How can you calculate the product of a set $A$, denoted by $\Pi A$ and defined by
$\forall z \in \Pi A(z \subseteq \bigcup A \wedge \forall y \in A (\exists x (z \cap y = \lbrace x \rbrace))) $
Also, is it different for finite and infinite sets?
How can you calculate the product of a set $A$, denoted by $\Pi A$ and defined by
$\forall z \in \Pi A(z \subseteq \bigcup A \wedge \forall y \in A (\exists x (z \cap y = \lbrace x \rbrace))) $
Also, is it different for finite and infinite sets?
Copyright © 2021 JogjaFile Inc.
If the sets in $A$ are not pairwise disjoint, then the product you define can be empty. This is not a standard situation, and certainly this definition (in general) is not the standard definition.
To see why this might be the case consider $A=\{\{1,2\},\{1,3\},\{2,3\}\}$, and you will see that there are no $z$ satisfying this definition.
The standard definition for the product of sets (we don't require them to be disjoint) is to say that $f\in\prod A$ is a function with domain $A$, and $f(a)\in a$ for all $a\in A$. So if $A$ is a collection of pairwise disjoint sets, then this is easily equivalent, simply take the range of $f$ to be $z$ as in your definition, and take $f(a)$ to be the unique $x$ such that $x\in z\cap a$.
As to how you calculate such a set (assuming that you require $A$ to be a set of pairwise disjoint sets)? You just go over all the possible sets which satisfy the definition, there's really no concrete algorithm. In the case of an infinite $A$, you need the axiom of choice to ensure that it isn't empty, and the result is an infinite set which is usually "very large".