How do I prove $\Omega^{0}(U)=C^{\infty}(U)?$

Question

How do I prove $\Omega^{0}(U)=C^{\infty}(U)?$

$\Omega^{0}(U)$ is the vector space of $C^{\infty}$ $0-$forms on $U$.

A $k-$ form $\omega$ on $U$ is a linear combination $$\omega =\sum_{I}a_I dx^I ,(\text{$I$ is a $k$ index set})$$ , with function, coefficients $a_I : U →\mathbb R$. We say that a $k$-form $\omega$ is $C^{\infty}$ on $U$ if all the coefficients $a_I$ are $C^{\infty}$ functions on $U$. For ordinary $k$-form there are $^nC_ k$ base elements exists. For $0$-form only one base element exists. $\omega=a(p)x$, where $x$ is the base element. If $a:U \to \mathbb R$ $C^{\infty}$ then $\omega \in C^{\infty} \implies $ $\Omega^{0}(U)\subset C^{\infty}(U)$

How do I prove the converse? Please help me.

Answer 1

The answer is right there in the definition of $\Omega^0$. It simply assigns every point $p \in U$ a real number, i.e. an element of $A_0(T_p \mathbb{R}^n)$. Such an object, that assigns every point in $U$ a unique value in $\mathbb{R}$ is just a $C^\infty(U)$ map by definition.