Indexed Families of Sets

Union

Definition 1. Let $\mathcal{F}$ be a family of sets. The union of the sets in $\mathcal{F}$, denoted by $\cup_{A \in \mathcal{F}} A$ is the set of all elements that are in $A$ for some $A \in \mathcal{F}$. That is, $$\bigcup_{A \in \mathcal{F}} A = \{ x \in U \, | \, \exists A \in \mathcal {F}, x \in A \}.$$

Intersection

Definition 2. Let $\mathcal{F}$ be a family of sets. The intersection of the sets in $\mathcal{F}$, denoted by $\cap_{A \in \mathcal{F}} A$ is the set of all elements that are in $A$ for all $A \in \mathcal{F}$. That is, $$\bigcap_{A \in \mathcal{F}} A = \{ x \in U \, | \, \forall A \in \mathcal {F}, x \in A \}.$$

Theorem 1

Theorem 1. $$\begin{align*} & \text{(a)} \,\, \bigcup_{\gamma \in \emptyset} A_{\gamma} = \emptyset \hspace{15cm} \\ & \text{(b)} \,\, \bigcap_{\gamma \in \emptyset} A_{\gamma} = U \end{align*}$$

Proof. (a) Suppose that there exists $x \in U$ such that $x \in \cup_{\gamma \in \emptyset} A_{\gamma}$. Then there is some $A_{\gamma}$ such that $x \in A_{\gamma}$ and this implies that $\exists \gamma \in \emptyset$ such that $x \in A_{\gamma}$. This assertation is a contradiction, so there does not exist $x \in U$ such that $x \in \cup_{\gamma \in \emptyset} A_{\gamma}$. 
(b) Let define $B_{\gamma} = A^c_{\gamma}$ for each $\gamma$. From (a) and De Morgan's Theroem, we have $$(\bigcup_{\gamma \in \emptyset} A_{\gamma})^c = \bigcap_{\gamma \in \emptyset} A^c_{\gamma} = \bigcap_{\gamma \in \emptyset} B_{\gamma} = \emptyset^c = U. \blacksquare$$

직관적으로 이해해보자. 합집합은 아무것도 없는 곳에서 하나씩 데리고 온다고 생각해야 한다. 따라서 아무것도 가져올 게 없으면 당연히 공집합이다. 반면 교집합은 이미 주어져 있는 전체에서 조금씩 긁어낸다고 생각해야 한다. 이때 아무것도 긁어낼 게 없다면 당연히 처음 있던 그대로, 즉 전체 집합이다.

De Morgan's Theorem

Theorem 2. $$\begin{align*} & \text{(a)} \,\, \left(\bigcup_{\gamma \in \Gamma} A_{\gamma} \right)^c = \bigcap_{\gamma \in \Gamma} A^c_{\gamma} \hspace{15cm} \\ & \text{(b)} \,\, \left(\bigcap_{\gamma \in \Gamma} A_{\gamma} \right)^c = \bigcup_{\gamma \in \Gamma} A^c_{\gamma} \end{align*}$$