Variation on Stokes Theorem for Manifolds (2) Let $\omega \in \Omega^0(\mathbb{R}^{2}\setminus\{0\})$ be a $0$-form such that $d\omega=0$. Is the following statement true:
For any compact, oriented, $0$-dimensional submanifold $M$ of $\mathbb{R}^{2}\setminus\{0\}$, we have $\displaystyle \int\limits_{M}j^*\omega = 0$ with $j:M \rightarrow \mathbb{R}^{2}\setminus\{0\}$ the inclusion map.
I was looking for a counterexample, more or less like the first example on http://en.wikipedia.org/wiki/Closed_and_exact_differential_forms. But I couldn't work it out exactly. How can a point be a boundary of something else (in order to apply Stokes)? 
 A: As Olivier suggests, if $f$ is a function (aka 0 form) on $\mathbb{R}^2-\{0\}$, and $df = 0$, then $f$ is a constant.
If you take a path in your space $\gamma$ from point $p$ to point $q$, then the boundary of this path is $q-p$ (i.e. the two endpoints, but with opposite signs).
Then we have 
$$
0 = \int_\gamma df = \int_{b\gamma} f = f(q)-f(p) = 0
$$
which verifies Stoke's theorem.
It seems like you were intrigued by the fact that "$d\theta$" is closed but not exact.  In other words, it represents a nontrivial cohomology class in $H^1(\mathbb{R}^2-\{0\})$.
You seem to be looking for nontrivial cohomology in $H^0$, in other words functions which are closed but not exact.  The problem is, there are no "-1 forms", so $H^0$ is just defined as the space of closed forms.  In other words the locally constant functions.  In your case, since your space is connected, this is one dimensional:  just the constant functions.  If your space were disconnected (say a union of $n$ disjoint disks) then functions with $df=0$ can be a different constant on each  connected component.  So $H^0 \cong \mathbb{R}^n$ in this case.
Does this all make sense?  Another interesting example to go hunting for is a closed but not exact $2$-form on $\mathbb{R}^3-\{0\}$.  This should look more like the wikipedia example, if you are just looking to play around with similar things to test your understanding.
