What shown below is a reference form Analysis on Manifolds by James Munkres
So I saw that some authors define the differential operetor $d$ as a function from $\bigcup_{k=0}^\infty\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)$ to $\bigcup_{k=1}^\infty\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)$ such that $$ d\omega\in\Omega^{k+1}(A) $$ for any $\omega\in\Omega^k(A)$ but anyway to give any definition I think that it is importat to prove that if $h$ and $k$ are two different naturals then $$ \Omega^k(A)\cap\Omega^h(A)=\emptyset $$ because otherwise we do not how calculate the quantity $d\omega$ if $\omega$ is a $h$-form and $k$-form. Well it is not hard to prove that if $h,k\ge 1$ and $h\neq k$ then the above identity holds. Indeed if $x$ and $y$ are two point of $\Bbb R^n$ then $$ \mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)\cap\mathcal{L}^h\big(\mathcal T_y(\Bbb R^n)\big)=\emptyset $$ when they are different and when they are not. However unfortunately I did not be able to prove that $$ \Omega^0(A)\cap\Omega^k(A)=\emptyset $$ for any $k>0$ so I ask to prove it. In particualr I know that if there exist a scalar function $f$ in $\Omega^0(A)\cap\Omega^k(A)$ for some $k>0$ then $$ f(x)\in\Bbb R\cap\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big) $$ for any $x\in A$ and thus the statement follows proving that $$ \Bbb R\cap\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)=\emptyset $$ but unfortunately I did not to prove it. Anyway if we define a $0$-tensor as an element of the set $\Bbb R^{\{0\}}$ where $0$ is the null vector of $\mathcal T_x(\Bbb R^n)$ then the above identity becames $$ \Bbb R^{\{0\}}\cap\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)=\emptyset $$ and it trivially holds because $\mathcal{L}^k\big(\mathcal T_x(\Bbb R^n)\big)\subseteq\Bbb R^{\big(\mathcal T_x(\Bbb R^n)\big)^k}$ and $$ \Bbb R^{\{0\}}\cap \Bbb R^{\big(\mathcal T_x(\Bbb R^n)\big)^k}=\emptyset $$ so that the above identity follows. So could someone help me, please?
In practice, we do not actually care whether $\Omega^k(\mathbb{R}^n)$ and $\Omega^h(\mathbb{R}^n)$ are literally disjoint sets or not when $k \ne h$. Instead, in a completely formal development, we would make it explicit that we are talking about a family of maps $d^{(k)} : \Omega^k(\mathbb{R}^n) \to \Omega^{k+1}(\mathbb{R}^n)$. However, in most situations it will be clear from context which map $d^{(k)}$ we must be thinking of applying so that putting in the explicit index would be unnecessary noise.
For a somewhat more complex example of what I mean by that last sentence, we will also have a wedge product $\wedge$ which formally is actually a family of maps ${\wedge}^{(k,\ell)} : \Omega^k(\mathbb{R}^n) \times \Omega^\ell(\mathbb{R}^n) \to \Omega^{k+\ell}(\mathbb{R}^n)$. Then one result you can state is: for $\omega \in \Omega^k(\mathbb{R}^n)$ and $\tau \in \Omega^\ell(\mathbb{R}^n)$, we have $$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^k \omega \wedge d\tau.$$ Now, from this context, it is straightforward to go through and assign what indices would have to be filled in for a more formal system to be able to accept the statement: $$d^{(k+\ell)}(\omega \wedge^{(k,\ell)} \tau) = (d^{(k)} \omega) \wedge^{(k+1,\ell)} \tau +^{(k+\ell+1)} [(-1)^k \cdot^{(k)} \omega] \wedge^{(k,\ell+1)} (d^{(\ell)} \tau).$$ (Note that here, I have also made it explicit that vector addition and scalar multiplication of differential forms are also formally indexed families, i.e. we will never try to add $\omega \in \Omega^k(\mathbb{R}^n)$ to $\tau \in \Omega^h(\mathbb{R}^n)$ unless $k = h$.)
As a side note, this point of view, that we don't really care in practice whether different sets might actually overlap or not, is part of the idea behind "type theory" as a language for mathematics. In most type theories, the question of whether elements of unrelated types are equal becomes something that it is not even meaningful to ask, in the sense that the corresponding formulas would not qualify as "well-formed formulas". And I'm personally convinced that outside of axiomatic set theory, a vast majority of mathematics is actually done and thought about in some version of type theory rather than in ZFC. Certainly, ZFC is useful from a theoretical point of view because of its restricted language and few axioms, and type theory is usually not too difficult to build "on top of" ZFC. A useful analogy to a computer science situation is to think of ZFC as a form of "machine code", and think of type theory as a higher-level language which can be translated to machine code.