Set Union is Self-Distributive/Sets of Sets

Theorem

Let $A$ and $B$ denote sets of sets.

Then:

$\ds \bigcup \paren {A \cup B} = \paren {\bigcup A} \cup \paren {\bigcup B}$

where $\ds \bigcup A$ denotes the union of $A$.

Proof

Let $\ds s \in \bigcup \paren {A \cup B}$.

Then by definition of union of set of sets:

$\exists X \in A \cup B: s \in X$

By definition of set union, either:

$X \in A$

or:

$X \in B$

If $X \in A$, then:

$s \in \set {x: \exists X \in A: x \in X}$

If $X \in B$, then:

$s \in \set {x: \exists X \in B: x \in X}$

Thus by definition of union of set of sets, either:

$\ds s \in \bigcup A$

or:

$\ds s \in \bigcup B$

So by definition of set union:

$\ds s \in \paren {\bigcup A} \cup \paren {\bigcup B}$

So by definition of subset:

$\ds \bigcup \paren {A \cup B} \subseteq \paren {\bigcup A} \cup \paren {\bigcup B}$

Now let $\ds s \in \paren {\bigcup A} \cup \paren {\bigcup B}$.

By definition of set union, either:

$\ds s \in \bigcup A$

or:

$\ds s \in \bigcup B$

That is, by definition of union of set of sets, either:

$s \in \set {x: \exists X \in A: x \in X}$

or:

$s \in \set {x: \exists X \in B: x \in X}$

Without loss of generality, let $s \in X$ such that $X \in A$.

Then by Set is Subset of Union:

$s \in X$ such that $X \in A \cup B$

That is:

$\ds s \in \bigcup \paren {A \cup B}$

Similarly if $x \in X$ such that $X \in B$.

So by definition of subset:

$\ds \paren {\bigcup A} \cup \paren {\bigcup B} \subseteq \bigcup \paren {A \cup B}$

Hence by definition of equality of sets:

$\ds \bigcup \paren {A \cup B} = \paren {\bigcup A} \cup \paren {\bigcup B}$

$\blacksquare$