Let $A_1,..., A_n$ be a family of sets of sets. I want to create a now set as the following:

The set $B$ is made of unions of all possible combinations of elements from any set.

For example: Let $A_1=\{\{1\},\{2\}\}$, $A_2 = \{\{3\}\}$ and $A_3 = \{\{4\}\}$. Then the set $B$ should be:

$$B=\{\{1\},\{2\}, \{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{2,3,4\},\{1,3,4\},\{1,2,3,4\}\}$$

My question is, how can I formally write this set?

My approach was the following:

  • First let's put all elements we want to combine in the same set: $\bigcup\limits_n A_n$

  • Then let's take it's power set: $\mathcal P\left(\bigcup\limits_n A_n\right)$
    In this power set we have all the combinations that we want:

Now we can define $B$ as:

$$B = \left\{ \bigcup_{a \in A} a : A \in \mathcal P\left(\bigcup\limits_n A_n\right)\right\}$$

My question is, Am I over complicating? Is there any other way of defining this set?

  • $\begingroup$ You mean $A_1 = \{1,2\}$, $A_2 = \{3\}$ and $A_3 = \{4\}$? $\endgroup$
    – azif00
    Commented Aug 19, 2020 at 18:49
  • $\begingroup$ No, those are supposed to be sets of sets. @Azif00 $\endgroup$ Commented Aug 19, 2020 at 18:50
  • 1
    $\begingroup$ So, you mean $\{A_1,\dots,A_n\}$ is a family of families of sets and $B$ is made of all possible unions of elements taken from each family $A_1,\dots,A_n$? $\endgroup$
    – azif00
    Commented Aug 19, 2020 at 18:53
  • $\begingroup$ In my approach incorrect? @Azif00? $\endgroup$ Commented Aug 19, 2020 at 19:02
  • $\begingroup$ @Azif00: You’re taking one union too many: you actually wnat the power set in your second version (or, to just by Eduardo’s example, the set of its non-empty elements). $\endgroup$ Commented Aug 19, 2020 at 19:04

1 Answer 1


$\bigcup_nA_n$ is the collection of all of the sets from which you can draw elements, so $\bigcup\bigcup_nA_n$ is the collection of all of the elements that you can use to form members of $B$; in your example




Apparently you want only the non-empty subsets of $B$, so


  • $\begingroup$ is my approach also correct? $\endgroup$ Commented Aug 19, 2020 at 19:14
  • 1
    $\begingroup$ @EduardoMagalhães: Yes, it appears to be. $\endgroup$ Commented Aug 19, 2020 at 19:20

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