Manifolds with boundaries and partitions of unity

How do I

1 show that $M=[0,3]\subset \mathbb{R}$ is a manifold with boundary?
2 find a $C^2$ partition of unity for the open cover $M=[0,2)\cup(1,3]$?
3 show that $\omega=(x-2)dx$ is/is not an orientation on $M$?

What I know:
Let $M$ be a manifold in vector space $V$. Then it is covered by coordinate patches $f:A\rightarrow B$, where $A$ is open in a vector space $W$ and $B\subset M$ is open in $V$.
Now for $a\in W^*$ we deine the halfspace $H_a=a^{-1}[0,\infty)$. Its boundary is $\partial H_a=\ker a=a^{-1}\{0\}$. Now a coordinate patch with boundary is $f:A\rightarrow B$ if $A$ is open in $H_a$ and $B$ is open in $M$. And a manifold with boundary is the union of such coordinate patches with boundary.

For a manifold $M=\bigcup_i A_i$ there exist smooth functions $g_i:M\rightarrow [0,1]$ such that for all $x\in M$ and all $i$:
1. $\sum_i g_i(x)=1$,
2. $\text{supp}(g_i)=\overline{\{x\in M:g_i(x)\neq 0\}}\subset A_i$
3. $x\in M$ has neighbourhood on which all but finitely many $g_i$ are zero.
This is a partition of unity.

An orientation on $n$-dimensional $M$ is a $C^1$ differential $n$-form $\omega\in \Omega^n(M)$ that is nowhere $0$.

• For 3): $\omega$ is not an orientation since it is zero at $x=2$. – studiosus Nov 16 '16 at 20:55
• @studiosus Is it that simple? – user389900 Nov 16 '16 at 20:56
• @DannyRandall yes it is. – user2520938 Nov 16 '16 at 20:58
• @user2520938 Could you maybe help me with the other questions as well? – user389900 Nov 16 '16 at 20:59
• Have you tried anything? – Pedro Nov 17 '16 at 4:56