In set theory, it is important to work with the notion of a family or indexed collection of sets. In the theory of stochastic processes, for example, this concept is widely used, mainly for filtering theory in sigma-algebras.

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Indexed collection

Definition (Indexed family of sets) Let $\Gamma$ be a set, we assume that for each element $\gamma \in \Gamma$ a set $A_{\gamma}.$ is associated. The collection of such sets $A_{\gamma}$ is said to be an indexed family of sets, indexed by the set $\Gamma$ and denoted by

$$\\{A_{\gamma}; \gamma \in \Gamma \\}$$
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If $\mathcal{C} = \\{ A_{\gamma}; \gamma \in \Gamma \\},$ we write

$$\bigcup \mathcal{C} = \bigcup_{\gamma \in \Gamma} A_{\gamma} = \\{x; x \in A_{\gamma} \text{ para algum } \gamma \in \Gamma \\}$$

e

$$\bigcap \mathcal{C} =\bigcap_{\gamma \in \Gamma} A_{\gamma} = \\{x; x \in A_{\gamma} \text{ para todo } \gamma \in \Gamma \\}.$$

Note that given any collection of sets, it is always possible to take an index set $\Gamma$ and make this collection an indexed family of sets, indexed by $\Gamma$.

Finite index set

In finite index sets, for example $\Gamma = \\{1,2,3,\dots, n\\}$ we have that:

  • Union of sets: $$\bigcup_{\gamma \in \Gamma} A_{\gamma} = \bigcup^n_{i=1} A_i = A_1 \cup A_2 \cup \dots \cup A_n$$
  • Intersection of sets: $$\bigcap_{\gamma \in \Gamma} A_{\gamma} = \bigcap^n_{i=1} A_i = A_1 \cap A_2 \cap \dots \cap A_n$$
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Set of enumerable indexes

If the set is enumerable, for example, $\Gamma = \mathbb{N}$ we have to

  • Union of sets: $$\bigcup_{\gamma \in \Gamma} A_{\gamma} = \bigcup^\infty_{i=1} A_i = A_1 \cup A_2 \cup \dots$$
  • Intersection of sets: $$\bigcap_{\gamma \in \Gamma} A_{\gamma} = \bigcap^\infty_{i=1} A_i = A_1 \cap A_2 \cap \dots $$

Result for empty family of subsets

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Theorem: Let $\\{A_{\gamma}; \gamma \in \Gamma\\}$ be an empty family of subsets of a set $U,$ i.e. $\Gamma = \emptyset.$ Then

  • (a) $\displaystyle\bigcup_{\gamma \in \Gamma} A_{\gamma} = \emptyset.$
  • (b) $\displaystyle\bigcap_{\gamma \in \Gamma} A_{\gamma} = U.$

Demo:

(a) Note that showing that $\displaystyle\bigcup_{\gamma \in \emptyset} A_{\gamma} = \emptyset$ is equivalent to showing that for all $x \in U,$ we have $x \notin \displaystyle\bigcup_{\gamma \in \emptyset} A_{\gamma}.$

For $x ∈ U,$ we have in notation:

$$x \notin \bigcup_{\gamma \in \emptyset} A_{\gamma} \Longleftrightarrow \sim \left(x \in \bigcup_{\gamma \in \emptyset} A_{\gamma} \right) $$

By definition of union we have:

$$\sim\left(x \in \bigcup_{\gamma \in \emptyset} A_{\gamma} \right) \Longleftrightarrow \sim\left(x \in A_{\gamma}, \text{ para algum } \gamma \in \emptyset \right)$$

Now let’s apply negation:

$$\sim\left(x \in A_{\gamma}, \text{ para algum } \gamma \in \emptyset \right) \Longleftrightarrow x \notin A_{\gamma}, \text{ para todo } \gamma \in \emptyset $$

Hence we have that $\gamma \in \emptyset$ implies that:

$$ x \notin A_{\gamma}$$

and this last proposition is true for all $x \in U,$ because $γ \in \emptyset$ is a contradiction. This completes the proof in part (a).

(b) In this item, we have to show that for all $x \in U,$ we have

$$x \in \bigcap_{γ\in \emptyset} A_\gamma.$$

Note that by definition

$$x \in \bigcap_{\gamma\in \emptyset} A_{\gamma} \Longleftrightarrow \\{x \in A_{\gamma}, \text{ para todo } \gamma \in \emptyset\\} $$

which is equivalent to the proposition $γ \in \emptyset$ implies $x \in A_{\gamma},$ which, as seen in the demonstration of item (a), is true for all $x \in U.$

$\square$

Generalized Morgan’s Laws

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Theorem (Generalized Morgan’s Laws) Let $\\{A_{\gamma}; \gamma \in \Gamma\\}$ be an arbitrary family of subsets of a set $U.$ Then

  • (a) $\displaystyle\left(\bigcup_{\gamma\in \Gamma} A_\gamma\right)^c = \bigcap_{γ\in \Gamma} A^c_\gamma$
  • (b) $\displaystyle\left(\bigcap_{\gamma\in \Gamma} A_\gamma\right)^c = \bigcup_{γ\in \Gamma} A^c_\gamma$

Demo:

a) For all $x \in U,$ we have that by the definition of complementary

$$x \in \left(\bigcup_{γ\in \Gamma} A_\gamma\right)^c \Longleftrightarrow \sim \left(x \in \bigcup_{γ\in \Gamma} A_\gamma\right)$$

By the definition of union we have:

$$\sim \left(x \in \bigcup_{γ\in \Gamma} A_\gamma\right) \Longleftrightarrow \sim (\text{existe } \gamma \in \Gamma \text{ tal que } x \in A_{\gamma})$$

By denying the last sentence we obtain that for all $\gamma \in \Gamma$ we have $x \notin A_{\gamma}$.

The next step is to apply the definition of complementary and we obtain $\gamma \in \Gamma$ we have to $x \in A^c_{\gamma}$.

By the definition of intersection

$$x \in \bigcap_{\gamma \in \Gamma} A^c_{\gamma}$$

Therefore, by definition of equality of sets we have the equality of item (a).

The demonstration of equality in item (b) remains as an exercise. 😂😂😂

$\square$

Generalized Distributive Laws

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Theorem (Generalized Distributive Laws) Let $A$ be a set and $\mathcal{C} = \\{B_{\gamma}; \gamma \in \Gamma\\}$ a family of subsets. Then

  • (a) $\displaystyle A \cap \left(\bigcup_{\gamma\in \Gamma} B_\gamma\right) = \bigcup_{\gamma\in \Gamma} A \cap B_\gamma$
  • (b) $\displaystyle A \cup \left(\bigcap_{\gamma\in \Gamma} B_\gamma\right) = \bigcap_{\gamma\in \Gamma} A \cup B_\gamma$

Demo:

(a) An element $x$ is in the set $\displaystyle A \cap \left(\bigcup_{\gamma\in \Gamma} B_\gamma\right)$ if and only if $x \in A$ and $x \in \bigcup_{\gamma\in \Gamma} B_\gamma$ by the definition of intersection.

Now from the definition of the union of any family of sets, we have that this proposition is equivalent to $x \in A$ and $x \in B_\gamma,$ for some $γ \in \Gamma,$ which can be expressed as $x \in A \cap B_\gamma,$ for some $γ \in \Gamma.$

By definition of union

$$x \in \bigcup_{\gamma\in \Gamma} A \cap B_\gamma,$$

which shows (a) by the definition of set equality.

The demonstration of equality in item (b) remains as an exercise. 😂😂😂

$\square$

This is the set family summary. Wait for the next posts.