# Naive Set Theory: Functions

## Indexed Set Operations

The basic union and intersection operations can be generalized to involve multiple sets. In order to do this, we introduce the concept of an indexed collection of sets.

Let $A$ be a set, let $\mathcal{B}$ be a collection of sets, and let $f$ be surjection from $A$ into $\mathcal{B}$. Then $A$ is an index of $\mathcal{B}$. We can write the members of $\mathcal{B}$ as $\{B_a\}_{a \in A}$, with the understanding that $B_a = f(a)$. We can also say that $\mathcal{B}$ is indexed by $A$.

We can now generalize our set operations. Let $\mathcal{B}$ be a collection of sets indexed by $A$. The arbitrary union of $\mathcal{B}$ is defined as

$$\bigcup\limits_{a \in A} B_a = \{ x : \text{there exists an } a \in A \text{ such that } x \in B_a \}.$$

Similarly, the arbitrary intersection of $\mathcal{B}$ is defined as

$$\bigcap\limits_{a \in A} B_a = \{ x : \text{ for all } a \in A, x \in B_a \}$$

## Problems

1. Show that $\left(\bigcup\limits_{a \in A} B_a\right)^c = \bigcap\limits_{a \in A}B_a^c$.

Let $x \in \left(\bigcup\limits_{a \in A} B_a\right)^c$. Then $x \notin \bigcup\limits_{a \in A} B_a$. Thus $x \notin B_a$ for all $B_a$. However, this means $x \in B_a^c$ for all $B_a$, and so $x \in \bigcap\limits_{a \in A}B_a^c$. Conversely, assume $x \in \bigcap\limits_{a \in A}B_a^c$. Then $x \in B_a^c$ for all $B_a^c$. Therefore $x \notin B_a$ for all $B_a$, and so $x \notin \bigcup\limits_{a \in A} B_a$. Thus $x \in \left(\bigcup\limits_{a \in A} B_a\right)^c$.

2. Show that $\bigcup\limits_{a \in A} B_a^c = \left(\bigcap\limits_{a \in A}B_a\right)^c$.

Let $x \in \bigcup\limits_{a \in A} B_a^c$. Then $x \in B_a^c$ for some $B_a^c$, and so $x \notin B_a$. Therefore $x \notin \bigcap\limits_{a \in A}B_a$, so $x \in \left(\bigcap\limits_{a \in A}B_a\right)^c$. Conversely, let $x \in \left(\bigcap\limits_{a \in A}B_a\right)^c$. Then $x \notin \bigcap\limits_{a \in A}B_a$. Then $x \notin B_a$ for some $B_a$. Therefore $x \in B_a^c$, and so $x \in \bigcup\limits_{a \in A} B_a^c$.