I Sets, Cartesian product

1. May 12, 2017

control

Hi guys,
I would like to ask if a set can contain coordinates of points, for example A={[1,3];[4,5];[4,7]} and if we can do Cartesian product of such sets, for example A={[1,3];[4,5]}, B={[7,8];[4,2]} A×B={[1,3][7,8];[1,3][4,2];[4,5][7,8];[4,5][4,2]} (is it correct to write it like that?). I am familiar with doing that when we have sets of numbers (A={1;2}, B={7;5} A×B={[1,7];[1,5];[2,7];[2,5]}). but I am not sure if it is correct with coordinates of points.
Mod note: Fixed typo "carthesian"

Last edited by a moderator: May 13, 2017
2. May 12, 2017

Staff: Mentor

You can write the elements whichever you want, e.g. $[1,3][7,8]$ or $[1,3;7,8]$ or $[1,3,7,8]$ or $\begin{bmatrix}1&3\\7&8\end{bmatrix}$. It is certainly useful not to mix them like $[1,7][3,8]$, because this would probably be harder to read, but as long as you're consistent, there is no rule.
I would probably write $A=\{(1,3),(4,5)\}\; , \;B=\{(7,8),(4,2)\}$ as round parenthesis are more common for tuples and commas as separators in a list, and then $A \times B = \{\; ((1,3),(7,8))\, , \, ((1,3),(4,2))\; , \;((4,5),(4,2))\, , \,((4,5),(4,2))\;\}$ but only in set theory. With different applications, this might change.

3. May 12, 2017

control

4. May 13, 2017

zwierz

Perhaps the definition of the Cartesian product would be of some use. Let $\Gamma$ be be an arbitrary nonvoid set, and a set $A_\gamma$ is putted in correspondence to each element $\gamma\in\Gamma$. Then by definition a set $\Pi_{\gamma\in \Gamma}A_\gamma$ consists of functions $f:\Gamma\to \bigcup_{\gamma\in \Gamma}A_{\gamma}$ such that $f(\gamma)\in A_\gamma$.
For example a set $\mathbb{R}\times\mathbb{N}$ consists of functions $f:\{1,2\}\to \mathbb{R}\cup\mathbb{N}$ (it looks little bit strange, obviously $\mathbb{R}\cup\mathbb{N}=\mathbb{R}$) such that $f(1)=a_1\in\mathbb{R},\quad f(2)=a_2\in\mathbb{N}$. This function is also presented as $(a_1,a_2)$.
Another example: a set $\mathbb{R}^\mathbb{N}$ consists of all functions $f:\mathbb{N}\to\mathbb{R}$ those functions can be presented as infinite sequences $(a_1,a_2,\ldots),\quad f(i)=a_i\in\mathbb{R}$.
If all the $A_\gamma$ are vector spaces over the same field then $\Pi_{\gamma\in \Gamma}A_\gamma$ is also a vector space. By the Choice axiom the set $\Pi_{\gamma\in \Gamma}A_\gamma$ is not empty as long as all the sets $A_\gamma$ are not empty

Last edited: May 13, 2017
5. May 13, 2017

Staff: Mentor

You forgot to mention that the Cartesian product solves a universal mapping problem.