Can sets contain coordinates of points and be used in Cartesian product?

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SUMMARY

This discussion confirms that sets can indeed contain coordinates of points, such as A = {(1,3), (4,5)}. The Cartesian product of such sets is valid, demonstrated through examples like A × B = {((1,3),(7,8)), ((1,3),(4,2)), ((4,5),(7,8)), ((4,5),(4,2))}. The notation for representing points can vary, but consistency is key. The Cartesian product also addresses universal mapping problems in set theory.

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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"
 
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control said:
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 carthesian 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?).
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 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.
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.
 
Thanks for answer.
 
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:
zwierz said:
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
You forgot to mention that the Cartesian product solves a universal mapping problem.
 

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