Problem 6 - Calculus for Biology and Medicine

Question

In Problems \(5-8\), assume that $$ \Omega=\\{1,2,3,4,5,6\\} $$ \(A=\\{1,3,5\\}\), and \(B=\\{1,2,3\\}\). Find \(A^{c}\) and show that \(\left(A^{c}\right)^{c}=A\).

Step-by-Step Solution

Answer

\(A^c = \{2,4,6\}\) and \( (A^c)^c = A = \{1,3,5\}\)

Step 1: Define the Complement of Set A

The complement of a set, say \( A \), is the set of all elements in the universal set \( \Omega \) that are not in \( A \). This is usually denoted as \( A^c \). In this problem, \( \Omega = \{1,2,3,4,5,6\} \) and \( A = \{1,3,5\} \). Therefore, calculate \( A^c \) by including all elements from \( \Omega \) that are not in \( A \).

Step 2: Calculate A Complement \( A^c \)

Using the definition, list all elements in \( \Omega \) that are not in \( A \). So, exclude \( 1, 3, \) and \( 5 \) from \( \Omega \) to get \( A^c = \{2,4,6\} \).

Step 3: Define the Complement of \( A^c \)

The complement of \( A^c \), noted as \( (A^c)^c \), is the set of elements in \( \Omega \) that are not in \( A^c \). So, calculate \( (A^c)^c \) by including all elements from \( \Omega \) that are not in \( A^c \).

Step 4: Calculate \((A^c)^c\)

To find \((A^c)^c\), list all elements in \( \Omega \) that are not in \( A^c = \{2,4,6\} \). Exclude the elements \( 2, 4, \) and \( 6 \) from \( \Omega \). This gives you \((A^c)^c = \{1,3,5\}\).

Step 5: Verify the Equality \((A^c)^c = A\)

Now that we calculated \((A^c)^c = \{1,3,5\}\), check if it matches the original set \( A \). Since both \((A^c)^c\) and \( A \) are \{1,3,5\}, this confirms them as equal.

Key Concepts

Complement of a Set

In set theory, the complement of a set is a fundamental concept that helps us understand which elements are outside a given set within a universal set.
Let's break it down with an example. Suppose you have a universal set, denoted as \( \Omega \), which contains all possible elements in your universe of discourse. Now, imagine you have a specific set \( A \) within \( \Omega \). The complement of \( A \), symbolized as \( A^c \), includes all the elements in \( \Omega \) that are not in \( A \).
To find the complement \( A^c \), you simply subtract the elements of \( A \) from \( \Omega \).

Here's how it's done:

• Identify all elements in the universal set \( \Omega \).
• Exclude elements that are in set \( A \).
• The remaining elements form the complement \( A^c \).

For example, if \( \Omega = \{1, 2, 3, 4, 5, 6\} \) and \( A = \{1, 3, 5\} \), the elements not in \( A \) are \( 2, 4, \) and \( 6 \). Therefore, the complement \( A^c = \{2, 4, 6\} \).
This operation is robust and helps in understanding set relationships and operations such as union, intersection, and difference.

Universal Set

The concept of a universal set is like having a complete menu from which you pick orders. It's the master list that consists of all elements relevant to a particular discussion or problem.
The universal set is typically denoted by the symbol \( \Omega \) and contains every object under consideration, which might belong to any subset you're working with.

Here's the process of using a universal set:

• Define the universe of discourse, the set \( \Omega \), encompassing all possible elements.
• Subclassifications of \( \Omega \) will be your sets (like set \( A \) or set \( B \)).

In exercises or problems, you're often asked to draw conclusions based on understanding these boundaries.

The universal set makes operations such as finding complements straightforward. By focusing on \( \Omega \), you know the totality from which smaller, specific sets derive. For example, if \( \Omega = \{1, 2, 3, 4, 5, 6\} \), within this, you can navigate easily to constructs like \( A = \{1, 3, 5\} \) and find its complement. Having a well-defined \( \Omega \) sets the stage for accurate assessments and logical operations.

Mathematical Proof

Mathematical proof is a logical argument that establishes the truth of a mathematical statement. Proofs are essential in mathematics as they ensure that conclusions drawn from sets, formulas, or equations stand on solid ground.

Let’s delve into an example of a proof related to the complement of a set. Consider proving that the complement of the complement of a set \( (A^c)^c \) equals the original set \( A \).

The steps to verify this would involve:

• First, find the complement \( A^c \), which includes all elements in \( \Omega \) not in \( A \).
• Then find \( (A^c)^c \), this set consists of elements in \( \Omega \) but not in \( A^c \).
• By definition, \( (A^c)^c = A \) because excluding elements not in \( A \) effectively returns us to the original set \( A \).

In \( \Omega = \{1, 2, 3, 4, 5, 6\} \), if \( A = \{1, 3, 5\} \), then \( A^c = \{2, 4, 6\} \). Thus, \( (A^c)^c = \{1, 3, 5\} = A \).

This logical sequence confirms the property and illustrates how proofs provide a structured method to validate statements in set theory. Thorough understanding and application of proofs is vital to mastering mathematical reasoning.