In the system of ODEs below, I have (if I'm correct) the function $$ f:\mathbb R^{n(k-1)}\times \mathbb R^n \times \mathbb R\rightarrow \mathbb R^n \tag 1 $$
From the cartesian product I can write $\mathbb R^{n(k-1)}=\mathbb R^{nk}\times \mathbb R^{-n}$.
But what is the meaning of the set $\mathbb R^{-n}$?
Bakground:
I found $f$ from the following system of ODEs: \begin{align} x_1^{(k)}&=f_1(t,x,x^{(1)},\dots, x^{(k-1)})\\ & \,\,\, \vdots \\ x_n^{(k)}&=f_n(t,x,x^{(1)},\dots, x^{(k-1)}),\\ \end{align} where $x:\mathbb R\rightarrow \mathbb R^n$ and $t\in \mathbb R$.
My work:
Because $x:\mathbb R\rightarrow \mathbb R^n$, I also have $x^{(1)},\dots, x^{(k-1)}, x^{(k)}:\mathbb R \rightarrow \mathbb R^n$.
And for each component $f_i$, $i=1, \dots, n$, I have:
$x^{(1)},\dots, x^{(k-1)}$ are $(k-1 )$ functions with $n$ components each, so $n(k-1)$ and therefore $\mathbb R^{n(k-1)}$.
$x$ is a function with $n$ components, so $\mathbb R^n$ and $t$ is a scalar, $t\in \mathbb R$.
Thus, $$ f:\mathbb R^{n(k-1)}\times \mathbb R^n \times \mathbb R\rightarrow \mathbb R^n $$
Is the domain of the function correct? If so, what is the meaning of $\mathbb R^{-n}$?
The equation $\mathbb{R}^{a} \times \mathbb{R}^b = \mathbb{R}^{a+b}$ is only defined for nonnegative $a$ and $b$. That doesn't automatically extend to negative “exponents” in a meaningful way.
If there were such a set $\mathbb{R}^{-n}$, then $\mathbb{R}^n \times \mathbb{R}^{-n}$ would have to be $\mathbb{R}^0$. But $\mathbb{R}^n$ is infinite, and $\mathbb{R}^0$ has cardinality one, making the equation impossible. For if $\mathbb{R}^{-n}$ is nonempty, then $\mathbb{R}^n \times \mathbb{R}^{-n}$ is infinite, and if $\mathbb{R}^n$ is empty, then $\mathbb{R}^n \times \mathbb{R}^{-n}$ is empty.