How can I prove $(0,2]$ is a smooth manifold? And something questions about the choice of the open set of a manifold?

Question

I am studying manifold, when I see the definition about the topological manifold, I am confused about the choice about the open set. For example, when we study the real line $\mathbb{R}$ we usually choose $(a,b)$ to be the open set we consider, but when it comes to the form $(a,b]$, we have to choose $(c,b] \subset (a,b]$ to be the open set to cover the $(a,b]$ (am I right?). But if we want to show that $(a,b]$ is a smooth manifold, we have to find the chart. So how can I find the proper homeomorphism which makes the open set of $(a,b]$ is homeomorphic to the open subset of $\mathbb{R}$?

Answer 1

It turns out that $M= (0, 2]$ is a topological manifold of dimension $1$ with boundary.

Here is an interior chart. Define $\phi : (0, 2) \subseteq M \to \mathbb{R}^1$ by $\phi(x) = x$, then $((0, 2), \phi)$ is an interior chart since $\phi$ is clearly a homeomorphism. So every point $p \in (0, 2) \subseteq M$ is contained in a neighborhood of $M$ homeomorphic to an open susbet of $\mathbb{R}^1$ and so all such $p$ are (manifold) interior points of $M$.

Now here is a boundary chart, define $\psi : (1, 2] \subseteq M \to \mathbb{H}^1 = [0, \infty)$ by $\psi(x)= -x+2$, then $\operatorname{Im}(\psi) = [0, 1)$ which is open in $\mathbb{H}^1$ and $\psi$ is clearly a homeomorphism. Thus $\{2\}$ is contained in a neighborhood of $M$ homeomorphic to an open subset of $\mathbb{H}^1$ and so $2$ is a (manifold) boundary point of $M$.

Thus $M$ is a topological manifold of dimension $1$ with boundary.