Union Distributes over Union

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Theorem

Set union is distributive over itself:

$\forall A, B, C: \left({A \cup B}\right) \cup \left({A \cup C}\right) = A \cup B \cup C = \left({A \cup C}\right) \cup \left({B \cup C}\right)$

where $A, B, C$ are sets.


Sets of Sets

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


Then:

$\displaystyle \bigcup \left({A \cup B}\right) = \left({\bigcup A}\right) \cup \left({\bigcup B}\right)$

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


Families of Sets

Let $I$ be an indexing set.

Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.


Then:

$\displaystyle \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$

where $\displaystyle \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.


General Result

Let $\left\langle{\mathbb S_i}\right\rangle_{i \in I}$ be an $I$-indexed family of sets of sets.

Then:

$\displaystyle \bigcup_{i \mathop \in I} \bigcup \mathbb S_i = \bigcup \bigcup_{i \mathop \in I} \mathbb S_i$


Proof

We have:

The result follows from Associative Commutative Idempotent Operation is Distributive over Itself.

$\blacksquare$