# Give a cell decomposition of $[0, 1] \subseteq \mathbb{R}$

I'm trying to come up with examples to better understand the definition of the cell decomposition of a topological space $X$.

The simplest example I could think of would be $X =[0, 1] \subseteq \mathbb{R}$.

This is the definition I'm working with.

If $X$ is a nonempty topological space, a cell decomposition of $X$ is a partition $\Gamma$ of $X$ into subspaces that are open cells of various dimensions, such that the following condition is satisfied: for each cell $e \in \Gamma$ of dimension $n \geq 1$ there exists a continuous map $\Phi$ from some closed $n$-cell $D$ into $X$ (called a characteristic map for $e$) that restricts to a homeomorphism from $\text{Int}(D)$ onto $e$ and maps $\text{Bd}(D)$ into the union of all cells of $\Gamma$ of dimensions strictly less than $n$.

So for $X = [0, 1]$ if I set $\Gamma = \{(0,1), \{0\}, \{1\}\}$ I run into a problem, because although $(0, 1)$ is homeomorphic to the open ball $\mathbb{B}^1$, $\{0\}$ and $\{1\}$ are both homeomorphic to the closed ball $\overline{\mathbb{B}^0}$, so unless there's a wilder decomposition, I can't partition $X$ into subspaces that are homeomorphic to open balls.

I'm guessing however that $[0, 1]$ does have a cell decomposition, so there must be a way to decompose $X$ into subspaces which are open cells, and where the union of those subspaces equals $X$, if so what would an example of a cell decomposition of $X$ be?

It has one $1$-cell and two $0$-cells. The continuous map $\Phi: \overline{e} \to \overline{e}$ is the identity map on $\overline{e}=[0,1]$. This certainly restricts to a homeomorphism (identity actually) on the open cell $e=(0,1)$ and maps the boundary to the $0$-cells $0$ and $1$.
Your idea actually does work, because $\mathbb{B}^0$ and $\overline{\mathbb{B}^0}$ are the same thing. In $\mathbb{R}^0$, any ball has just one point, whether it is open or closed.