathcal{E}$ of dimension $n \ge 1$, exists a continuous map $\Phi$ from closed $n$-cell $D$ into $X$, the characteristic map for $e$ that restriction a homeomoprhism from $Int \, D $ into $e$ and maps $\partial D$ into the union of all cells of $\mathcal{E}$ of dimension strict less than $n$.

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  • $\begingroup$ By assumption $X_n\cong X_{n-1}\bigcup_\varphi\left(\bigsqcup_\alpha D_\alpha\right)$. It follows that there is a homeomorphism $X_n\mbox{-}X_{n-1}\cong \bigsqcup_\alpha \mathring{D^n}$ with adisjoint union of open n-disks, indexed over the n-cells of $X$. $\endgroup$ – Tyrone Oct 29 '17 at 13:51
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Every open $n$-cell $U$ is an open subset of $X_n$. To prove this, we use the fact that $X_n$ is obtained from $X_{n-1}$ by attaching $n$-cells, so a subset of $X_n$ is open iff its intersection with $X_{n-1}$ is open in $X_{n-1}$ and its inverse image under the characteristic map of each $n$-cell is open. In this case, $U\cap X_{n-1}$ is empty, as is the inverse image of $U$ under the characteristic map of any $n$-cell other than $U$ itself. The inverse image of $U$ under its own characteristic map $D\to X_n$ is the interior of $D$, which is open in $D$. Thus $U$ is open in $X_n$.

In particular, then, the open $n$-cells form a partition of $X_n\setminus X_{n-1}$ by open sets. Since each one is connected, they are the connected components of $X_n\setminus X_{n-1}$.

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