Problem 54 - Discrete Mathematics with Applications

Question

Prove each, where \(A, B,\) and \(C\) are any sets. $$A-B=A \cap B^{\prime}$$

Step-by-Step Solution

Answer

To prove that \(A-B = A \cap B'\), we showed that if \(x \in A-B\), then \(x \in A \cap B'\), and if \(x \in A \cap B'\), then \(x \in A-B\). Since the elements of both sets are the same, we concluded that \(A-B = A \cap B'\).

Step 1: Define the sets and operations

Let's start by defining the sets and operations involved in the given expression. We have sets \(A, B,\) and \(C\). We need to work with the difference between sets \(A\) and \(B\), denoted as \(A-B\), and the intersection of sets \(A\) and the complement of set \(B\), denoted as \(A \cap B'\). Recall that: - The difference \(A-B\) contains all elements of \(A\) that are not in \(B\). - The complement of set \(B\), denoted as \(B'\), contains all elements not belonging to \(B\). - The intersection of sets \(A\) and \(B'\), denoted as \(A \cap B'\), contains all elements that belong both to \(A\) and to \(B'\).

Step 2: Show that an element in \(A-B\) belongs to \(A \cap B'\)

Let's consider an element \(x \in A - B\). It means that \(x\) belongs to set \(A\), but does not belong to set \(B\). Since \(x\) is not in \(B\), it is in the set \(B'\) by definition of the complement. Therefore, \(x\) also belongs to the intersection \(A \cap B'\), as it is in both sets \(A\) and \(B'\). This shows that if \(x \in A-B\), then \(x \in A \cap B'\).

Step 3: Show that an element in \(A \cap B'\) belongs to \(A-B\)

Now, let's consider an element \(x \in A \cap B'\). By definition of the intersection, \(x\) belongs to sets \(A\) and \(B'\). Since it belongs to \(B'\), the element \(x\) is not in set \(B\). Therefore, \(x\) is in set \(A - B\), because it belongs to set \(A\), but does not belong to set \(B\). This shows that if \(x \in A \cap B'\), then \(x \in A-B\).

Step 4: Conclude the proof

We have shown that if \(x \in A-B\), then \(x \in A \cap B'\), and if \(x \in A \cap B'\), then \(x \in A-B\). This means that the elements of both sets are the same. Therefore, we can conclude that \(A-B = A \cap B'\), proving the given expression.

Key Concepts

Set Difference

In set theory, the concept of set difference is a fundamental operation. It is denoted as the difference between two sets, such as \(A-B\). This operation represents all elements that are in set \(A\) but not in set \(B\). When you think about it, set difference is like subtracting everything in one set from another.

Imagine having two circles representing sets, where one overlaps the other. The set difference includes only the part of the first circle that doesn’t touch the second circle.

• Example: If \(A = \{1, 2, 3\}\) and \(B = \{3, 4, 5\}\), then \(A-B = \{1, 2\}\).
• Visualization: Picture it as taking away the part where the two circles overlap.

Set Intersection

Set intersection deals with elements that are common to both sets. Denoted as \(A \cap B\), it includes all elements that are present in both \(A\) and \(B\).

This operation is like finding a common ground between two lists. It shows what they share.

• Example: If \(A = \{2, 3, 4\}\) and \(B = \{3, 4, 5\}\), then \(A \cap B = \{3, 4\}\).
• Visualization: Think of the overlapping part of two circles.

When using set intersection with set complements, like \(A \cap B'\), it finds elements in \(A\) that aren’t in \(B\). It filters \(A\) using the outside of \(B\).

This is why \(A - B\) is the same as \(A \cap B'\). Both describe the elements that belong to \(A\) but not \(B\).

Set Complement

The complement of a set represents elements not in that set. If you have a universal set, the complement consists of everything outside the specified set.

Notated as \(B'\), it includes all items not found in set \(B\). It’s the reverse image of \(B\).

• Example: In a universal set \(U = \{1, 2, 3, 4, 5\}\), if \(B = \{3, 4\}\), then \(B' = \{1, 2, 5\}\).
• Visualization: Consider shading everything outside one circle in a universal space of sets.

Using complements simplifies operations like set intersection with a complement. For instance, combining \(A\) with \(B'\) (i.e., \(A \cap B'\)) captures all of \(A\) that is not in \(B\), mimicking the process of subtraction in set difference (\(A - B\)).

Understanding complements is key to mastering set operations as it allows for efficient structuring of elements outside given criteria.