How can I define this set?

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?

• You mean $A_1 = \{1,2\}$, $A_2 = \{3\}$ and $A_3 = \{4\}$? Commented Aug 19, 2020 at 18:49
• No, those are supposed to be sets of sets. @Azif00 Commented Aug 19, 2020 at 18:50
• 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 $A_1,\dots,A_n$? Commented Aug 19, 2020 at 18:53
• In my approach incorrect? @Azif00? Commented Aug 19, 2020 at 19:02
• @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). Commented Aug 19, 2020 at 19:04

$$\bigcup_nA_n$$ 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

$$\bigcup_nA_n=\big\{\{1\},\{2\},\{3\},\{4\}\big\}\,,$$

and

$$\bigcup\bigcup_nA_n=\{1,2,3,4\}\,.$$

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

$$B=\wp\left(\bigcup\bigcup_nA_n\right)\setminus\{\varnothing\}\,.$$

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