Is the support of the differential of a k-form a subset of the support of the k-form? I'm finishing up a self-study of Munkres's Analysis on Manifolds and ran across this claim:

Let $\omega$ be a $(k-1)$-form defined on an open set of $R^{n}$ containing a compact oriented $k$-manifold $M$ (for $k>1$). Then the support of $d\omega$ is contained in the support of $\omega$.

Is there a simple way to prove this?
Having a hard time treating the case where $x \in R^{n}$ is a limit point of the support of $d\omega$, rather than simply a point of $R^{n}$ at which $d\omega(x)$ is non-zero.
Thanks in advance.
 A: Here is a proof.
Note that if $\omega$ is a differential form, which is $0$ on some open set $U$, then $d\omega=0$ on $U$ as well (the answer by Albert essentially shows this).
Now let $p$ be a point outside $\text{supp}(\omega)$, and $U$ an open set around $p$ that is disjoint from $\text{supp}(\omega)$.  Then $\omega=0$ on $U$, and by the previous paragraph, $d\omega=0$ on $U$ as well. Thus $p$ is also not in $\text{supp}(d\omega)$.  It's easy to see this implies $\text{supp}(d\omega) \subset \text{supp}(\omega)$.
A: Generally, any (k-1)-form $\omega$ can be represented as

(1) $\omega$ = $\sum_{[I]}^{}$ $f_{I} dx_{I}$,

for I = {$i_{1}$, … , $i_{k-1}$} an ascending (k-1)-tuple from the set {1, … , n}; $f_{I}$ a $0$-form defined on $R^{n}$; and $dx_{I}$ = $dx_{i_{1}} \wedge ... \wedge dx_{i_{k-1}}$, where $dx_{i}$ is the elementary 1-form $\tilde{{\phi}_{i}}$ in $R^{n}$. The symbol [I] indicates that the summation extends over all ascending (k-1)-tuples from the set {1, … , n}.
Suppose x $\notin$ Support $\omega$. Then $\omega(x) = 0$ on all (k-1)-tuples ((x; $v_{1}$), ...,(x; $v_{k-1}$)), $v_{i} \in R^{n}$. It's easy to show that $\omega(x) = 0$ implies each of the $f_{I}$ in (1) equals zero. For example, suppose k = 4 and n = 5. To show that $f_{(1,2,5)}= 0$, evaluate the right-hand side of (1) on the 3-tuple of vectors ((x; $e_{1}$), (x; $e_{2}$), (x; $e_{5}$)).
That each of the $f_{I}$ equals $0$ implies that

(2) d$\omega$ =  $\sum_{[I]}^{}$ $df_{I} \wedge $ $dx_{I}$

also equals zero for this choice of x.
We've established that

(3) d$\omega$(x) $\neq  0 \Rightarrow\omega(x) \neq 0$.

From (3) it follows directly that Closure $\big(\{x|d\omega(x)\neq 0\}\big)$ $\subset$ Closure $\big(\{x|\omega(x) \neq 0\}\big)$.
