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In the question unordered cartesian product an shorthand notation for the unordered cartesian product was discussed but without any standard notation. So my question is what would be the explicit definition of all different sets consisting of exactly one element of each subset of a set S. There can be an arbitrary number of sets in S. $\{\{a, b\}, \{c\}\}$ should lead to $\{\{a, c\}, \{b, c\}\}$

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It's unclear the question but i think i can give an answer:

You can think "the set of all different sets consisting on exactly one element of each subset of a a set $S$" as the cartesian product: \begin{equation} \prod\mathcal{P}(X)\setminus\{\emptyset\}=\{f\mid f:I\to\bigcup_{i\in I}X_i, \text{ }f(i)\in X_i\} \end{equation} where I is an index set for $\mathcal{P}(X)\setminus\{\emptyset\}$, i.e. $\mathcal{P}(X)\setminus\{\emptyset\}=\{X_i\mid i\in I\}$.

(You must sustract the empty set beacause you cannot take the cartesian product of a family which contains an empty set).

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    $\begingroup$ You can take the Cartesian product of a family which contains the empty set. It's $\emptyset$. $\endgroup$ – DanielWainfleet Jun 30 '16 at 6:43
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I have not seen any special symbol for this. I would write $S=\{\{a,b\}:a\in A\land b\in B\}.$ Of course if $x\in A\cap B$ then $\{x\}$ belongs $S.$ If you wish to exclude one-member sets you can write $T=\{\{a,b\}:a\in A\land b\in B\land a\ne b\}. $.... However there is are symbols $[A]^2$ for the set of all subsets of $A$ that have exactly $2$ members each, and $[A]^1=\{\{a\}:a\in A\}.$ So $T=S\backslash [A\cup B]^1.$ But this seems cumbersome.

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