Associative Laws of Sets

# Associative Laws of Sets

We will now look at two very important associative laws of sets. The proofs are relatively simple and straightforward.

Theorem 1 (Associative Law for the Union of Three Sets): If $A$, $B$, and $C$ are sets, then $(A \cup B) \cup C = A \cup (B \cup C)$. |

**Proof:**Suppose that $x \in (A \cup B) \cup C$. Then it follows that $x \in A \cup B$ (which means that $x \in A$ or $x \in B$) or $x \in C$. Therefore we can say $x \in A$ or $x \in B \cup C$ or rather $x \in A \cup (B \cup C)$. Therefore $(A \cup B) \cup C \subseteq A \cup (B \cup C)$.

- To show that $(A \cup B) \cup C \supseteq A \cup (B\cup C)$ follows the exact same procedure so we'll omit it. $\blacksquare$

Theorem 2 (Associative Law for the Intersection of Three Sets): If $A$, $B$, and $C$ are sets, then $(A \cap B) \cap C = A \cap (B \cap C)$. |

**Proof:**Suppose that $x \in (A \cap B) \cap C$. Then $x \in A \cap B$ and $x \in C$. So $x \in A$ and $x \in B \cap C$ or rather $x \in A \cap (B \cap C)$. Therefore $(A \cap B) \cap C \subseteq A \cap (B \cap C)$.

- Once again to showing that $(A \cap B) \cap C \supseteq A \cap (B \cap C)$ follows the same procedure and it will be omitted. $\blacksquare$