2
$\begingroup$

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 $t=[0,\infty[$?

$\endgroup$
  • 2
    $\begingroup$ Look up cartesian product $\endgroup$ – D_S Jun 23 '17 at 12:27
2
$\begingroup$

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 $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$.

$\endgroup$
  • $\begingroup$ Thank you! I updated my question regarding the cartesian product. $\endgroup$ – JDoeDoe Jun 23 '17 at 17:43
2
$\begingroup$

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.

Regarding your update:

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.

$\endgroup$
  • $\begingroup$ Thank you! I updated my question regarding the cartesian product. $\endgroup$ – JDoeDoe Jun 23 '17 at 17:43
  • $\begingroup$ 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? $\endgroup$ – JDoeDoe Jun 24 '17 at 8:53
  • $\begingroup$ 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. $g(t_1,x_1,x_2,x_3,\xi_1,\xi_2,\xi_3)=f(t_1,(x_1,x_2,x_3),(\xi_1,\xi_2,\xi_3))$ $\endgroup$ – mvw Jun 24 '17 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.