How to prove that $M = [0,1] \subset \mathbb{R}$ is an analytic $1$-manifold with boundary? I would like to prove that the set $M = [0,1] \subset \mathbb{R}$ is a $1$-manifold with boundary. Therefore I need to show:
1) There exist a collection of open sets $V^{a} \subset \mathbb{R}$ with $U^{a} = V^{a} \cap M$ such that $M = \cup U^{a}$.
2) There exists an analytic diffemorphism $x^{a}$ defined on each $U^{a}$ that maps $U^{a}$ onto some set $W \cap H^{1}$ where $W$ is an open set in $\mathbb{R}$ and $H^{1}$ is closed half-space defined as $H^{1} = \{ x_{1} \in \mathbb{R}: x_{1} \geq 0 \}$
For 1) I take $U^{a} = \mathbb{R} \cap M$, i.e. $U^{a} = M$. So I just need one coordinate chart.
Now for 2) I don't know how to find a suitable $x^{a}$ and $W$. I am thinking to define $x^{a}$ as identity map, but then I don't know what $W$ should be. I was thinking to choose $W = (-\epsilon, 1+ \epsilon)$ for arbitrary small $\epsilon > 0$. Then $W \cap H^{1} = [0, 1+\epsilon)$ and of course an identity map does not work.
Edit. Also, by definition, the boundary $\partial M = \{0, 1\}$ should be mapped onto the boundary $\partial H^{1} = \{0\}$. How can this be satisfied if $x^{a}$ is a diffeomorphism?
Any help is appreciated!
 A: Notice that the condition in (2) imposes further conditions on the sets $V^a$ chosen in (1). In particular, one cannot choose arbitrary sets $V^a$ satisfying the condition in (1) alone and expect to be able to choose $x^a$ in (2) that work for those. In other words, in principle you should be choosing the family $\{(V^a, x^a)\}$ all at once.
In particular, we can see immediately that your choice of $\{V^a\}$ (namely, taking a single open set $V$ containing $M$, so that $U = V \cap M = M$) cannot work: The boundary of $U$ in $\Bbb R$ is $\{0, 1\}$, and this set should be mapped diffeomorphically via the chosen map $x : U \to W \cap H^1$, but the boundary of the latter has zero or one elements, according to whether $0 \in W$.
The idea here is that each of the constituent sets $V^a$ in your choice should only contain one piece of the boundary---after all, each set $V^a$ should look like some open subset of the half-plane $H^1$. If we take, for example, some collection $\{V^a\}$ with $V^1 = (-\epsilon, 1)$ for some $\epsilon > 0$, then $U^1 = [0, 1)$, so that the boundary of $U^1$ just has one piece, namely, $\partial U^1 = \{0\}$. Like you hint, you can choose $x^1$ to be the identity map and $W$ to be a suitable set ($W = V^1$ works) so that $(V^1, x^1)$ satisfies the criterion in (2).
Now, our $U^1$ misses precisely the point $1 \in M$, so to finish producing a collection $\{(V^a, x^a)\}$ satisfying the criteria (such a collection is called a (real-)analytic atlas on $M$), we need to pick some $V^2 \ni 1$ and a suitable analytic diffeomorphism $x^2$.
