# Meaning of $\times$ in this definition of a function?

What is the meaning of writing $\times$ here?

For $(t,x,\xi)\in[0,\infty)\times\mathbb{R}^3 \times \mathbb{R}^3$ we consider the function $f(t,x,\xi)$.

Can't we just write $t\in[0,\infty)$, $x\in\mathbb{R}^3$ and $\xi\in\mathbb{R}^3$?

Update:

Using the cartesian product, does "$(t,x,\xi)\in[0,\infty)\times\mathbb{R}^3 \times \mathbb{R}^3$" actually mean

\begin{align*} [0,\infty[ \times \mathbb{R}^3\times \mathbb{R}^3= \bigg\{ (t,x,\xi):t&\in[0,\infty[,\\ x&=(x_1,x_2,x_3)\in \mathbb{R}^3,\\ \xi&=(\xi_1,\xi_2,\xi_3)\in \mathbb{R}^3 \bigg\} \quad ? \end{align*} And also, should I write $t\in[0,\infty[$ or $t=[0,\infty[$?

• Look up cartesian product – D_S Jun 23 '17 at 12:27

You CAN just write that. Although in certain contexts you may find it convenient to refer to all pairs $(x, y)$ where $x \in \mathbb{R}$ and $y \in \mathbb{R}$, which the Cartesian cross product gives you as $\mathbb{R} \times \mathbb{R}$.

RESPONSE TO THE UPDATE

Yes, the cross product does mean exactly that. $\mathbb{R}^3$ is short for $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$.

• Thank you! I updated my question regarding the cartesian product. – JDoeDoe Jun 23 '17 at 17:43

You can. The grouping of the three variables and domains each into a tuple adds no extra information here, I believe.

It would save some effort if such an aggregate was named and referenced further on, like in

$$(t,x,\xi)\in S = [0,\infty)\times\mathbb{R}^3 \times \mathbb{R}^3 \\ \dotsb \\ T \subset S, \forall u \in T: f(u)\dotsb$$ but here it is not named.

Seems fine. You should write $t \in [0,\infty)$, because $t$ is some element from the set $[0,\infty)$, thus a non-negative real number, not the set itself.
• Great! And also, from the cartesian product can we conclude that $f$ is a scalar function of seven variables? I.e. $f(t,x,\xi)=f(t,x_1,x_2,x_3,\xi_1,\xi_2,\xi_3)$, such that $\mathbb{R}^7\rightarrow\mathbb{R}$. Is it correct? – JDoeDoe Jun 24 '17 at 8:53
• No, your definition says nothing about the value set (codomain) of that function. About the arguments: $f$ has three arguments. While you can bijectively map those three arguments to seven arguments and can come up with a function $g$ which consumes the seven mapped arguments and will yield the same values as $f$, that function would likely be considered a different function. $g(t_1,x_1,x_2,x_3,\xi_1,\xi_2,\xi_3)=f(t_1,(x_1,x_2,x_3),(\xi_1,\xi_2,\xi_3))$ – mvw Jun 24 '17 at 13:44