Question

Let \(\mathcal{F}\) be a collection of sets and define $$ I=\cap\\{F \mid F \in \mathscr{F}\\} \quad \text { and } \quad U=\cup\\{F \mid F \in \mathcal{F}\\} $$ Prove that $$ \text { (a) } I^{c}=\cup\left\\{F^{c} \mid F \in \mathcal{F}\right\\} \text { and (b) } U^{c}=\left\\{\cap F^{c} \mid F \in \mathscr{F}\right\\} \text { . } $$

Short answer

Question: Prove the following properties of intersections and unions of sets' complements: (a) \(I^c = \cup\{F^c \mid F \in \mathcal{F}\}\) (b) \(U^c = \cap\{F^c \mid F \in \mathcal{F}\}\) where \(I\) is the intersection of all sets in \(\mathcal{F}\), \(U\) is the union of all sets in \(\mathcal{F}\), and \(A^c\) is the complement of set \(A\). Answer: We have proved both properties as follows: (a) We showed that for any element \(x\): - if \(x \in I^c\), then \(x \in \cup\{F^c \mid F \in \mathcal{F}\}\) - if \(x \in \cup\{F^c \mid F \in \mathcal{F}\}\), then \(x \in I^c\) Thus, \(I^c = \cup\{F^c \mid F \in \mathcal{F}\}\). (b) We showed that for any element \(x\): - if \(x \in U^c\), then \(x \in \cap\{F^c \mid F \in \mathcal{F}\}\) - if \(x \in \cap\{F^c \mid F \in \mathcal{F}\}\), then \(x \in U^c\) Therefore, \(U^c = \cap\{F^c \mid F \in \mathcal{F}\}\).

Step-by-step solution

Definitions and notations

Before we start proving the properties, let's establish the definitions and notations that we will be using: - \(\mathcal{F}\): A collection of sets - \(I\): Intersection of all sets in \(\mathcal{F}\), \(I = \cap\{F \mid F \in \mathcal{F}\}\) - \(U\): Union of all sets in \(\mathcal{F}\), \(U = \cup\{F \mid F \in \mathcal{F}\}\) - \(A^c\): Complement of set \(A\)

Proving property (a)

To prove property (a), we need to show that \(I^c = \cup\{F^c \mid F \in \mathcal{F}\}\). Let's take an element \(x\) outside the intersection \(I\) of \(\mathcal{F}\) (i.e., \(x \in I^c\)). This means that \(x\) doesn't belong to at least one \(F_i \in \mathcal{F}\). So, \(x\) belongs to the complement of \(F_i\), \(F_i^c\). Thus, \(x \in \cup\{F^c \mid F \in \mathcal{F}\}\). We have shown that if \(x \in I^c\), then \(x \in \cup\{F^c \mid F \in \mathcal{F}\}\). On the other hand, let's assume \(x\) belongs to the union of the complements of the sets in \(\mathcal{F}\), which means \(x \in F_i^c\) for some \(F_i \in \mathcal{F}\). Therefore, x doesn't belong to \(F_i\) and hence is not in \(I\). Thus, \(x \in I^c\). Since we have proved both directions, we have shown that \(I^c = \cup\{F^c \mid F \in \mathcal{F}\}\) for property (a).

Proving property (b)

Next, we will prove property (b), which states that \(U^c = \cap\{F^c \mid F \in \mathcal{F}\}\). First, let's consider an element \(x\) outside the union \(U\) of \(\mathcal{F}\) (i.e., \(x \in U^c\)). Since \(x\) doesn't belong to any set \(F_i \in \mathcal{F}\), it must belong to the complements of all sets \(F_i\) in \(\mathcal{F}\). Hence, \(x \in \cap\{F^c \mid F \in \mathcal{F}\}\). We have shown that if \(x \in U^c\), then \(x \in \cap\{F^c \mid F \in \mathcal{F}\}\). Now, we will consider the other direction. Let's assume \(x\) belongs to the intersection of the complements of the sets in \(\mathcal{F}\) (i.e., \(x \in \cap\{F^c \mid F \in \mathcal{F}\}\)). This means that \(x\) doesn't belong to any \(F_i \in \mathcal{F}\). Since \(x\) doesn't belong to any set in \(\mathcal{F}\), it is outside the union, so \(x \in U^c\). Since we have proved both directions, we have shown that \(U^c = \cap\{F^c \mid F \in \mathcal{F}\}\) for property (b). Conclusion: We have successfully proved both properties (a) and (b) for the intersections and unions of sets' complements.

Key concepts

Intersection

In set theory, the intersection of a collection of sets refers to the set of elements that are present in all of those sets. We denote the intersection of sets in \(\mathcal{F}\) by \(I = \cap\{F \mid F \in \mathcal{F}\}\). This means an element \(x\) is in \(I\) if and only if \(x\) is in every set \(F\) within the collection \(\mathcal{F}\).

This concept is helpful when we want to find common elements shared among different sets. For instance, if we have a group of sets in a database, each representing customers who made purchases, the intersection would show the customers who made purchases in every category.

In our solution, we demonstrated that the complement of the intersection \(I^c\) is equivalent to the union of the complements of the individual sets in \(\mathcal{F}\), represented as \(\cup\{F^c \mid F \in \mathcal{F}\}\). This means an element is outside all sets if it is outside any one of those sets individually.

Union

The union of a collection of sets combines all the elements from the sets into a single set. In terms of notation, if \(\mathcal{F}\) is a collection of sets, then \(U = \cup\{F \mid F \in \mathcal{F}\}\) represents the union of all sets in \(\mathcal{F}\).

An element \(x\) is in the union \(U\) if it is present in at least one of the sets \(F\). The union is significant when we want to aggregate comprehensive data or cover all possible cases, like gathering all topics available in a curriculum from different schools.

In the solution, the equivalence \(U^c = \cap\{F^c \mid F \in \mathcal{F}\}\) was shown, demonstrating that being outside the union \(U\) means being outside each set individually in \(\mathcal{F}\).

Complement of a Set

The complement of a set, denoted by \(A^c\), consists of all the elements not present in the set \(A\). Understanding the complement is essential in determining what is missing or excluded from a given set.

In the context of our problem, taking the complement of an intersection or a union involves flipping the perspective: it shows what isn’t common or isn’t included across the sets. For the intersection \(I\), the complement \(I^c\) included those elements not present in every set. Similarly, for the union \(U\), the complement \(U^c\) contained elements not found in any of the sets in \(\mathcal{F}\).

Being familiar with complements helps in scenarios like identifying which items were not bought by any customer if analyzing a grocery store’s sales data. The derived relationship between complements, intersection, and union aids in constructing or breaking down comprehensive sets based on their constituent parts.